8 &

GALOISIAN GALOIS THEORY

Lecture Notes in Computational Algebra

ABSTRACT.  These notes are an exposition of Galois Theory from the original
Lagrangian and Galoisian point of view. A particular effort was made here to
better understand the connection between Lagrange’s purely combinatorial
approach and Galois algebraic extensions of the latter. Moreover, stimulated
by the necessities of present day computer explorations, the algorithmic
approach has been given priority here over every other aspect of presentation.
In particular, you may not find here the clean simplistic look characteristic
of the classical exposition of E. Artin. In contrast these notes should
provide a good starting point in attempting constructions in this most
difficult computational arena.

1\. Symmetric Functions

Unless otherwise specified all fields we shall work with here will be assumed
to have zero characteristic. If ${\cal F}$ is such a field and
$x_{1},x_{2},\ldots,x_{n}$ are indeterminates the expression

${\cal F}[x_{1},x_{2},\ldots,x_{n}]$

will denote the ring of polynomials in $x_{1},x_{2},\ldots,x_{n}$ with
coefficients in ${\cal F}$. In contrast to customary notation, the field of
rational functions of $x_{1},x_{2},\ldots,x_{n}$ with coefficients in ${\cal
F}$ will be denoted by

$Rat[{\cal F};x_{1},\ldots,x_{n}]$

We may also write

$Rat[{\cal F};f_{1},f_{2},\ldots,f_{m}]$

to represent all rational expressions in $f_{1},f_{2},\ldots,f_{n}$ with
coefficients in ${\cal F}$, whatever $f_{1},f_{2},\ldots,f_{m}$ may be in any
particular situation.

As customary, $S_{n}$ denotes the group of all permutations of $1,2,\ldots,n$.

If $\sigma=(\sigma_{1},\sigma_{2},\ldots,\sigma_{n})\in S_{n}$ and

$\Phi(x_{1},x_{2},\ldots,x_{n})\in Rat[{\cal F};x_{1},\ldots,x_{n}]$

we set $\enskip x\sigma\enskip=\enskip
x_{\sigma_{1}},x_{\sigma_{2}},\ldots,x_{\sigma_{n}}$ and

$\sigma\Phi\enskip=\enskip\Phi(x\sigma)\enskip=\enskip\Phi(x_{\sigma_{1}},x_{\sigma_{2}},\ldots,x_{\sigma_{n}})$

We shall say that $\Phi(x_{1},x_{2},\ldots,x_{n})$ is $k-valued$ if and only
if the collection

$\\{\hskip 2.5pt\sigma\hskip 2.5pt\Phi\hskip 2.5pt:\hskip 2.5pt\sigma\in
S_{n}\hskip 2.5pt\\}$

has cardinality $k$. Of course $1$-valued functions are usually called
symmetric. It will be convenient to denote here by $Sym[{\cal
F};x_{1},x_{2},\ldots,x_{n}]$ and $Ratsym[{\cal F};x_{1},x_{2},\ldots,x_{n}]$
the collections of symmetric elements of ${\cal F}[x_{1},x_{2},\ldots,x_{n}]$
and $Rat[{\cal F};x_{1},\ldots,x_{n}]$ respectively.

The combinatorial study of $k$-valued function is properly the domain of
Lagrange theory and will be carried out in the next section. In this section
we will limit ourselves to establishing the results on symmetric functions
that are needed in our further developments.

We recall that the symmetric polynomial

$e_{k}(x_{1},x_{2},\ldots,x_{n})\enskip=\enskip\sum_{1\leq
i_{1}<i_{2}<\cdots<i_{k}\leq n}\hskip 2.5ptx_{i_{1}}x_{i_{2}}\cdots
x_{i_{k}}\enskip$

is usually referred to as the $k^{th}\enskip elementary$ symmetric function.

A vector of integers

$\lambda\enskip=\enskip(\lambda_{1}\geq\lambda_{2}\geq\cdots\geq\lambda_{k}>0)$

is said to be a partition of $n$ and we write $\lambda\vdash n$ if and only if

$\lambda_{1}+\lambda_{2}+\cdots+\lambda_{k}=n$

We shall also say that $k$ is the number of parts of $\lambda$. If $\lambda$
and $\mu$ are partitions of the same number with $k$ and $h$ parts
respectively, we shall say that $\lambda$ dominates $\mu$ and write
$\lambda\hskip 2.5pt\geq\mu$ if and only if

$\lambda_{1}+\lambda_{2}+\cdots+\lambda_{s}\geq\mu_{1}+\mu_{2}+\cdots+\mu_{s}\hskip
36.135pt(\enskip for\enskip s=1,2,\ldots,min(h,k)\enskip)$

It easy to see that this partial order is linearly extended by the
lexicographic order of integer vectors. Finally, given a partition $\mu$ the
partition $\mu^{\prime}$ whose parts $\mu_{s}^{\prime}$ are given by

$\mu_{s}^{\prime}\enskip=\enskip\\#\\{i\hskip 2.5pt:\hskip 2.5pt\mu_{i}\geq
s\\}$

is usually referred to as the conjugate  of $\mu$. A simple combinatorial
argument shows that we have $\lambda\geq\mu$ if and only if
$\lambda^{\prime}\leq\mu^{\prime}$.

Given a vector

$p=(p_{1},p_{2},\ldots,p_{n})$

of non-negative integers, the weakly decreasing rearrangement of the positive
components of $p$ will referred to as the shape of $p$ and denoted by
$\lambda(p)$. Given a partition $\lambda$ the symmetric polynomial

$m_{\lambda}(x)\enskip=\enskip\sum_{\lambda(p)=\lambda}\hskip 2.5ptx^{p}\hskip
36.135pt(\enskip x^{p}=x_{1}^{p_{1}}x_{1}^{p_{2}}\cdots x_{n}^{p_{n}}\enskip)$

is usually referred to as the monomial symmetric function indexed by
$\lambda$. The following fact is immediate.

Theorem 1.1

Every symmetric polynomial $\Phi(x_{1},x_{2},\ldots,x_{n})$ has a unique
expansion of the form

$\Phi(x)\enskip=\enskip\sum_{\lambda}\hskip 2.5ptc_{\lambda}\hskip
2.5ptm_{\lambda}(x)$ $None$

where the $c_{\lambda}$ are integers if $\Phi$ has integer coefficients, and
$c_{\lambda}$ is in ${\cal F}$ if $\Phi\in{\cal F}[x_{1},x_{2},\ldots,x_{n}]$.
Proof

The symmetry of $\Phi$ implies that the coefficients $c_{p}$ and $c_{q}$ of
any two monomials $x^{p}$ and xq appearing in $\Phi$ must be the same if $p$
and $q$ have the same shape. Thus 1.1 is obtained by collecting terms of
$\Phi$ according to shape.

Let now $\mu=(\mu_{1},\mu_{2},\ldots,\mu_{h})$ be a partition and set

$e_{\mu}(x)\enskip=\enskip e_{\mu_{1}}(x)e_{\mu_{2}}(x)\cdots e_{\mu_{h}}(x)$
$None$

Since this polynomial is clearly symmetric in $x_{1},x_{2},\ldots,x_{n}$ it
must have an expansion of the form

$e_{\mu}(x)\enskip=\enskip\sum_{\lambda}\hskip 2.5ptm_{\lambda}(x)\hskip
2.5ptc_{\lambda\mu}\enskip.$ $None$

It develops that the coefficients $c_{\lambda\mu}$ have a suggestive
combinatorial interpretation.

Proposition 1.1

If $\mu$ has $h$ parts and $\lambda$ has $k$ parts then $c_{\lambda\mu}$ gives
the number of $h\times k$ matrices with $0,1$-entries and row and column sums
given by $\mu$ and $\lambda$ respectively. Proof

There is a one-to-one correspondence between these matrices and the monomials
obtained by expanding the product in 1.3. In fact, if
$m_{1}(x),m_{2}(x),\ldots,m_{h}(x)$ are monomials coming out of
$e_{\mu_{1}}(x),e_{\mu_{2}}(x),\ldots,e_{\mu_{h}}(x)$ respectively, then
$m_{i}(x)$ corresponds to a subset of $1,2,\ldots,n$ of cardinality $\mu_{i}$,
thus it may be represented by a $0,1$-vector with $n$ components in an obvious
manner. Putting together these vectors as the rows of an $h\times n$ matrix
$M$ we see that these monomials multiply to
$x^{p}=x_{1}^{p_{1}}x_{2}^{p_{2}}\cdots x_{n}^{p_{n}}$ if and only if the
columns of $M$ add up to $p_{1},p_{2},\ldots,p_{n}$ respectively. This is our
desired correspondence. Since, since monomials of the same shape have the same
coefficient, the assertion follows by taking
$x^{p}=x_{1}^{\lambda_{1}}x_{2}^{\lambda_{2}}\cdots x_{k}^{\lambda_{k}}$.

Remark 1.1

Let $M$ be one of the $0,1$-matrices with row and column sums given by $\mu$
and $\lambda$. Note that the number of $1\hskip 2.5pt^{\prime}s$ in the first
$s$ columns of $M$ is equal to

$\lambda_{1}+\lambda_{2}+\cdots+\lambda_{s}$

On the other hand by moving all the $1\hskip 2.5pt^{\prime}s$ along their rows
until they are bumper to bumper to the left (and all the $0^{\prime}s$
similarly to the right) we obtain a matrix $M^{\prime}$ whose column sums are
the parts of the partition $\mu^{\prime}$ conjugate to $\mu$. This not only
gives the inequality

$\mu^{\prime}_{1}+\mu^{\prime}_{2}+\cdots+\mu^{\prime}_{s}\hskip
2.5pt\geq\lambda_{1}+\lambda_{2}+\cdots+\lambda_{s}\hskip 36.135pt(\enskip
for\enskip s=1,2,\ldots,min(h,k)\enskip)$

but also assures that when $\lambda=\mu$ there can only be one matrix with the
desired row and column sums.

We can thus obtain

Theorem 1.2

Every homogeneous symmetric polynomial $\Phi(x_{1},x_{2},\ldots,x_{n})$ of
degree $m$ has a unique expansion of the form

$\Phi(x)\enskip=\enskip\sum_{\mu\vdash m}\hskip 2.5ptd_{\mu}\hskip
2.5pte_{\mu}(x)$

where the coefficients $d_{\mu}$ are integers if $\Phi$ has integer
coefficients and they are elements of ${\cal F}$ if $\Phi\in Sym[{\cal
F};x_{1},x_{2},\ldots,x_{n}]$. In particular, $Sym[{\cal
F};x_{1},x_{2},\ldots,x_{n}]$ is the polynomial ring generated by the
elementary symmetric functions $e_{1}(x),e_{2}(x),\ldots,e_{n}(x)$.

Proof

Let $D=\|d_{\lambda\mu}\|_{\lambda,\mu\vdash m}$ denote the matrix obtained
when the partitions $\lambda$ are in lexicographic order and the partitions
$\mu$ are in reverse lexicographic order. This done we see that an immediate
consequence of the observations made in the Remark above is that $D$ must be
unitriangular and therefore invertible over the integers. This shows that the
collection $\\{e_{\mu}(x)\\}_{\mu\vdash m}$ must also give a basis and that
the elements of the basis $\\{m_{\lambda}(x)\\}_{\lambda\vdash m}$ have
integral linear expansions in terms of the $e_{\mu}(x)\hskip 2.5pt^{\prime}s$.
This establishes our assertions.

Corollary 1.2

$Ratsym[{\cal F};x_{1},x_{2},\ldots,x_{n}]\enskip=\enskip Rat[{\cal
F};e_{1},e_{2},\ldots,e_{n}]$ $None$

Proof

Note that every element $\Phi\in Rat[{\cal F};x_{1},\ldots,x_{n}]$ can be
expressed in the form

$\Phi(x)={P(x)\over Q(x)}$

with $P,Q\in{\cal F}[x_{1},x_{2},\ldots,x_{n}]$. Multiplying numerator and
denominator by the polynomial

$\prod_{\sigma\in A_{n}\atop\sigma\neq id}\hskip 2.5pt\sigma Q$

we can write

$\Phi(x)={P^{*}(x)\over Q^{*}(x)}$

where

$P^{*}(x)=P(x)\prod_{\sigma\in A_{n}\atop\sigma\neq id}\hskip 2.5pt\sigma
Q\enskip\enskip\enskip\enskip,\enskip\enskip\enskip\enskip
Q^{*}(x)=\prod_{\sigma\in A_{n}}\hskip 2.5pt\sigma Q\enskip.$

Since by its very construction $Q^{*}(x)$ is already symmetric, we see that
$\Phi$ will be symmetric if and only if $P^{*}$ is. Thus our assertion follows
immediately from Theorem 1.2.

2\. The Euclidean algorithm the Resultant and the Discriminant.

If $A,B$ are polynomials in ${\cal F}[t]$, with $degree\hskip
2.5ptB<degree\hskip 2.5ptA$, then the quotient  and the remainder anof the
division of $A$ by $B$ are respectively the unique polynomials $Q$ and $R$
satisfying the requirements

$\eqalign{&(1)\enskip\enskip A\enskip=\enskip B\hskip 2.5ptQ\enskip+\enskip
R\cr&(2)\enskip\enskip degree(R)\hskip 2.5pt<\hskip
2.5ptdegree(B)\cr}\enskip.$

To construct these two polynomials we can proceed as follows. Set

$\Delta(A/B)\enskip=\enskip{leading\enskip coeff\enskip A\over leading\enskip
coeff\enskip B}\enskip\enskip t^{degree(A)-degree(B)}$

Clearly we have

$degree\bigl{(}A-\Delta(A/B)\hskip 2.5ptB\bigr{)}\enskip<\enskip degree(A).$
$None$

Now set

$R^{(o)}=A\enskip\enskip{\rm and}\enskip\enskip
R^{(i)}=R^{(i-1)}-\Delta(R^{(i-1)}/B)\hskip 2.5ptB\hskip 36.135pt(\enskip
i=1,2,\ldots\enskip)$ $None$

Since, by 2.1, degrees are decreasing at least by one at each step of the
recursion, after $k\leq degree(A)-degree(B)$ steps we shall have
$degree(R^{(k)})<degree(B)$. At this point we stop the recursion. By adding
the identities in 2.1 we easily derive that

$A\enskip=\enskip\Bigl{(}\sum_{i=0}^{k-1}\hskip 2.5pt\Delta(R^{(i)}/B)\hskip
2.5pt\Bigr{)}\enskip B\enskip+\enskip R^{(k)}$ $None$

Thus we may take

$Q=\sum_{i=0}^{k-1}\hskip
2.5pt\Delta(R^{(i)}/B)\enskip\enskip\enskip,\enskip\enskip\enskip
R=R^{(k)}\enskip.$ $None$

The important fact to note is that from 2.3 we deduce that if $A,B\in{\cal
F}[t]$ then $R(t)$ is in ${\cal F}[t]$ as well. We shall refer to the process
above as the division algorithm.

The Euclidean algorithm is the process which yields the greatest common
divisor $D(t)$ of two polynomials $A$ and $B$. It is shown by Berlekamp [1]
that $D$ may be computed by the following process. Set

$r_{-2}=A\enskip\enskip,\enskip\enskip r_{-1}=B\enskip\enskip,\enskip\enskip
p_{-2}=0\enskip\enskip,\enskip\enskip p_{-1}=1\enskip\enskip,\enskip\enskip
q_{-2}=1\enskip\enskip,\enskip\enskip q_{-1}=0$

Then compute $a_{k},r_{k},p_{k},q_{k}$ according to the recursions

$\eqalign{&(1)\enskip\enskip r_{k-2}\enskip=\enskip a_{k}\hskip
2.5ptr_{k-1}\enskip+\enskip r_{k}\enskip\enskip(division)\cr&(2)\enskip\enskip
p_{k}\enskip=\enskip a_{k}\hskip 2.5ptp_{k-1}\enskip+\enskip
p_{k-2}\cr&(3)\enskip\enskip q_{k}\enskip=\enskip a_{k}\hskip
2.5ptq_{k-1}\enskip+\enskip q_{k-2}\cr}\enskip.\hskip 36.135pt(\enskip
k=0,1,2,\ldots\enskip)$

since $degree(r_{k})$ dcreases at least by one at each step after
$n<degree(B)$ steps we shall have $r_{n}=0$. It is shown in [] that these
recursion force the following basic identities

$\eqalign{&(1)\enskip\enskip q_{n}\hskip 2.5ptp_{n-1}\enskip-\enskip
p_{n}\hskip 2.5ptq_{n-1}\enskip=\enskip=(-1)^{n}\cr&(2)\enskip\enskip
A\enskip=\enskip r_{n-1}\hskip 2.5ptp_{n}\cr&(3)\enskip\enskip
B\enskip=\enskip r_{n-1}\hskip 2.5ptq_{n}\cr}\enskip.$

Since equation (1) here yields that $p_{n}$ and $q_{n}$ are relatively prime,
we see that (2) and (3) yield that the greatest common divisor $D$ of $A$ and
$B$ is necessarily given by $r_{n-1}$.

As pointed out in [] the advantage of this process over the one that is
usually described in most textbooks is that it provides the final answer
without excessive storage of partial results. In fact, only $7$ results need
to be stored at any particular time, a number that is independent of the
choice of $A$ and $B$.

If the roots of a polynomial

$P(t)\enskip=\enskip a_{o}+a_{1}t+a_{2}t^{2}+\cdots+a_{m}t^{m}$

are $x_{1},x_{2},\ldots,x_{n}$ then when $a_{n}=1$ we may write it in the form

$P(t)\enskip=\enskip(t-x_{1})(t-x_{2})\cdots(t-x_{m})$ $None$

Thus, we see that we must have

$a_{n-k}\enskip=\enskip(-1)^{k}\hskip 2.5pte_{k}(x_{1},x_{2},\ldots,x_{n}).$
$None$

Clearly, if $P(t)$ and

$Q(t)\enskip=\enskip
b_{o}+a_{1}t+b_{2}t^{2}+\cdots+b_{m}t^{m}\enskip=\enskip(t-y_{1})(t-y_{2})\cdots(t-y_{n})\hskip
36.135pt(\enskip b_{n}=1\enskip)$

have a root in common, the expression

$\prod_{i=1}^{m}\prod_{j=1}^{n}(x_{i}-y_{j})$

will necessarily vanish. It develops that a multiple of this expression may be
written as a polynomial in the coefficients of $P$ and $Q$. This polynomial is
usually referred to as the Resultant of $P$ and $Q$ and will be denoted here
by $R[P,Q]$. The case $m=2$, $n=3$ is sufficient to get accross the idea and
avoids excessive notation.

Theorem 2.1

$R[P,Q]=det\enskip\pmatrix{a_{o}&a_{1}&a_{2}&0&0\cr 0&a_{o}&a_{1}&a_{2}&0\cr
0&0&a_{o}&a_{1}&a_{2}\cr b_{o}&b_{1}&b_{2}&b_{3}&0\cr
0&b_{o}&b_{1}&b_{2}&b_{3}\cr}\enskip=\enskip(-1)^{2\times 3}\hskip
2.5pta_{2}^{3}\hskip
2.5ptb_{3}^{2}\enskip\prod_{i=1}^{2}\prod_{j=1}^{3}(x_{i}-y_{j})$ $None$

Proof

Note that since the division by $a_{2}$ and $b_{3}$ does not change the roots
of $P$ and $Q$, we can divide both sides of 2.7 by $a_{2}^{3}\hskip
2.5ptb_{3}^{2}$ and reduce ourselves to the case $a_{2}=b_{3}=1$. This given,
note that we have the following matrix multiplication identity:

$\pmatrix{a_{o}&a_{1}&a_{2}&0&0\cr 0&a_{o}&a_{1}&a_{2}&0\cr
0&0&a_{o}&a_{1}&a_{2}\cr b_{o}&b_{1}&b_{2}&b_{3}&0\cr
0&b_{o}&b_{1}&b_{2}&b_{3}\cr}\times\pmatrix{1&1&1&0&0\cr
y_{1}&y_{2}&y_{3}&0&0\cr y_{1}^{2}&y_{2}^{2}&y_{3}^{2}&0&0\cr
y_{1}^{3}&y_{2}^{3}&y_{3}^{3}&1&0\cr
y_{1}^{4}&y_{2}^{4}&y_{3}^{4}&0&1\cr}\enskip=\enskip\pmatrix{P(y_{1})&P(y_{2})&P(y_{3})&0&0\cr
y_{1}P(y_{1})&y_{2}P(y_{2})&y_{3}P(y_{3})&0&0\cr
y_{1}^{2}P(y_{1})&y_{2}^{2}P(y_{2})&y_{3}^{3}P(y_{3})&0&0\cr 0&0&1&b_{3}&0\cr
0&0&0&0&b_{3}\cr}$

and 2.7 follows immediately by equating determinants of boths sides and
cancelling the common factor.

Remark 2.1

Note that the vanishing of the determinant in 2.7 assures that we can find a
non trivial solution to the corresponding homogeneous system. Now a simple
computation shows that we have

$\pmatrix{\alpha_{o}&\alpha_{1}&\alpha_{2}&-\beta_{o}&-\beta_{1}}\enskip\times\enskip\pmatrix{a_{o}&a_{1}&a_{2}&0&0\cr
0&a_{o}&a_{1}&a_{2}&0\cr 0&0&a_{o}&a_{1}&a_{2}\cr b_{o}&b_{1}&b_{2}&b_{3}&0\cr
0&b_{o}&b_{1}&b_{2}&b_{3}\cr}\enskip=0$

if and only if

$(\alpha_{o}+\alpha_{1}t+\alpha_{2}t^{2})\hskip
2.5ptP(t)\enskip-\enskip(\beta_{o}+\beta_{1}t)\hskip 2.5ptQ(t)\enskip=\enskip
0$

and this is equivalent to the statement that $P$ and $Q$ have a non trivial
common factor.

In the same vein we see that the expression

$\Delta(x)\enskip=\enskip\prod_{1\leq i<j\leq n}(x_{i}-x_{j})$

vanishes if and only if $P(x)$ as given by 2.5 has multiple roots. Since,
$\Delta(x)^{2}$ is clearly symmetric in $x_{1},x_{2},\ldots,x_{n}$, Theorem
1.2 and 2.6 guarantee that the latter polynomial should be expressible as a
polynomial in the coefficients $a_{o},a_{1},\ldots,a_{n}$ of $P(t)$. In fact,
we need only replace $Q$ by the derivative $P^{\prime}(t)={d\over dt}P(t)$ in
$R[P,Q]$ to obtain

Theorem 2.2

$R[P,P^{\prime}]\enskip=\enskip a_{m}^{m(m-1)}\hskip
2.5pt\prod_{i=1}^{n}\prod_{j=1}^{n}\ {}^{(i)}\hskip
2.5pt(x_{i}-x_{j})=(-1)^{n(n-1)/2}\hskip 2.5pta_{m}^{m(m-1)}\hskip
2.5pt\Delta(x)^{2}$ $None$

where the superscript ‘‘$\ {}^{(i)}$’’ in the product is to indicate that the
factor corresponding to $j=i$ is to be omitted.

Proof

It is easily seen that the general form of 2.7 may also be written as

$R[P,Q]\enskip=\enskip\hskip 2.5pt(-1)^{nm}a_{m}^{n}\prod_{i=1}^{n}\hskip
2.5ptQ(x_{1})Q(x_{2})\cdots Q(x_{n})\enskip$ $None$

Now we easily see that

$P^{\prime}(x_{i})\enskip=\enskip a_{n}\hskip 2.5pt\prod_{j=1}^{n}\
{}^{(i)}\hskip 2.5pt(x_{i}-x_{j})$ $None$

and 2.8 follows by setting $Q=P^{\prime}$ in 2.9.

The polynomial $R[P,P^{\prime}]$ is usually referred to as the discriminant of
the equation

$P(t)=0\enskip.$

We shall terminante with a simple fact that will play a crucial role in the
sequel. Recall that a polynomial $B(t)$ with coefficients in a field ${\cal
F}$ is said to be irreducible over ${\cal F}$ if it does not admit a
factorization

$B(t)=P(t)Q(t)$

into two polynomials $P,Q\in{\cal F}[t]$ of strictly lesser degree than $B$.
The Euclidean Algorithm immediately yields that if $B(t)$ is irreducible and
$A(t)$ is a polynomial in ${\cal F}[t]$ which has a root in common with $A(t)$
then $B(t)$ must be a factor of $A(t)$. The reason for this is that if $A(t)$
and $B(t)$ share a root (a fact which may be verified by computing $R[A,B]$)
then the greatest common divisor $D(t)$ of $A$ and $B$ (as yielded by the
process descrbed above), is in ${\cal F}[t]$ as well and we would be led to a
contradiction unless $D$ is equal to a constant multiple of $B$. The important
conclusion we draw from this is that if a polynomial $A\in{\cal F}[t]$ shares
a root with an irreducible polynomial $B(t)\in{\cal F}[t]$ then it must vanish
for all the other roots of $B(t)$. This fact has the following immediate
extension.

Proposition 2.1

Let $B(t)\in{\cal F}[t]$ be irreducible in ${\cal F}$ and let $\Phi(t)\in
Rat[{\cal F};t]$ vanish at one of the roots of $B(t)$ then $\Phi(t)$ vanishes
at all the other roots of $B(t)$. Proof

By hypothesis $\Phi(t)=P(t)/Q(t)$ with $P,Q\in{\cal F}$ using the Euclidean
Algorithm we can cancel out (if necessary) the greatest common divisor of $P$
and $Q$ assure that $P$ and $Q$ have no common root. But then $\Phi(t)$
vanishes if and only if $P(t)$ does, and we are thus reduced to the case
discussed above.

3\. The cubic and the quartic

Formulas giving the general solution of the cubic equation

$E_{3}(t)\enskip=\enskip(t-x_{1})(t-x_{2})(t-x_{3})\enskip=\enskip
t^{3}-e_{1}t^{2}+e_{2}t-e_{3}\enskip=\enskip 0$ $None$

where first discovered by Ferreo (sometimes before 1505) rediscoverd by
Tartaglia and published by Cardano in 1545. Setting $\omega=e^{{2\pi\hskip
1.75pti}/3}$ the three roots of 3.1 may be written as follows

$\eqalign{x_{1}&\enskip=\enskip{e_{1}\over
3}\enskip+\enskip\enskip\enskip{\root 3 \of{-{\scriptstyle{1\over 2}}q+{\root
2 \of{R}}}}\enskip\enskip+\enskip\enskip{\root 3 \of{-{\scriptstyle{1\over
2}}q-{\root 2 \of{R}}}}\enskip,\cr x_{2}&\enskip=\enskip{e_{1}\over
3}\enskip+\enskip\omega\hskip 2.5pt{\root 3 \of{-{\scriptstyle{1\over
2}}q+{\root 2 \of{R}}}}\enskip+\enskip\omega^{2}\hskip 2.5pt{\root 3
\of{-{\scriptstyle{1\over 2}}q-{\root 2 \of{R}}}}\enskip,\cr
x_{3}&\enskip=\enskip{e_{1}\over 3}\enskip+\enskip\omega^{2}\hskip 2.5pt{\root
3 \of{-{\scriptstyle{1\over 2}}q+{\root 2 \of{R}}}}\enskip+\enskip\omega\hskip
2.5pt{\root 3 \of{-{\scriptstyle{1\over 2}}q-{\root 2 \of{R}}}}\enskip,\cr}$
$None$

where

$p\enskip=\enskip e_{2}-e_{1}^{2}/3\enskip\enskip,\enskip\enskip
q=-e_{3}+{{\scriptstyle{1\over 3}}}e_{1}e_{2}-{\scriptstyle{2\over 27}}\hskip
2.5pte_{1}^{3}\enskip\enskip and\enskip\enskip
R\enskip=\enskip(q/2)^{2}+(p/3)^{3}\enskip.$ $None$

These formulas are usually derived by the following process, apparently due to
Hudde (1650). We start by making the substitution $t=y+{\scriptstyle{1\over
3}}e_{1}$ in 3.1 and transform it to

$y^{3}+py+q=0$

This given the further substitution

$y=z-{p\over 3z}$ $None$

brings us to the equation

$z^{3}-{p^{3}\over 27z^{3}}+q\enskip=\enskip 0$

or better yet

$z^{6}+qz^{3}-{p^{3}/27}\enskip=\enskip 0\enskip.$ $None$

Since this is a quadratic equation for $z^{3}$ we immediately derive the two
solutions

$z_{1}^{3}\enskip=\enskip-{\scriptstyle{1\over 2}}\hskip
2.5ptq\enskip+\enskip{\root 2 \of{R}}\enskip\enskip or\enskip\enskip
z_{2}^{3}\enskip=\enskip-{\scriptstyle{1\over 2}}\hskip
2.5ptq\enskip-\enskip{\root 2 \of{R}}$

since

$(-{\scriptstyle{1\over 2}}\hskip 2.5ptq\enskip+\enskip{\root 2
\of{R}})(-{\scriptstyle{1\over 2}}\hskip 2.5ptq\enskip-\enskip{\root 2
\of{R}})\enskip=\enskip-(p/3)^{3}$

we may extract cube roots so that

$z_{2}\enskip=\enskip\enskip-\enskip{p\over 3z_{1}}$ $None$

This is given the six roots of 3.5 are

$\matrix{z_{1}&z_{2}\cr\omega z_{1}&\omega^{2}z_{2}\cr\omega^{2}z_{1}&\omega
z_{2}\cr}$

Where each of these pairs multiplies to $-p/3$.

We can now use 3.4 and derive that the three roots of 3.3 may be written in
the form

$\eqalign{y_{1}&\enskip=\enskip\enskip z_{1}\enskip+\enskip\enskip z_{2}\cr
y_{2}&\enskip=\enskip\omega z_{1}\enskip+\enskip\omega^{2}z_{2}\cr
y_{3}&\enskip=\enskip\omega^{2}z_{1}\hskip 2.5pt+\hskip 2.5pt\omega z_{2}\cr}$
$None$

From which the formulas in 3.2 can be immediately obtained.

The quartic equation was treated in a similar manner. That is “ad hoc”
manipulations were used to transform it to equations which could be solved by
extraction of roots. To give a brief idea of the process in this case we start
with

$E_{4}(t)\enskip=\enskip(t-x_{1})(t-x_{2})(t-x_{3})(t-x_{4})\enskip=\enskip
t^{4}-e_{1}t^{3}+e_{2}t^{2}-e_{3}t+e_{4}\enskip=\enskip 0\enskip.$ $None$

Completing the square suggested by the first two terms we can rewrite this
equation in the form

$(t^{2}-{\scriptstyle{1\over 2}}e_{1}\hskip
2.5ptt)^{2}\enskip=\enskip({\scriptstyle{1\over 4}}e_{1}^{2}-e_{2})\hskip
2.5ptt^{2}\enskip+\enskip e_{3}\hskip 2.5ptt-e_{4}\enskip.$

We then add $(t^{2}-{\scriptstyle{1\over 2}}\hskip
2.5pte_{1}t)y+{{\scriptstyle{1\over 4}}}y^{2}$ to both sides and get

$(t^{2}-{\scriptstyle{1\over 2}}\hskip 2.5pte_{1}t+{\scriptstyle{1\over
2}}y)^{2}\enskip=\enskip({{\scriptstyle{1\over 4}}}e_{1}^{2}-e_{2}+y)\hskip
2.5ptt^{2}+(e_{3}-{\scriptstyle{1\over 2}}e_{1}y)\hskip
2.5ptt+{{\scriptstyle{1\over 4}}}y^{2}-e_{4}\enskip.$ $None$

Next, $y$ is determined so that also the term on the right becomes a perfect
square. This requires the coefficients

$A={\scriptstyle{1\over
4}}e_{1}^{2}-e_{2}+y\enskip\enskip\enskip,\enskip\enskip\enskip
B=e_{3}-{\scriptstyle{1\over 2}}\hskip 2.5pte_{1}\hskip
2.5pty\enskip\enskip\enskip,\enskip\enskip\enskip C={\scriptstyle{1\over
4}}\hskip 2.5pty^{2}-e_{4}$

of the quadratic on the right hand side of 3.9 satisfy the equation

$B^{2}-4AC\enskip=\enskip 0\enskip.$

This leads to a cubic equation for $y$. To see what are the roots of this
equation, we should try to factor $B^{2}-4AC$. Nowdays this is easily done,
using any of the available computer algebra packages. In this manner we
discover the pleasing fact that $B^{2}-4AC$ factors beautifully in terms of
the roots of $E_{4}(t)$. Namely, we have

$B^{2}-4AC\enskip=\enskip(y-x_{2}x_{4}-x_{1}x_{3})\enskip(y-x_{1}x_{2}-x_{3}x_{4})\enskip(y-x_{1}x_{4}-x_{2}x_{3})\enskip.$

Thus the three roots of this cubic when expressed in terms of the roots of 3.8
are none other than

$y_{1}\enskip=\enskip x_{1}x_{2}+x_{3}x_{4}\enskip\enskip,\enskip\enskip
y_{2}\enskip=\enskip x_{1}x_{3}+x_{2}x_{4}\enskip\enskip,\enskip\enskip
y_{3}\enskip=\enskip x_{1}x_{4}+x_{2}x_{3}$

We should also note that setting $y=y_{1}$ we have

$\eqalign{&A\enskip=\enskip{\scriptstyle{1\over
4}}e_{1}^{2}-e_{2}+y_{1}\enskip=\enskip{\scriptstyle{1\over 4}}\hskip
2.5pt(x_{1}+x_{2}-x_{3}-x_{4})^{2}\cr&B\enskip=\enskip
e_{3}-{\scriptstyle{1\over 2}}\hskip 2.5pte_{1}\hskip
2.5pty_{1}\enskip=\enskip-{\scriptstyle{1\over 2}}\hskip
2.5pt(x_{1}x_{2}-x_{3}x_{4})(x_{1}+x_{2}-x_{3}-x_{4})\cr&C\enskip=\enskip{\scriptstyle{1\over
4}}\hskip 2.5pty_{1}^{2}-e_{4}\enskip=\enskip{\scriptstyle{1\over 4}}\hskip
2.5pt(x_{1}x_{2}-x_{3}x_{4})^{2}\cr}$ $None$

which yield that the right hand side of 3.9 when $y=y_{1}$ reduces to

${\scriptstyle{1\over 4}}\hskip 2.5pt\bigl{(}\hskip
2.5pt-(x_{1}+x_{2}-x_{3}-x_{4})\hskip 2.5ptt+x_{1}x_{2}-x_{3}x_{4}\hskip
2.5pt\bigr{)}^{2}\enskip.$

This allows us to rewrite 3.9 in the form

$L^{2}-R^{2}\enskip=\enskip(L\enskip-\enskip R)\enskip(L\enskip+\enskip
R)\enskip=\enskip 0\enskip,$ $None$

with

$L\enskip=\enskip t^{2}-{\scriptstyle{1\over 2}}\hskip
2.5pte_{1}t+{\scriptstyle{1\over 2}}y_{1}\enskip\enskip\enskip{\rm
and}\enskip\enskip\enskip R\enskip=\enskip{\scriptstyle{1\over 2}}\hskip
2.5pt(-(x_{1}+x_{2}-x_{3}-x_{4})\hskip 2.5ptt+x_{1}x_{2}-x_{3}x_{4})$

Now the factorization of $A$ in 3.10 suggests setting

$z\enskip=\enskip x_{1}+x_{2}-x_{3}-x_{4}\enskip=\enskip{\root 2
\of{e_{1}^{2}-4e_{2}+4y_{1}}}\enskip,$ $None$

and then the factorization of $B$ in 3.10 gives that

$-{\scriptstyle{1\over
2}}(x_{1}x_{2}-x_{3}x_{4})\enskip=\enskip(e_{3}-{\scriptstyle{1\over
2}}e_{1}y_{1})/z\enskip.$

This given, we may write

$-R\enskip=\enskip{\scriptstyle{1\over 2}}z\hskip
2.5ptt\enskip+\enskip(e_{3}-{\scriptstyle{1\over 2}}\hskip 2.5pte_{1}y_{1})/z$

and the equation in 3.11 may yet be rewritten as

$\bigl{(}t^{2}-{\scriptstyle{1\over 2}}(e_{1}+z)\hskip
2.5ptt+{\scriptstyle{1\over 2}}y_{1}-(e_{3}-{\scriptstyle{1\over
2}}e_{1}y_{1})/z\bigr{)}\hskip 2.5pt\bigl{(}t^{2}-{\scriptstyle{1\over
2}}(e_{1}-z)\hskip 2.5ptt+{\scriptstyle{1\over
2}}y_{1}+(e_{3}-{\scriptstyle{1\over 2}}e_{1}y_{1})/z\bigr{)}=0\enskip.$

Now this is none other than factoring $E_{4}(t)$ in the form

$E_{4}(t)\enskip=\enskip\bigl{(}t^{2}-(x_{1}+x_{2})\hskip 2.5ptt-
x_{1}x_{2}\bigr{)}\enskip\bigl{(}t^{2}-(x_{3}+x_{4})\hskip 2.5ptt-
x_{3}x_{4}\bigr{)}\enskip.$

In fact, it may be easily verified that

$\eqalign{&x_{1}+x_{2}\enskip=\enskip{\scriptstyle{1\over
2}}e_{1}+{\scriptstyle{1\over 2}}z\enskip\enskip,\enskip\enskip
x_{1}x_{2}\enskip=\enskip{\scriptstyle{1\over
2}}y_{1}-(e_{3}-{\scriptstyle{1\over
2}}e_{1}y_{1})/z\enskip,\cr&x_{3}+x_{4}\enskip=\enskip{\scriptstyle{1\over
2}}e_{1}-{\scriptstyle{1\over 2}}z\enskip\enskip,\enskip\enskip
x_{3}x_{4}\enskip=\enskip{\scriptstyle{1\over
2}}y_{1}+(e_{3}-{\scriptstyle{1\over 2}}e_{1}y_{1})/z\enskip.\cr}$

Thus we may obtain the desired expressions for the pairs $x_{1},x_{2}$ and
$x_{3},x_{4}$ by solving the two quadratic equations

$\eqalign{&t^{2}-{\scriptstyle{1\over 2}}(e_{1}+z)\hskip
2.5ptt+{\scriptstyle{1\over 2}}y_{1}-(e_{3}-{\scriptstyle{1\over
2}}e_{1}y_{1})/z\enskip=\enskip 0\enskip,\cr&t^{2}-{\scriptstyle{1\over
2}}(e_{1}-z)\hskip 2.5ptt+{\scriptstyle{1\over
2}}y_{1}+(e_{3}-{\scriptstyle{1\over 2}}e_{1}y_{1})/z\enskip=\enskip
0\enskip.\cr}$

This given, in the $17^{th}$ and $18^{th}$ centuries it was natural to assume
that the solution of the general polynomial equation should be obtainable by
similar manipulations and successive root extractions. This was the motivating
force [] in Lagrange’s investigations in the 1770’s that led him to his
historic paper Réflections sur la résolution algébrique des équations .

To be precise Lagrange was investigating the possibility of finding closed
form expressions for the roots $x_{1},x_{2},\ldots,x_{n}$ which (like those
appearing in 3.2) only involved the elementary symmetric functions
$e_{1},e_{2},\ldots,e_{n}$, roots of unity and radicals. We shall refer to
this as ‘‘solving the general equation by radicals’’.

His point of departure was a close examination of the solutions of the cubic
and the quartic. Remarkably, he was able to sort out of those simingly ad hoc
manipulations a unifying general mechanism of solution. As we shall see
Lagrange discovered that in both cases the final formulas could be reached by
a sequence of identical, purely combinatorial, steps.

This done, he tried to apply this mechanism to the quintic only to discover
that the possibility of pushing it through to the production of general
formulas for the roots of the quintic appeared to lead to a contradiction!

In fact, he was (and he knew he was) within reach of proving the impossibility
of solving the quintic equation by radicals.

Around 1799 Ruffini tried to complete Lagranges proof and although he was able
to push the argument quite a bit further he nevertheless was left with a
hypothesis which he could not remove.

The glory of proving the impossibility of solving the general equation by
radicals was bestowed to Abel (for his 1826 paper) (see []) even though he was
only concerned with the quintic and, as in Ruffini’s work, there were still a
number of gaps in his arguments. We shall not deal with Abel’s work here since
it it departs from the combinatorial approach proposed by Lagrange and later
completed by Galois. In fact, the missing step needed to complete Lagrange
argument and obtain the unsolvability of the quintic by radicals can be
supplied by one single idea of Galois.

To appreciate the beauty of Lagrange’s discoveries we should view his results
in the original 1771 form. Unfortunately, for clarity we must deviate a bit
from Lagrange’s terminology. For instance, although Lagrange proved that the
order of a subgroup of a group is a divisor of the order of the group, he had
to do so in an indirect manner, since the notion of a group in its present
form really started with Galois. Although using modern terminology distorts
somewhat the historical perspective, we will try as much as possible to keep
unchanged the contents of Lagrange’s discoveries. Our main goal in the next
two sections is to present the basic theorems of what is now referred to as
Galois theory in a sequence that makes the transition from Lagrange to Galois
as natural and effortless as possible.

4\. Lagrange’s “Galois” Theory

Throughout Lagrange’s work the roots of an equation

$E_{n}(t)\enskip=\enskip
a_{o}+a_{1}t+a_{2}t^{2}+\cdots+a_{n}t^{n}\enskip=\enskip=0$

are assumed to be independent variables $x_{1},x_{2},\ldots,x_{n}$ and
$E_{n}(t)$ is written in the form

$E_{n}(t)\enskip=\enskip(t-x_{1})(t-x_{2})\cdots(t-x_{n})$ $None$

The basic idea that led Lagrange to an understanding  of the classical
solutions of the cubic and the quartic is a careful analysis of the effect
that permutations of the roots have on various rational functions of the
roots. To make precise what we mean by this we need some notation.

We are given a field ${\cal F}$ which remains unchanged throughout, and for a
function $\Phi$ of the roots which may be in ${\cal
F}[x_{1},x_{2},\ldots,x_{n}]$ or in $Rat[{\cal F};x_{1},\ldots,x_{n}]$ as
needed, we set

$G_{\Phi}\enskip=\enskip\\{\hskip 2.5pt\sigma\in S_{n}\hskip 2.5pt:\hskip
2.5pt\sigma\Phi=\Phi\hskip 2.5pt\\}\enskip.$ $None$

Although Lagrange did not realize (nor did he need) that $G_{\Phi}$ is a
group, we shall not ignore this fact here and obtain Lagrange’s results by
standard present day techniques. We recall that $G_{\Phi}$ is usually referred
to as the stabilizer of $\Phi$.

The first basic result of Lagrange can be stated as follows

Theorem 4.1

For $\Phi,\Psi\in Rat[{\cal F};x_{1},\ldots,x_{n}]$ we have

$G_{\Psi}\subseteq G_{\Phi}$ $None$

If and only if

$\Phi\in Rat[{\cal F},e_{1},e_{2},\ldots,e_{n},\Psi]$

Proof

If

$\theta(y_{1},y_{2},\ldots,y_{n},t)\enskip\in\enskip Rat[{\cal
F},y_{1},y_{2},\ldots,y_{n},t]$

and

$\Phi(x_{1},x_{2},\ldots,x_{n})\enskip=\enskip\theta(e_{1},e_{2},\ldots,e_{n},\Psi)\enskip.$
$None$

Then we clearly have 4.3 since every permutation $\sigma$ leaves
$e_{1},e_{2},\ldots,e_{n}$ unchanged and if $\sigma\in G_{\Psi}$ then also
$\Phi$ does not change. So the condition in 4.3 is trivially necessary.

To show the converse, we resort to the left coset decomposition

$S_{n}\enskip=\enskip\tau_{1}\hskip 2.5ptG_{\Psi}\enskip+\enskip\tau_{2}\hskip
2.5ptG_{\Psi}\hskip 2.5pt+\cdots+\hskip 2.5pt\tau_{k}\hskip
2.5ptG_{\Psi}\enskip.\hskip 36.135pt(\enskip\tau_{1}=identity\enskip)$ $None$

(which by the way, was a Lagrange invention) and set

$Q(t)\enskip=\enskip\sum_{i=1}^{k}\hskip 2.5pt\tau_{i}\Phi\hskip
2.5pt\prod_{j=1}^{k}\hskip-0.72229pt\ {}^{(i)}\hskip
2.5pt(t-\tau_{j}\Psi)\enskip.$

Since any $\sigma\in S_{n}$ permutes the left cosets of $G_{\Psi}$ we may
write

$\sigma\hskip 2.5pt\tau_{i}\enskip=\enskip\tau_{\pi_{i}}h_{i}\hskip
36.135pt(\hskip 2.5pth_{i}\in G_{\Psi}\hskip 2.5pt)$

where the map $i\rightarrow\pi_{i}$ is a permutation of $(1,2,\ldots,k)$. In
particular, from 4.3 we deduce that $\sigma\hskip
2.5pt\Phi=\tau_{\pi_{i}}\Phi$, and thus we must have

$\sigma Q(t)\enskip=\enskip\sum_{i=1}^{k}\hskip 2.5pt\tau_{\pi_{i}}\Phi\hskip
2.5pt\prod_{j=1}^{k}\hskip-0.72229pt\ {}^{(\pi_{i})}\hskip
2.5pt(t-\tau_{j}\Psi)\enskip=\enskip Q(t)\enskip.$

This implies that the coefficients of $Q(t)$ are in $Ratsym[{\cal
F};x_{1},x_{2},\ldots,x_{n}]$, so by Theorem 1.2 they are in $Rat[{\cal
F},e_{1},e_{2},\ldots,e_{n}]$. The same can be said about the polynomial

$P(t)\enskip=\enskip\prod_{i=1}^{k}(t-\tau_{i}\Psi)\enskip.$ $None$

Now setting $t=\Psi$ in $Q(t)$ we get

$Q(\Psi)\enskip=\enskip\Phi\hskip 2.5pt\prod_{j=1}^{k}\hskip-0.72229pt\
{}^{(1)}\hskip 2.5pt(\Psi-\tau_{j}\Psi)\enskip=\enskip\Phi\hskip
2.5ptP^{\prime}(\Psi)\enskip.$ $None$

Since, by construction, the values $\tau_{i}\Psi$ (for $i=1,2,\ldots,k$) are
all distinct we shall have $P^{\prime}(\Psi)\neq 0$ and we can divide it out
in 4.7 to obtain

$\Phi\enskip=\enskip{Q(\Psi)\over
P^{\prime}(\Psi)}\enskip=\enskip\theta(\Psi)$

with

$\theta(t)\enskip=\enskip{Q(t)\over P^{\prime}(t)}\enskip\in\enskip Rat[{\cal
F};e_{1},e_{2},\ldots,e_{n},t]$

as desired.

Remark 4.1

If $G$ is a group and $H\subseteq G$ is a subgroup then the left coset
decomposition

$G\enskip=\enskip\tau_{1}\hskip 2.5ptH\enskip+\enskip\tau_{2}\hskip
2.5ptH\enskip+\enskip\cdots\enskip+\enskip\tau_{m}\hskip 2.5ptH\hskip
36.135pt(\enskip\tau_{1}=identity\enskip)$ $None$

yields that

If $H\subseteq G$ then the order of $H$ divides the order of $G$

4.9

In fact, the equation above gives that $|G|/|H|=m$. This result, which
appeared for the first time in the work of Lagrange, was formulated there as a
statement concerning the number of different values taken by rational
functions of the roots. More precisely, we can derive from 4.5 that $\Psi$ is
a $k$-valued rational function of the roots if and only if
$|S_{n}|/|G_{\Psi}|=k$. Similarly, $\Phi$ is $h$-valued if and only if
$|S_{n}|/|G_{\Phi}|=h$. This gives that when $G_{\Psi}\subseteq G_{\Phi}$ we
have $k/h=|G_{\Phi}|/|G_{\Psi}|$. We shall refer to $h$ and $k$ respectively
as the multiplicities of $\Psi$ and $\Phi$. So taking $H=G_{\Psi}$ and
$G=G_{\Phi}$ in 4.8 we get that $k=hm$. We see then that in Lagrange’s
language the statement in 4.9 becomes

$If\enskip\Phi\enskip is\enskip a\enskip rational\enskip function\enskip
of\enskip\Psi\enskip then\enskip the\enskip multiplicity\enskip
of\enskip\Phi\enskip divides\enskip that\enskip of\enskip\Psi\enskip.$

Lagrange’s proof was based precisely on the coset decomposition. Only he did
not have to call it that way. Indeed to get 4.8 for $H=G_{\Psi}$ and
$G=G_{\Phi}$ all he had to do was bunch together the elements of $G_{\Phi}$
that yielded the same value of $\Psi$.

It will be good here and after for $H$ a subgroup of $G$ to express the fact
that $m=|G|/|H|$ by writing $H\subseteq_{m}G$. This given, these observations
can be sharpened into the following corollary of Theorem 4.1.

Theorem 4.2

If for $\Phi,\Psi\in Rat[{\cal F};x_{1},\ldots,x_{n}]$ we have

$G_{\Psi}\subseteq_{k}G_{\Phi}$ $None$

then $\Psi$ satisfies an equation of degree $k$ with coefficients in
$Rat[{\cal F};e_{1},e_{2},\ldots,e_{n},\Phi]$ which is irreducible in
$Rat[{\cal F};e_{1},e_{2},\ldots,e_{n},\Phi]$.

Proof

Let

$G_{\Phi}\enskip=\enskip\tau_{1}\hskip
2.5ptG_{\Psi}\enskip+\enskip\tau_{2}\hskip
2.5ptG_{\Psi}\enskip+\enskip\cdots\enskip+\enskip\tau_{k}\hskip
2.5ptG_{\Psi}\hskip 36.135pt(\enskip\tau_{1}=identity\enskip)$ $None$

and set

$Q(t)\enskip=\enskip\prod_{i=1}^{k}\hskip 2.5pt(t-\tau_{i}\hskip
2.5pt\Psi)\enskip=\enskip q_{o}(x)+q_{1}(x)\hskip 2.5ptt+\cdots+q_{k}(x)\hskip
2.5ptt^{k}\enskip.$ $None$

Since any $\sigma\in G_{\Phi}$ permutes the left cosets of $G_{\Psi}$ in 4.11
we may write

$\sigma\hskip 2.5pt\tau_{i}\enskip=\enskip\tau_{\pi_{i}}h_{i}\hskip
36.135pt(\hskip 2.5pth_{i}\in G_{\Psi}\hskip 2.5pt)$

where the map $i\rightarrow\pi_{i}$ is again a permutation of
$(1,2,\ldots,k)$. This gives that for all $\sigma\in G_{\Phi}$ we have

$\sigma Q(t)\enskip=\enskip\prod_{i=1}^{n}\hskip
2.5pt(t-\tau_{\pi_{i}}\Psi)\enskip=\enskip Q(t)\enskip.$

Consequently each of the coefficients $q_{i}(x)$ is left unchanged by the
elements of $G_{\Phi}$. From Theorem 4.1 we then derive that each $q_{i}(x)\in
Rat[{\cal F},e_{1},e_{2},\ldots,e_{n},\Phi]$. On the other hand from 4.12 we
get that

$Q(\Psi)\enskip=\enskip
q_{o}(x)+q_{1}(x)\Psi+\cdots+q_{k}(x)\Psi^{k}\enskip=\enskip 0\enskip.$

Now suppose, if possible, that $Q(t)$ has a factorisation
$Q(t)=Q_{1}(t)Q_{2}(t)$ where both polynomials $Q_{1}(t)$ and $Q_{2}(t)$ have
coefficients in $Rat[{\cal F};e_{1},e_{2},\ldots,e_{n},\Phi]$ It will then
follow that both of them will be invariant under the action of $G_{\Phi}$. In
particular, we must have

$\tau_{i}(Q_{1}(\Psi))\enskip=\enskip Q_{1}(\tau_{i}\Psi)\hskip
36.135pt(\enskip for\enskip any\enskip\enskip i=1,2,\ldots,k\enskip)$

So if $\Psi$ is a root of the equation $Q_{1}(t)=0$ then all the other roots
of $Q(t)$ must satisfy it as well and $Q_{2}(t)$ must reduce to a constant in
$Rat[{\cal F},e_{1},e_{2},\ldots,e_{n},\Phi]$. The analogous conclusion holds
if $Q_{2}(\Psi)=0$. Thus $Q(t)$ is irreducible as asserted. This completes our
proof.

Remark 4.2

Here and after, if $G_{\Psi}\subseteq_{k}G_{\Phi}$ and we have the left coset
decomposition in 4.11, then functions

$\Psi_{1}=\tau_{1}\Psi\hskip 2.5pt,\hskip 2.5pt\Psi_{2}=\tau_{2}\Psi\hskip
2.5pt,\hskip 2.5pt\ldots,\Psi_{k}=\tau_{k}\Psi$

will be referred to as the conjugates of $\Psi$ in $G_{\Phi}$. Note that if
$\Phi$ is in $Rat[{\cal F},e_{1},e_{2},\ldots,e_{n},\Psi]$, Theorem 4.1
assures that $G_{\Psi}\in G_{\Phi}$. So in any case we must have 4.11 for some
$k$. Now suppose that $\Psi$ is a root of the equation $R(t)=0$ where $R(t)$
is a polynomial of degree $h$ with coefficients in $Rat[{\cal
F},e_{1},e_{2},\ldots,e_{n},\Phi]$. Since this polynomial is then invariant
under the action of $G_{\Phi}$ all the conjugates of $\Psi$ in $G_{\Phi}$ must
also be roots of $R(t)=0$. This implies that the polynomial $Q(t)$ in 4.12
must be a factor of $R(t)$. However, if $R(t)$ is also irreducible in
$Rat[{\cal F},e_{1},e_{2},\ldots,e_{n},\Phi]$, then $R(t)$ and $Q(t)$ can only
differ by a factor in $Rat[{\cal F},e_{1},e_{2},\ldots,e_{n},\Phi]$ and we
must also $h=k$. This should explain why we call
$PS_{1},\Psi_{2},\ldots,\Psi_{k}$ the “conjugates” of $\Psi$.

We have reached a point where to proceed further we need to make more precise
what we mean by solving the general equation by radicals. To begin with we
shall assume that the given field ${\cal F}$ (nowdays referred to as the
ground field) contains all the roots of unity of any order $\leq n$. This
given, solving by radicals the $n^{tic}$ in 4.1, in the Lagrange setting is to
mean that we can find a sequence of rational functions $\Phi_{i}\in Rat[{\cal
F};x_{1},x_{2},\ldots,x_{n}]$ ($i=0,1,2,\ldots$) such that

$\eqalign{\Phi_{o}&\hskip 2.5pt\in\hskip 2.5ptRat[{\cal
F};e_{1},e_{2},\ldots,e_{n}]\cr\Phi_{i}&\enskip=\enskip{\root p_{i}
\of{\theta_{i}(e_{1},e_{2},\ldots,e_{n},\Phi_{i-1})}}\cr}$ $None$

where each $\theta_{i}$ is a rational function

$\theta_{i}\enskip=\enskip\theta_{i}(y_{1},y_{2},\ldots,y_{n},t)\enskip\in\enskip
Rat[{\cal F},y_{1},y_{2},\ldots,y_{n},t]\enskip.$ $None$

Finally, we shall require that the end function of this sequence say
$\Phi_{d}$ be one of the roots or better yet (as we shall see) a function from
which all the roots may be derived by rational operations.

Note first that since for any two integers $p$ and $q$ and for any $\Phi$ we
have

${\root pq \of{\Phi}}\enskip=\enskip{\root p \of{\root q \of{\Phi}}}$

there is no loss in requiring that the integers $p_{i}$ in 4.13 are all
primes. Finally we can simplify the convoluted form of the recursion in 4.14
by rewriting it in the form

$\eqalign{&a)\enskip\enskip\Phi_{o}\enskip\in Rat[{\cal
F};e_{1},e_{2},\ldots,e_{n}]\cr&b)\enskip\enskip\Phi_{i}^{p_{i}}\in Rat[{\cal
F},e_{1},e_{2},\ldots,e_{n},\Phi_{i-1})\cr}$ $None$

Remark 4.3

Note further that as long as the coefficients $m_{1},m_{2},\ldots,m_{n}$ are
all distinct the function

$v(x)\enskip=\enskip v(x_{1},x_{2},\ldots,x_{n})\enskip=\enskip
m_{1}x_{1}+m_{2}x_{2}+\cdots+m_{n}x_{n}$ $None$

will necessarily be $n!$-valued. Since its stabilizer consists of just the
identity permutation, the hypotheses of Theorem 4.1 are satisfied for any
rational function of the roots $x_{1},x_{2},\ldots,x_{n}$. Thus for any
$\Phi\in Rat[{\cal F};x_{1},\ldots,x_{n}]$ we can construct a rational
function $\theta_{\Phi}(t)\in Rat[{\cal F};e_{1},e_{2},\ldots,e_{n},t]$ giving

$\Phi(x_{1},x_{2},\ldots,x_{n})\enskip=\enskip\theta_{\Phi}\bigl{(}v(x_{1},x_{2},\ldots,x_{n})\bigr{)}\enskip.$
$None$

We should note here for further reference that the proof of Theorem 4.1 yields
that

$\theta_{\Phi}\enskip=\enskip{Q_{\Phi}(t)\over P^{\prime}(t)}$ $None$

where

$Q_{\Phi}(t)\enskip=\enskip\sum_{\sigma\in S_{n}}\hskip
2.5pt\sigma\Phi\prod_{\tau\in S_{n}}\hskip-0.72229pt^{(\sigma)}\hskip
2.5pt(t-\tau v)$ $None$

and

$P(t)\enskip=\enskip\prod_{\tau\in S_{n}}(t-\tau v)\enskip.$ $None$

It is important to notice that the denominator of $\theta_{\Phi}$ in 4.18 is
independent of $\Phi$ itself. 5\. Lagrange’s derivation of the roots of the
cubic and the quartic.

Armed with this information, Lagrange could then come up with the following
a-priori reconstruction of the solutions of the cubic and the quartic.

a) The cubic

Let $\omega=e^{2\pi i/3}$ and note that, since $1,\omega,\omega^{2}$ are
distinct, the expression

$(x_{1}+\omega x_{2}+\omega^{2}x_{3})/3$

is necessarily a $6$-valued function of the roots of $E_{3}(t)$. Its values
are

$\matrix{z_{1}=(x_{1}+\omega
x_{2}+\omega^{2}x_{3})/3\enskip\enskip&z_{2}=(x_{1}+\omega
x_{3}+\omega^{2}x_{2})/3\enskip\enskip\cr z_{3}=(x_{2}+\omega
x_{3}+\omega^{2}x_{1})/3\enskip\enskip&z_{4}=(x_{2}+\omega
x_{1}+\omega^{2}x_{3})/3\enskip\enskip\cr z_{5}=(x_{3}+\omega
x_{1}+\omega^{2}x_{2})/3\enskip\enskip&z_{6}=(x_{3}+\omega
x_{2}+\omega^{2}x_{1})/3\enskip\enskip\cr}\enskip.$ $None$

Now we see that

$z_{3}=\omega^{2}z_{1}\enskip\hskip 2.5pt,\hskip 2.5pt\enskip z_{5}=\omega
z_{1}\enskip\enskip\enskip{\rm and}\enskip\enskip\enskip z_{4}=\omega
z_{2}\enskip\hskip 2.5pt,\hskip 2.5pt\enskip z_{6}=\omega^{2}z_{2}\enskip.$

Thus

$\eqalign{(z-z_{1})(z-z_{3})(z-z_{5})&\enskip=\enskip
z^{3}-z_{1}^{3}\cr(z-z_{2})(z-z_{4})(z-z_{6})&\enskip=\enskip
z^{3}-z_{2}^{3}\cr}$

This implies that the $6$-degree equation

$(z-z_{1})(z-z_{2})(z-z_{3})(z-z_{4})(z-z_{5})(z-z_{6})\enskip=\enskip 0$
$None$

must take the form

$(z^{3}-z_{1}^{3})(z^{3}-z_{2}^{3})\enskip=\enskip
z^{6}-(z_{1}^{3}+z_{2}^{3})\hskip 2.5ptz^{3}\enskip+\enskip
z_{1}^{3}z_{2}^{3}\enskip=\enskip 0\enskip.$ $None$

Lagrange, on the basis that 5.2 is symmetric in the roots of $E_{3}(t)$, could
now predict that the expressions

$z_{1}^{3}+z_{2}^{3}\enskip\enskip{\rm and}\enskip\enskip z_{1}^{3}z_{2}^{3}$

must necessarily be polynomials in $e_{1},e_{2},e_{3}$. And indeed it can
directly be checked from 5.1 that

$z_{1}^{3}+z_{2}^{3}\enskip=\enskip-q\enskip\enskip{\rm and}\enskip\enskip
z_{1}^{3}z_{2}^{3}\enskip=\enskip-(p/3)^{3}$ $None$

with $p$ and $q$ given by 3.3. This immediately leads to the bicubic equation

$z^{6}+qz^{3}-(p/3)^{3}\enskip=\enskip 0\enskip.$ $None$

Now its solution leads to the extraction of the square root of

$\Phi_{o}=q^{2}+4(p/3)^{3}=4\hskip 2.5ptR\enskip.$

This expression is essentially the discriminant of $E_{3}(t)$. More precisely
we have

$\Phi_{o}(x)\enskip=\enskip-{\scriptstyle{1\over
27}}(x_{1}-x_{2})^{2}(x_{1}-x_{3})^{2}(x_{2}-x_{3})^{2}$

and its square root may be chosen to be

$\Phi_{1}(x)\enskip=\enskip i({x_{1}-x_{2}\over{\root 2
\of{3}}})({x_{1}-x_{3}\over{\root 2 \of{3}}})({x_{2}-x_{3}\over{\root 2
\of{3}}})\enskip.$

This is a polynomial in the roots of $E_{3}(t)$ whose stabilizer
$G_{\Phi_{1}}$ is the group of even permutations of $S_{3}=G_{\Phi_{o}}$. Now
this is in perfect agreement with Theorem 4.2.

Of course, in view of 5.4, we also have

$\Phi_{o}(x)\enskip=\enskip(z_{1}^{3}+z_{2}^{3})^{2}-4(z_{1}^{3}z_{2}^{3})\enskip=\enskip(z_{1}^{3}-z_{2}^{3})^{2}$

and it can be easily verified that

$\Phi_{1}(x)\enskip=\enskip z_{1}^{3}-z_{2}^{3}\enskip.$

In other words $\Phi_{1}(x)$ is the solution of

$\Phi_{1}^{2}(x)\enskip=\enskip\Phi_{o}\enskip.$

Using 5.4 again we deduce that

$z_{1}^{3}\enskip=\enskip-{\scriptstyle{1\over
2}}q\enskip+\enskip{\scriptstyle{1\over 2}}\Phi_{1}(x)\enskip.$

This given, the final step is the construction of the $6$-valued function

$\Phi_{2}(x)=z_{1}=(x_{1}+\omega x_{2}+\omega^{2}x_{3})/3$

whose stabilizer $G_{\Phi_{2}}$ is trivial and is the solution of

$\Phi_{2}(x)^{3}\enskip=\enskip-{\scriptstyle{1\over 2}}\hskip
2.5ptq\enskip+\enskip{\scriptstyle{1\over 2}}\Phi_{1}(x)\enskip.$

In summary, this construction of a $3!$-valued function of the roots of
$E_{3}(t)$ has led us to the following scheme:

$\matrix{function&&expression&group\cr\enskip&\enskip&\enskip&\enskip\cr\Phi_{o}&\enskip=\enskip&-{\scriptstyle{1\over
27}}(x_{1}-x_{2})^{2}(x_{1}-x_{3})^{2}(x_{2}-x_{3})^{2}&S_{3}\cr\downarrow&\enskip&\enskip&3\downarrow\cr\Phi_{1}&\enskip=\enskip&i({x_{1}-x_{2}})({x_{1}-x_{3}})({x_{2}-x_{3}})/3{\root
2
\of{3}}&\\{id,(1,2,3),(1,3,2)\\}\cr\downarrow&\enskip&\enskip&2\downarrow\cr\Phi_{2}&\enskip=\enskip&(x_{1}+\omega
x_{2}+\omega^{2}x_{3})/3&\\{id\\}\cr}$

Moreover we have that

$\Phi_{2}^{3}\enskip=\enskip-{\scriptstyle{1\over 2}}\hskip
2.5ptq+{\scriptstyle{1\over 2}}\hskip
2.5pt\Phi_{1}\enskip\enskip\enskip,\enskip\enskip\enskip\Phi_{1}^{2}\enskip=\enskip\Phi_{o}$

This shows that the solution of the cubic can be obtained the succession of
steps

$\matrix{\Phi_{2}^{3}\in Rat[{\cal
F},e_{1},e_{2},e_{3},\Phi_{1}]&\longleftarrow&\Phi_{1}^{2}\in Rat[{\cal
F},e_{1},e_{2},e_{3},\Phi_{o}]&\longleftarrow&\Phi_{o}\in Rat[{\cal
F},e_{1},e_{2},e_{3}]\cr G_{\Phi_{2}}\subseteq
G_{\Phi_{1}}&&G_{\Phi_{1}}\subseteq G_{\Phi_{o}}&&G_{\Phi_{o}}=S_{3}\cr}$

b) The quartic

We can proceed in the same manner as for the cubic and construct a sequence of
rational functions of the roots according to the scheme expressed in 4.15,
terminating again with a $4!$-valued function

$v(x)\enskip=\enskip m_{1}x_{1}+m_{2}x_{2}+m_{3}x_{3}+m_{4}x_{4}\enskip.$

We may choose here

$v(x)\enskip=\enskip x_{1}-x_{2}+i(x_{3}-x_{4})\enskip.$ $None$

As in Section 3, we set $\omega=e^{2\pi i/3}$ and

$y_{1}=x_{1}x_{2}+x_{3}x_{4}\enskip\enskip,\enskip\enskip
y_{2}=x_{1}x_{3}+x_{2}x_{4}\enskip\enskip,\enskip\enskip
y_{3}=x_{1}x_{4}+x_{2}x_{3}\enskip.$ $None$

Moreover we let

$w_{1}=x_{1}+x_{2}-x_{3}-x_{4}\enskip\enskip,\enskip\enskip
w_{2}=x_{1}+x_{3}-x_{2}-x_{4}\enskip\enskip,\enskip\enskip
w_{3}=x_{1}+x_{4}-x_{2}-x_{3}\enskip.$ $None$

This given, we find that in this case repetitive uses of Lagrange’s Theorem
4.1 naturally leads us to the the following scheme:

$\matrix{function&&expression&group\cr\enskip&\enskip&\enskip&\enskip\cr\Phi_{o}&\enskip=\enskip&(x_{1}-x_{2})^{2}(x_{1}-x_{3})^{2}\cdots(x_{3}-x_{4})^{2}&S_{4}\cr\downarrow&\enskip&\enskip&2\downarrow\cr\Phi_{1}&\enskip=\enskip&(x_{1}-x_{2})(x_{1}-x_{3})\cdots(x_{3}-x_{4})&A_{4}\cr\downarrow&\enskip&\enskip&3\downarrow\cr\Phi_{2}&\enskip=\enskip&y_{1}+\omega
y_{2}+\omega^{2}y_{3}&\\{id,(1,2)(3,4),(1,3)(2,4),(1,4)(2,3)\\}\cr\downarrow&\enskip&\enskip&2\downarrow\cr\Phi_{3}&\enskip=\enskip&w_{2}w_{3}+i(y_{2}-y_{3})&\\{id,(1,2)(3,4)\\}\cr\downarrow&\enskip&\enskip&2\downarrow\cr\Phi_{4}&\enskip=\enskip&x_{1}-x_{2}+i(x_{3}-x_{4})&\\{id\\}\cr}$
$None$

The fact that $G_{\Phi_{o}}=S_{4}$ is immediate since $\Phi_{o}$ is a
symmetric function of the roots whose expression in terms of
$e_{1},e_{2},e_{2},e_{4}$ is given by the discriminant formula 2.8. Clearly
$\Phi_{1}$ is invariant only under even permutations of the roots. So
$G_{\Phi_{1}}$ is simply the alternating group $A_{4}$. To obtain
$G_{\Phi_{2}}$, we simply observe that, since $1,\omega,\omega^{2}$ are
distinct, $\Phi_{2}$ is invariant only under those permutations of the roots
that leave $y_{1},y_{2},y_{3}$ individually invariant. In other words
$G_{\Phi_{2}}=G_{y_{1}}\cap G_{y_{3}}\cap G_{y_{3}}$. This gives us the third
entry in the fourth column of 5.8. To obtain $G_{\Phi_{3}}$ we note that for
$y_{2}-y_{3}$ not to change we need each of $y_{1},y_{2},y_{3}$ to remain
unchanged, thus $G_{y_{2}-y_{3}}=G_{\Phi_{2}}$. Now it is easily seen that

$G_{w_{2}w_{3}}\enskip=\enskip\\{id,(1,2),(3,4),(1,2)(3,4)\\}$

and thus we must have

$G_{\Phi_{3}}\enskip=\enskip G_{w_{2}w_{3}}\cap
G_{\Phi_{2}}\enskip=\enskip\\{id,(1,2)(3,4)\\}\enskip$

as asserted in 5.8.

We can easily see that

$\Phi_{1}^{2}\enskip=\enskip\Phi_{o}\enskip\enskip{\rm
and}\enskip\enskip\Phi_{4}^{2}\enskip=\enskip\Phi_{3}$ $None$

We can painlessly check on the computer that

$\Phi_{2}^{3}\enskip=\enskip{\scriptstyle{3\over 2}}{\root 2 \of{3}}\hskip
2.5pt\Phi_{1}\enskip+\enskip J\enskip,$ $None$

where

$J\enskip=\enskip{\scriptstyle{1\over
2}}(2y_{1}-y_{2}-y_{3})(2y_{2}-y_{1}-y_{3})(2y_{3}-y_{1}-y_{2})\enskip.$

Note next that we have the coset decomposition

$G_{\Phi_{2}}\enskip=\enskip
G_{\Phi_{2}}\enskip+\enskip(1,3)(2,4)G_{\Phi_{2}}$

Thus the conjugate of $\Phi_{3}$ in $G_{\Phi_{2}}$ is

$\Phi_{3}^{\prime}\enskip=\enskip(1,3)(2,4)\Phi_{3}\enskip=\enskip-
w_{2}w_{3}+i(y_{2}-y_{3})\enskip.$

We can now immediately conclude from Theorem 4.2 that the coefficients of the
polynomial

$(t-\Phi_{3})(t-\Phi_{3}^{\prime})=\big{(}t-i(y_{2}-y_{3})\big{)}^{2}-w_{2}^{2}w_{3}^{2}=t^{2}-4i(y_{2}-y_{3})t-4(y_{2}-y_{3})^{2}-w_{2}^{2}w_{3}^{2}\enskip=\enskip$
$None$

must be in $Rat[{\cal F};e_{1},e_{2},e_{3},e_{4},\Phi_{2}]$. It is interesting
to see what they actually turn out to be. For instance we can write

$w_{2}^{2}w_{3}^{2}\enskip=\enskip{(w_{1}w_{2}w_{3})^{2}\over
w_{1}^{2}}\enskip.$ $None$

Now from 3.10 we get that

$w_{1}^{2}\enskip=\enskip e_{1}^{2}-4e_{2}+4y_{1}\enskip.$ $None$

Since $G_{\Phi_{2}}\subseteq G_{y_{1}}$, we know that $y_{1}$ should be in
$Rat[{\cal F};e_{1},e_{2},e_{3},e_{4},\Phi_{2}]$. In fact, we can easily
verify that

$y_{1}\enskip=\enskip{\scriptstyle{1\over 3}}\hskip
2.5pt(e_{2}+\Phi_{2}+{\overline{\Phi}}_{2})\enskip.$ $None$

In case we might worry that the complex conjugate ${\overline{\Phi}}_{2}$ may
not be in $Rat[{\cal F};e_{1},e_{2},e_{3},e_{4},\Phi_{2}]$, I will quickly
point out that we have the identity

${\overline{\Phi}}_{2}\enskip=\enskip{e_{1}^{2}-3e_{2}\over\Phi_{2}}\enskip.$

Similarly, we can easily derive that

$i(y_{2}-y_{3})\enskip=\enskip{1\over{\root 2 \of{2}}}\hskip
2.5pt(\Phi_{2}-{\overline{\Phi}}_{2})\enskip.$

Finally combining 5.11,5.12,5.13,5.14 we get that the function

$\Phi_{3}^{*}\enskip=\enskip\Phi_{3}-i(y_{2}-y_{3})$

satisfies the equation

$(\Phi_{3}^{*})^{2}\enskip=\enskip{w_{1}^{2}w_{2}^{2}w_{3}^{2}\over
e_{1}^{2}-4e_{2}+4(e_{2}+\Phi_{2}+{\overline{\Phi}}_{2})/3}\enskip.$

Thus we have again obtained the desired $n!$-valued function by constructing a
sequence of functions $\Phi_{i}(x)\in Rat[{\cal F};x_{1},x_{2},\ldots,x_{n}]$
satisfying the recurrence

$\Phi_{i}(x)^{p_{i}}\in Rat[{\cal
F};e_{1},e_{2},\ldots,e_{n},\Phi_{i-1}]\hskip 36.135pt(\enskip{\rm
with}\enskip\enskip\Phi_{o}\in Rat[{\cal F};e_{1},e_{2},\ldots,e_{n}]\enskip)$
$None$

Where, exponents $p_{1},p_{2},p_{3}\ldots$ give the prime factorization of
$n!$. Moreover we also have verified that the additional condition

$G_{\Phi_{i}}\subseteq G_{\Phi_{i-1}}$ $None$

holds throughout.

This given, Lagrange was convinced (and, with hindsight, he was right), that
if the quintic should be solvable by radicals, then the $5!$-valued function
of its roots must be reachable by a sequence of steps as given in 5.14 and
5.15. However in trying to reproduce the same scheme for the quintic equation
he run into insurmountable difficulties. His final efforts were towards
constructing a rational function of the roots $x_{1},x_{2}\,\ldots,x_{5}$ that
took less than $5$ values or one that took exactly five values but was the
solution of a binomial quintic. The reason for this search was his need to
reduce the construction of the roots to the solution of an equation of degree
less than five or to an equation of degree $5$ he could solve by taking
radicals. Of course he was allowing the possibility, of having to solve an
equation of degree greater than five as long as this equation, as in the case
of the cubic, could be reduced to the solution of an equation of lesser degree
for a power of one of its roots. However, this power would then have to be a
$k$-valued function for some $k<5$. The best he could do was to produce a
$6$-valued function. He concludes his work (see []) by saying that although he
had not tried every possibility, the search (without MAPLE or MATHEMATICA) was
considerably time consuming… and he had no further time to spend in the search
for something

‘‘whose existence is very much in doubt’’

He had again the correct suspicion! The non existence of what he was looking
for was proved (30 years later) by Ruffini and generalized for $n>5$ by
Cauchy.

Remark 5.1

It is interesting in this connection to observe that there is a neat
representation theoretical reason why there is no $k$-valued function of the
roots of $E_{n}(t)$ when $2<k<n$ and $n\geq 5$. In fact, the action of $S_{n}$
on the stabilizer of any $k$-valued function $\Phi$ induces a permutation
representation with only one occurrence of the trivial. If the remaining
irreducible constituents are all sign representations then $\Phi$ is only $2$
valued. So for $k>2$ this representation would have to have a irreducible
constituent of degree $\geq 2$. But for $n\geq 5$ except for the trivial and
the sign representation all the other irreducible representations have
dimension $\geq n-1$. So $k\leq 2$ or $k\geq n$. At any rate we can also give
this result an elementary proof.

It is important to see at this point what are the implications of the
existence of a sequence of functions $\Phi_{i}$ satisfying 5.14 and 5.15. To
this end we terminate this section with a collection of results which shed
considerable light on the difficulties encountered by Lagrange in completing
his program. We should note that although some of the arguments that follow
use the “group” structure of stabilizers, which is one of the main discoveries
of Galois, most of the calculations are actually due to Lagrange. This
apparent paradox illustrates in a remarkable way how close Lagrange got to
discover some of the main points of “Galois” theory.

Theorem 5.1

Let $\Phi$ and $\Psi$ be in $Rat[{\cal F};e_{1},e_{2},\ldots,e_{n}]$. Suppose
that

$\eqalign{&a)\enskip\enskip G_{\Psi}\subseteq G_{\Phi}\enskip\enskip
but\enskip\enskip G_{\Psi}\neq
G_{\Phi}\enskip\enskip,and\cr&b)\enskip\enskip\Psi^{p}\in Rat[{\cal
F};e_{1},e_{2},\ldots,e_{n},\Phi]\cr}$ $None$

where $p$ is prime. Then the decomposition of $G_{\Phi}$ into left cosets of
$G_{\Psi}$ may be written in the form

$G_{\Phi}\enskip=\enskip G_{\Psi}+\gamma
G_{\Psi}+\gamma^{2}G_{\Psi}+\cdots+\gamma^{p-1}G_{\Psi}$ $None$

with $\gamma$ a $p$-cycle that commutes with $G_{\Psi}$. Moreover, the
conjugates of $\Psi$ in $G_{\Phi}$ can be written in the form

$\Psi_{i}=\gamma^{i-1}\Psi=\omega^{i-1}\Psi\hskip 36.135pt(\enskip
with\enskip\enskip\omega=e^{2\pi i/p}\enskip)$ $None$

Proof

Since by assumption $G_{\Psi}\neq G_{\Phi}$ we have

$G_{\Phi}\enskip=\enskip\tau_{1}G_{\Psi}+\tau_{2}G_{\Psi}+\cdots+\tau_{k}G_{\Psi}\enskip.$

with $k>1$. Set $\omega=e^{2\pi i/p}$ and let

$Q(t)\enskip=\enskip\prod_{i=0}^{p-1}\hskip 2.5pt(t-\omega^{i}\Psi)\enskip.$
$None$

From 5.16 b) we deduce that

$Q(t)\enskip=\enskip t^{p}\enskip-\enskip R(x)$

with $R(x)\in Rat[{\cal F};e_{1},e_{2},\ldots,e_{n},\Phi]$. In particular for
any $\sigma\in G_{\Phi}$ we must have

$\sigma Q(t)\enskip=\enskip\prod_{i=0}^{p-1}\hskip
2.5pt(t-\omega^{i}\sigma\Psi)\enskip=\enskip
Q(t)\enskip=\enskip\prod_{i=0}^{p-1}\hskip 2.5pt(t-\omega^{i}\Psi)\enskip.$

and this gives

$\sigma\Psi\enskip=\enskip\omega^{i(\sigma)}\hskip 2.5pt\Psi$ $None$

for some $0\leq i(\sigma)\leq p-1$. This implies that the conjugates of $\Psi$
are all multiples of $\Psi$ by powers of $\omega$. Thus

$k\leq p\enskip.$

Now $k>1$ gives that $\tau_{2}\Psi\neq\Psi$ so $\enskip i(\tau_{2})\neq
0\enskip(mod\enskip p)$. But since $p$ is prime, the successive powers
$(\omega^{i(\tau_{2})})^{s}$ for $s=0,1,\ldots,p-1$ are simply
$1,\omega,\ldots,\omega^{p-1}$ in some order. Thus

$\Psi,\tau_{2}\Psi,\tau_{2}^{2}\Psi\ldots,\tau_{2}^{p-1}\Psi$

are all distinct. This gives

$p\leq k\enskip$

Thus $k$ must be equal to $p$ and

$\tau_{2}^{p}=id\enskip.$

Moreover, we could have indexed our coset representatives $\tau_{i}$ to be
successive powers of an element $\gamma\in G_{\Phi}$ for which
$\gamma\Psi=\omega\Psi$. This gives 5.17. With this choice we have

$\Psi_{i}\enskip=\enskip\gamma^{i}\Psi\enskip=\enskip\omega^{i-1}\hskip
2.5pt\Psi\hskip 36.135pt(\enskip i=1,2,\ldots,p\enskip)$

Finally, we see that $\gamma\Psi=\omega\Psi$ gives

$\gamma^{-1}h\gamma\hskip 2.5pt\Psi\enskip=\enskip\Psi\hskip
36.135pt(\enskip\forall\enskip\enskip h\in G_{\Psi})\enskip.$

In other words

$\gamma^{-1}\hskip 2.5ptG_{\Psi}\gamma\enskip=\enskip G_{\Psi}$ $None$

This completes our proof.

Remark 5.2

In modern terminology this theorem simply says that 5.16 a) and b) imply that
$G_{\Psi}$ is a normal subgroup $G_{\Phi}$ and the quotient
$G_{\Phi}/G_{\Psi}$ is isomorphic to the group of integers $mod\hskip 2.5ptp$.
To express these two properties we shall here and after write

$G_{\Psi}\triangleleft_{p}G_{\Phi}\enskip.$ $None$

Thus the possibility of solving the general equation $E_{n}(t)$ by radicals as
was done with the cubic and the quartic according to the scheme given by 5.14
and 5.15 requires that we should have a sequence of subgroups $G_{i}\subseteq
S_{n}$ (for $i=0,1,\ldots,k)$ satisfying the conditions

$\\{id\\}=G_{m}\triangleleft_{p_{m}}G_{m-1}\triangleleft_{p_{m-1}}\cdots
G_{2}\triangleleft_{p_{2}}G_{1}\triangleleft_{p_{1}}G_{o}=S_{n}$ $None$

We shall soon see that this is impossible for $n\geq 5$. However, we must
first establish some basic facts about $S_{n}$ and its subgroups.

Proposition 5.1

Let $G$ be a subgroup of $S_{n}$ which contains all $3$-cycles and let $H$ be
a subgroup $G$. Suppose that for some $\gamma\in G$ we have

$\eqalign{&a)\enskip\enskip G=H+\gamma
H+\cdots+\gamma^{p-1}H\enskip,\cr&b)\enskip\enskip\gamma H\enskip=\enskip
H\gamma\enskip.\cr}$ $None$

Then for $n\geq 5$ also $H$ contains all $3$-cycles.

Proof

Condition a) in 5.24 gives that we can write

$(1,2,3)\enskip=\enskip\gamma^{i}a\enskip\enskip\enskip,\enskip\enskip\enskip(3,4,5)\enskip=\enskip\gamma^{j}b$

for some $0\leq i,j<p$ and $a,b\in H$. On the other hand b) in 5.24 gives that
we can write

$(1,2,3)(3,4,5)\enskip=\enskip\gamma^{i}a\gamma^{j}b=\gamma^{i+j}\hskip
2.5ptc\enskip\enskip\enskip,\enskip\enskip\enskip(3,2,1)(5,4,3)\enskip=\enskip
a^{-1}\gamma^{-i}b^{-1}\gamma^{-j}=c^{\prime}\gamma^{-i-j}$

for some $c,c^{\prime}\in H$. Thus

$(2,5,3)\enskip=\enskip(3,2,1)(5,4,3)(1,2,3)(3,4,5)\enskip=\enskip
c^{\prime}\gamma^{-i-j}\gamma^{i+j}\hskip 2.5ptc\in H\enskip.$

Since the same manipulations can be carried out when $1,2,3,4,5$ are replaced
by any $5$ distinct integers $i_{1},i_{2},i_{3},i_{4},i_{5}$, we see that $H$
must contain all $3$ cycles as asserted.

Proposition 5.2

If a subgroup $G\subseteq S_{n}$ contains all $3$-cycles then $G=S_{n}$ or $G$
is the alternating group $A_{n}$

Proof

Note that we have

$\eqalign{(1,2)(2,3)&\enskip=\enskip(1,2,3)\cr(1,2)(3,4)&\enskip=\enskip(1,2)(2,3)(2,3)(2,4)\enskip=\enskip(1,3,2)(2,4,3)\cr}$

Thus the product of two $2$-cycles can always be expressed as either a
$3$-cycle or as a product of two $3$-cycles. This implies that every even
permutation is a product of $3$-cycles. Thus under our hypothesis $G\supseteq
A_{n}$. Now if $G$ contains a single permutation $\sigma$ not in $A_{n}$ (that
is an odd permutation), the identity

$S_{n}\enskip=\enskip A_{n}\enskip+\enskip\sigma A_{n}$

yields that $G=S_{n}$ as desired.

Combining these two propositions we derive that we cannot have 5.23 when
$n\geq 5$. In fact the string of inclusions in 5.23 must stop with
$G_{1}=A_{n}$ and can go no further! We can thus state

Theorem 5.2 ( Ruffini)

For $n\geq 5$ the general equation $E_{n}(t)=0$ cannot be solved by radicals
by successive adjunctions of rational functions $\Phi_{i}\in Rat[{\cal
F},x_{1},x_{2},\ldots,x_{n}]$ satisfying the recursions in 5.14 and 5.15 as
was possible for the cubic and the quartic.

We should note that this result doesn’t quite put to rest the possibility of
solving the general equation by adjoining “radicals”. This is because there
are still some unjustified hypotheses in the present setting. First of all, in
our original definition (see 4.15) of solution by radicals we did not have the
extra condition 5.15. As we shall see this is a relatively minor point, easily
fixed in the Galois setting. Considerably more troubling is that in this
definition we require that each newly constructed $\Phi_{i}$ should turn out
to be a rational function in the roots. What if we allow such steps as taking
a $p^{th}$ root of $x_{1}x_{2}\cdots x_{n}$? Can we then construct a solution?
These doubts would be removed if we could show that the existence of a
sequence of steps as in 5.14, 5.15 without the further assumption that
$\Phi_{i}\in Rat[{\cal F},x_{1},x_{2},\ldots,x_{n}]$ implies the existence of
the same sequence of steps with this assumption satisfied. This is in essence
what Abel succeeded in showing. We shall not present Abel’s arguments here
since what must be proved to remove the remaining doubts can be done in an
easier way in the Galois setting. We shall thus terminate our treatment of
Lagrange’s “Galois” Theory with one final result which in some sense reverses
Theorem 5.1.

Theorem 5.3

If $\Psi,\Phi\in Rat[{\cal F};x_{1},x_{2},\ldots,x_{n}]$ and

$G_{\Phi}\enskip=\enskip G_{\Psi}+\gamma
G_{\Psi}+\gamma^{2}G_{\Psi}+\cdots+\gamma^{p-1}G_{\Psi}$ $None$

with $\gamma^{p}=id$ and

$\gamma G_{\Psi}\enskip=\enskip G_{\Psi}\gamma\enskip.$ $None$

Then we can find $\Theta\in Rat[{\cal F};x_{1},x_{2},\ldots,x_{n}]$ such that

$\eqalign{&1)\enskip\enskip G_{\Theta}\supseteq
G_{\Psi}\enskip\enskip(\Theta\in Rat[{\cal
F};e_{1},e_{2},\ldots,e_{n},\Psi]),\cr&2)\enskip\enskip\Theta^{p}\in Rat[{\cal
F};e_{1},e_{2},\ldots,e_{n},\Phi].\cr}$ $None$

Proof

Set

$\Theta\enskip=\enskip\Psi+\omega\gamma\Psi+(\omega\gamma)^{2}\Psi+\cdots+(\omega\gamma)^{p-1}\Psi\enskip.\hskip
36.135pt(\enskip\omega=e^{2\pi i/3}\enskip)$ $None$

It is easily verified that

$\gamma\Theta\enskip=\enskip\omega^{-1}\Theta\enskip.$ $None$

Moreover, 5.26 gives that each of the conjugates $\Psi_{i}=\gamma^{i-1}\Psi$
has the same stabilizer as $\Psi$. Thus from the definition 5.28 we get that

$G_{\Theta}\supseteq G_{\Psi}$ $None$

and therefore (again by Theorem 4.1) we must have

$\Theta\in Rat[{\cal F};e_{1},e_{2},\ldots,e_{n},\Psi]$

Finally, 5.30 together with 5.25 and 5.26 implies that the polynomial

$Q(t)\enskip=\enskip\prod_{i=1}^{p}(t-\gamma^{i-1}\Theta)\enskip=\enskip\prod_{i=1}^{p}(t-\Theta/\omega^{i-1})\enskip=\enskip
t^{p}-\Theta^{p}$ $None$

is left invariant by every element of $G_{\Phi}$. Thus from Theorem 4.1 (with
$\Psi\rightarrow\Phi$ and $\Phi\rightarrow\Theta$) we derive that

$\Theta^{p}\in Rat[{\cal F};e_{1},e_{2},\ldots,e_{n},\Phi]\enskip,$

as desired.

Remark 5.3

We should point out that the inequality in 5.30 cannot in general be sharpened
to an equality. Indeed, in our construction of the roots of the quartic, the
function

$\Phi_{3}=w_{2}w_{3}+i(y_{2}-y_{3})$

satisfies the conditions in 5.25 and 5.26 with $\Psi\rightarrow\Phi_{3}$,
$\Phi\rightarrow\Phi_{2}$ $p=2$. In this case we have

$\gamma=(1,3)(2,4)\enskip\enskip,\enskip\enskip\omega=-1$

and

$\Theta=\Phi_{3}+\omega\gamma\Phi_{3}=2w_{2}w_{3}\enskip.$

As we have seen

$G_{w_{2}w_{3}}\enskip=\enskip\\{id,(1,2),(3,4),(1,2)(3,4)\\}$

while

$G_{\Phi_{3}}\enskip=\enskip\\{id,(1,2)(3,4)\\}\enskip.$

Nevertheless, in a typical application of Theorem 5.3 we may want to obtain
$\Theta$ by extracting a $p^{th}$ root of a function in $Rat[{\cal
F};e_{1},e_{2},\ldots,e_{n},\Phi]$ then recover $\Psi$ by rational operations
involving $\Theta$ and possibly other known functions. Now this is provided by
the following beautiful identity.

Theorem 5.4 (Lagrange)

If $\Psi,\Phi\in Rat[{\cal F};x_{1},x_{2},\ldots,x_{n}]$ and we have 5.25,
5.26 with $\gamma^{p}=id$ and $p$ a prime. Then

$\Psi\enskip=\enskip(c_{o}+c_{1}\Theta+\cdots+c_{p-1}\Theta^{p-1})/p$ $None$

where all the coefficients $c_{i}$ are in $Rat[{\cal
F};e_{1},e_{2},\ldots,e_{n},\Phi]$.

Proof

For convenience set $\Psi_{s}=\gamma^{s}\Psi$

$\Theta_{i}\enskip=\enskip\sum_{s=0}^{p-1}\hskip
2.5pt\omega^{si}\Psi_{s}\enskip.$

Then it is easily verified that

$(1-\omega^{i}\gamma)\Theta_{i}\enskip=\enskip 0\enskip.$ $None$

Moreover, since for $p$ a prime and any $0<s<p$ we have

$1\enskip+\enskip\omega^{s}\enskip+\enskip\omega^{2s}+\cdots+\omega^{(p-1)s}\enskip=\enskip
0\enskip,$

we immediately derive that

$\Psi\enskip=\enskip(\Theta_{1}+\Theta_{2}+\cdots+\Theta_{p})/p\enskip.$
$None$

On the other hand, 5.33 gives that the ratios

$c_{i}\enskip=\enskip\Theta_{i}/\Theta^{i}$ $None$

are all invariant under $\gamma$. Since as we have seen 5.26 gives that all
the conjugates $\Psi_{i}$ of $\Psi$ in $G_{\Phi}$ have $G_{\Psi}$ as
stabilizer, we deduce that these ratios are stable under $G_{\Phi}$ and (again
from Thorem 4.1) we must conclude that they are all in $Rat[{\cal
F};e_{1},e_{2},\ldots,e_{n},\Phi]$. This given, we see that by substituting
5.35 in 5.34 we get 5.32 with the desired properties.

6\. Galoisian “Galois Theory.

Lagrange’s pursuits were brought to a conclusion by Abel around 1829 when
Galois started his investigations. Galois’ starting point may have been the
idea that although there may be no general formulas for the roots of
$E_{n}(t)$ for $n\geq 5$ which involved only extraction of roots, it may still
be possible to find them for some special equations. This possibility had
already emerged in the work of Gauss and Abel but it is not clear to what
extent Galois had been aware of their work. We may say that his discoveries
stemmed from a natural extension of Lagrange’s original methods. To present
Galois’ results, we shall need some additional notation and definitions.

Our basic ingredients here will be two fields ${\cal F}$ and ${\cal E}$, with
${\cal F}$ a proper subfield of ${\cal E}$. The equation to be solved will be
written as

${\tilde{E}}_{n}(t)\enskip=\enskip(t-\alpha_{1})(t-\alpha_{2})\cdots(t-\alpha_{n})\enskip=\enskip
0\enskip.$ $None$

where $\alpha_{1},\alpha_{2},\ldots,\alpha_{n}$ are distinct and, unless
explicitely mentioned, will remain unchanged throughout our presentation. We
shall also assume hereafter that

$\eqalign{&1)\enskip\enskip\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\enskip\in\enskip{\cal
E}\enskip\enskip{\rm and}\cr&2)\enskip\enskip
e_{1}(\alpha_{1},\alpha_{2},\ldots,\alpha_{n}),e_{2}(\alpha_{1},\alpha_{2},\ldots,\alpha_{n}),\ldots,e_{n}(\alpha_{1},\alpha_{2},\ldots,\alpha_{n})\enskip\in\enskip{\cal
F}\cr}$ $None$

We see that we shall have to work here with functions
$\Phi(x_{1},x_{2},\ldots,x_{n})$ of the independent variables
$x_{1},x_{2},\ldots,x_{n}$ and at the same study their values when when
$x_{1},x_{2},\ldots,x_{n}$ are replaced by
$\alpha_{1},\alpha_{2},\ldots,\alpha_{n}$. As clarity requires, these values
will be represented by any of the symbols below

$\Phi(\alpha_{1},\alpha_{2},\ldots,\alpha_{n})\enskip=\enskip\Phi(\alpha)\enskip=\enskip{\tilde{\Phi}}\enskip.$
$None$

For a given $\sigma\in S_{n}$ we shall also use the symbols

$\sigma\Phi(\alpha_{1},\alpha_{2},\ldots,\alpha_{n})\enskip=\enskip\sigma\Phi(\alpha)\enskip=\enskip\sigma{\tilde{\Phi}}\enskip$

to denote the value
$\Phi(\alpha_{\sigma_{1}},\alpha_{\sigma_{2}},\ldots,\alpha_{\sigma_{n}})$.

This given, one of the fundamental differences between the Galoisian and
Lagrangian setups is that although a given $\Phi\in Rat[{\cal
F};x_{1},x_{2},\ldots,x_{n}]$ and all its images $\sigma\Phi$ are well defined
as elements of $Rat[{\cal F};x_{1},x_{2},\ldots,x_{n}]$ some of the values
$\sigma{\tilde{\Phi}}$ may make no sense at at all. An example in point is the
rational function

$\Phi(x)\enskip=\enskip{1\over x_{1}^{2}+x_{2}^{2}-x_{3}^{2}}$

when $n=3$ and $\alpha_{2}^{2}+\alpha_{3}^{2}-\alpha_{1}^{2}=0$.

In summary in the Galois setting, we have to be careful with denominators! We
shall avoid the problem by dealing hereafter only with polynomials functions
of the roots. As we shall see this is not a serious restriction, and with it,
most of the results of Lagrange Theory can be extended to the Galois setting
with nearly identical proofs.

Another important difference is that for some $\Phi\in{\cal
F}[x_{1},\ldots,x_{n}]$ there may be more permutations of the roots of
${\tilde{E}}_{n}(t)$ that leave
$\Phi(\alpha_{1},\alpha_{2},\ldots,\alpha_{n})$ unchanged than there are in
$G_{\Phi}$. We shall take account of this difference by setting for
$\Phi\in{\cal F}[x_{1},\ldots,x_{n}]$

$S_{\Phi}\enskip=\enskip\\{\sigma\in
S_{n}\enskip:\enskip\Phi(\alpha_{\sigma_{1}},\alpha_{\sigma_{2}},\ldots,\alpha_{\sigma_{n}})=\Phi(\alpha_{1},\alpha_{2},\ldots,\alpha_{n})\enskip\\}$
$None$

Note that $S_{\Phi}$ should be considered a property of $\Phi$ and not a
property of the value $\Phi(\alpha)$. Moreover we should point out that in
general this collection of permutations may not even be a group! For instance
for the equation

${\tilde{E}}_{3}(t)\enskip=\enskip\enskip=\enskip(t+1)(t+i)(t-i)\enskip=\enskip
t^{3}+t^{2}+t+1\enskip,$

we can take ${\cal F}$ and ${\cal E}$ to be the fields of rational and complex
numbers respectively. Now, if we label the roots by setting

$-1=\alpha_{1}\enskip\enskip,\enskip\enskip
i=\alpha_{2}\enskip\enskip,\enskip\enskip-i=\alpha_{3}$

then for $\Phi(x_{1},x_{2},x_{3})\enskip=\enskip x_{2}^{2}$ we have (in cycle
notation)

$G_{\Phi}\enskip=\enskip\\{id\hskip 2.5pt,\hskip 2.5pt(1,3)\\}\enskip.$

On the other hand the permutations of $\alpha_{1},\alpha_{2},\alpha_{3}$ that
leave ${\tilde{\Phi}}$ unchanged form the set

$\\{id\hskip 2.5pt,\hskip 2.5pt(1,3)\hskip 2.5pt,\hskip 2.5pt(2,3)\hskip
2.5pt,\hskip 2.5pt(1,2,3)\\}\enskip$

This is not a group since $(1,2)=(1,2,3)(2,3)$, yet

$\Phi(\alpha_{1},\alpha_{2},\alpha_{3})=\alpha_{2}=-1\enskip\enskip{\rm
and}\enskip\enskip\Phi(\alpha_{2},\alpha_{1},\alpha_{3})=\alpha_{1}^{2}=1\enskip.$

To take account of this possibility we shall say that a given $\Phi\in{\cal
F}[x_{1},\ldots,x_{n}]$ is Galois if and only if

$S_{\Phi}\enskip=\enskip G_{\Phi}$

We are now in a position to proceed with our treatment.

Theorem 6.1

For every subgroup $G\subseteq S_{n}$ we have a Galois $\Phi\in{\cal
F}[x_{1},x_{2},\ldots,x_{n}]$ such that

$S_{\Phi}=G\enskip.$ $None$

Proof

We start with the case $G=S_{n}$. Here can take again a linear function

$v(x_{1},x_{2},\ldots,x_{n})\enskip=\enskip
m_{1}x_{1}+m_{2}x_{2}+\cdots+m_{n}x_{n}$ $None$

as in 4.16, but we must be a bit more careful in choosing the coefficients
$m_{i}$. To this end note that if $m_{1},m_{2},\ldots,m_{n}$ are chosen to be
integers in the interval $[0,M]$, then for any given pair of distinct
permutations $\sigma,\tau\in S_{n}$ the equation

$m_{1}\alpha_{\sigma_{1}}+m_{2}\alpha_{\sigma_{2}}+\cdots+m_{n}\alpha_{\sigma_{n}}\enskip=\enskip
m_{1}\alpha_{\tau_{1}}+m_{2}\alpha_{\tau_{2}}+\cdots+m_{n}\alpha_{\tau_{n}}$

can have at most $(M+1)^{n-1}$ distinct solution vectors
$(m_{1},m_{2},\ldots,m_{n})$. Thus to assure that

$\sigma v(\alpha)\neq\tau
v(\alpha)\enskip\enskip\enskip(\enskip\forall\enskip\enskip i\neq j\enskip)$

we need to avoid at most ${n!\choose 2}\times(M+1)^{n-1}$ vectors. However,
when $M+1>{n!\choose 2}$ there will remain some $n^{tuples}$ for us to choose
and satisfy our requirement that $S_{v}=S_{n}$. Having made one such choice of
$(m_{1},m_{2},\ldots,m_{n})$, the desired $\Phi$ for any given subgroup $G$
can be readily produced. In fact, we can show that we can set

$\Phi\enskip=\enskip\prod_{\sigma\in G}\hskip 2.5pt(N-\sigma v(x))$ $None$

where $N$ is a suitably chosen integer. To see this note first that the form
of 6.7 guarantees that whatever $N$ we choose we shall have at least
$S_{\Phi}\supseteq G$. Now let

$S_{n}\enskip=\enskip\tau_{1}G\enskip+\enskip\tau_{2}G\enskip+\enskip\cdots\enskip+\enskip\tau_{k}G$

be the decompostion of $S_{n}$ into left cosets of $G$. This given, our choice
of $m_{1},m_{2},\ldots,m_{n}$ assures that the polynomials

${\tilde{P}}_{i}(t)\enskip=\enskip\tau_{i}\prod_{\sigma\in G}\hskip
2.5pt(t-\sigma v(\alpha))\enskip=\enskip\prod_{\sigma\in G}\hskip
2.5pt(t-\tau_{i}\sigma v(\alpha))$

have no roots in common. Since they all have degree $|G|$, the equation

${\tilde{P}}_{i}(t)={\tilde{P}}_{j}(t)$

for $i\neq j$ can then have at most $|G|$ solutions. Thus if we want an
integer $N$ which gives ${\tilde{P}}_{i}(N)\neq{\tilde{P}}_{j}(N)$ for all
$i\neq j$ we need avoid at most ${k\choose 2}\times|G|$ values. Clearly we can
find such an $N$ in the interval $[0,M]$ as soon as $M>{k\choose 2}\times|G|$.
This completes our argument.

Remark 6.1

All the constructions, proofs and definitions in this section will use an
$n!$-valued Galois function

$v(x)\enskip=\enskip m_{1}x_{1}+m_{2}x_{2}+\cdots+m_{n}x_{n}$

which must remain unchanged throughout the rest of the section. We must
therefore make sure that some of the objects we introduce, such as for
instance the “Galois Group” of our equation ${\tilde{E}}_{n}(t)$ do not depend
on the choice of $m_{1},m_{2},\ldots,m_{n}$. This is one of the prices we have
to pay for not following the abstract approach. However, we believe that this
will be well compensated by the additional insights that our insistence on
explicit constructions will provide.

We begin by showing that the values of every polynomial $\Phi(x)$ are in fact
polynomials in the values of $v(x)$. More precisely we have

Proposition 6.1

For any $\Phi\in{\cal F}[x_{1},\ldots,x_{n}]$ we can construct a polynomial
$\Theta_{\Phi}(t)\in{\cal F}[t]$ such that for any $\gamma\in S_{n}$ we have

$\gamma{\tilde{\Phi}}\enskip=\enskip\Theta_{\Phi}\big{(}\gamma{\tilde{v}}\big{)}\enskip.$
$None$

Proof

As in section 4 (see 4.19 and 4.20) we let

$Q(t)\enskip=\enskip\sum_{\sigma\in S_{n}}\hskip 2.5pt\sigma\Phi\prod_{\tau\in
S_{n}}\hskip-0.72229pt^{(\sigma)}\hskip 2.5pt(t-\tau
v)\enskip\enskip,\enskip\enskip P(t)\enskip=\enskip\prod_{\tau\in
S_{n}}(t-\tau v)\enskip.$ $None$

Since both $Q$ and $P$ are by construction $S_{n}$-invariant, their
coefficients are polynomials in ${\cal F}[e_{1},e_{2},\ldots,e_{n}]$. The
hypothesis in 6.3 2) then yields that the polynomials

${\tilde{Q}}(t)\hskip 2.5pt,\hskip 2.5pt{\tilde{P}}(t)\hskip 2.5pt,\hskip
2.5pt{\tilde{P}}^{\prime}(t)$

have coefficients in ${\cal F}$. Setting $t=\gamma{\tilde{v}}$ in

${\tilde{Q}}(t)\enskip=\enskip\sum_{\sigma\in S_{n}}\hskip
2.5pt\sigma{\tilde{\Phi}}\prod_{\tau\in
S_{n}}\hskip-0.72229pt^{(\sigma)}\hskip 2.5pt(t-\tau{\tilde{v}})$

gives

${\tilde{Q}}(\gamma{\tilde{v}})\enskip=\enskip\gamma{\tilde{\Phi}}\hskip
2.5pt{\tilde{P}}^{\prime}(\gamma{\tilde{v}})\enskip.$ $None$

Note further that since by construction $v$ takes $n!$ distinct values, the
polynomials ${\tilde{P}}$ and ${\tilde{P}}^{\prime}$ have no common root. We
can thus apply the Berlekamp algorithm and construct a pair of polynomials
$p(t),q(t)\in{\cal F}[t]$ such that

$p(t){\tilde{P}}^{\prime}(t)\enskip+\enskip q(t){\tilde{P}}(t)\enskip=\enskip
1$

Setting $t=\gamma{\tilde{v}}$ in this equation yields

$p(\gamma{\tilde{v}}){\tilde{P}}^{\prime}(\gamma{\tilde{v}})\enskip=\enskip
1\enskip.$

Multiplying both sides of 6.10 by $p(\gamma{\tilde{v}})$ and using this
equation we finally get

$p(\gamma{\tilde{v}}){\tilde{Q}}(\gamma{\tilde{v}})\enskip=\enskip\gamma{\tilde{\Phi}}\enskip,$

and this gives 6.8 with

$\Theta_{\Phi}(t)\enskip=\enskip p(t){\tilde{Q}}(t)\enskip.$

It develops that the polynomial

${\tilde{P}}(t)\enskip=\enskip\prod_{\tau\in S_{n}}\hskip
2.5pt(t-\tau{\tilde{v}})$ $None$

plays a crucial role in our development. It may or may not be reducible in
${\cal F}[t]$. If it is, we can write its factorization into irreducibles in
the form

${\tilde{P}}(t)\enskip=\enskip\prod_{\tau\in
T_{1}}(t-\tau{\tilde{v}})\enskip\prod_{\tau\in
T_{2}}(t-\tau{\tilde{v}})\cdots\prod_{\tau\in
T_{m}}(t-\tau{\tilde{v}})\enskip=\enskip{\tilde{P}}_{1}(t){\tilde{P}}_{2}(t)\cdots{\tilde{P}}_{m}(t)\enskip,$

where $T_{1},T_{2},\cdots,T_{m}$ are disjoint subsets, and $T_{1}$ is the
subset that contains the identity permutation. Now we have the following
crucial fact.

Proposition 6.2

The subset $T_{1}$ is a group

Proof

Since by definition $T_{1}$ contains the identity, we only need to show that
if $\eta,\xi\in T_{1}$ then their product $\eta\xi$ is also in $T_{1}$. To
this end, we use Proposition 6.1 and construct the polynomial
$\Theta(t)\in{\cal F}[t]$ that gives 6.8 for $\Phi=\xi v$. We shall thus have

$\gamma\xi{\tilde{v}}\enskip=\enskip\Theta(\gamma{\tilde{v}})\hskip
36.135pt(\enskip\forall\enskip\enskip\gamma\in S_{n})\enskip.$ $None$

Now note that by hypothesis

${\tilde{P}}_{1}(\xi{\tilde{v}})\enskip=\enskip\prod_{\tau\in
T_{1}}(\xi{\tilde{v}}-\tau{\tilde{v}})\enskip=\enskip 0.$

In particular, using 6.12 for $\gamma=id$ we can rewrite this in the form

${\tilde{P}}_{1}\big{(}\Theta({\tilde{v}})\big{)}\enskip=\enskip 0\enskip.$

But this says that the polynomials ${\tilde{P}}_{1}(t)$ and
${\tilde{P}}_{1}\big{(}\Theta(t)\big{)}$ have a root in common. Since they are
both in ${\cal F}[t]$ and ${\tilde{P}}_{1}(t)$ is irreducible in ${\cal
F}[t]$, we deduce that ${\tilde{P}}_{1}\big{(}\Theta(t)\big{)}$ must vanish
for all the other roots of ${\tilde{P}}_{1}(t)$. In particular we must have

${\tilde{P}}_{1}\big{(}\Theta(\eta{\tilde{v}})\big{)}\enskip=\enskip 0$

Now using 6.12 with $\gamma=\eta$ this may yet be rewritten as

${\tilde{P}}_{1}(\eta\xi{\tilde{v}})\enskip=\enskip 0\enskip.$

But this implies that the permutation $\eta\xi$ lies in $T_{1}$ as well.

Q.E.D.

We shall hereafter denote $T_{1}$ by $G$ and refer to it as the Galois Group
of ${\tilde{E}}_{n}(t)$. The polynomial ${\tilde{P}}_{1}(t)$ itself will be
referred to as a Galois resolvent of ${\tilde{E}}_{n}(t)$. For instance when
$n=3$ and

${\tilde{E}}_{3}(t)\enskip=\enskip t^{3}+t^{2}+t+1\enskip,$

we may take ${\cal F}$ to be the field of rational numbers and
$v=x_{2}-x_{1}$. This gives

$P(t)\enskip=\enskip\bigl{(}t^{2}-(x_{2}-x_{1})^{2}\bigr{)}\bigl{(}t^{2}-(x_{2}-x_{3})^{2}\bigr{)}\bigl{(}t^{2}-(x_{3}-x_{1})^{2}\bigr{)}\enskip.$

Now this can be rewritten as

$P(t)\enskip=\enskip
t^{6}+(6e_{2}-2e_{1}^{2})+(e_{1}^{2}-3e2)^{2}t^{2}-(x_{1}-x_{2})^{2}(x_{1}-x_{3})^{2}(x_{2}-x_{3})^{2}$

and substituting $e_{1}=1$, $e_{2}=1$ and $e_{3}=-1$ we get (using formula
2.7)

${\tilde{P}}(t)\enskip=\enskip t^{6}+4t^{4}+4t^{2}+16\enskip.$

Its irreducible factorization is

${\tilde{P}}(t)\enskip=\enskip(t^{2}-2t+2)(t^{2}+2t+2)(t^{2}+4)\enskip.$

Now the roots of $t^{3}+t^{2}+t+1$ are $-1,i,-i$ so if we label them
$\alpha_{1},\alpha_{2},\alpha_{3}$ respectively, then the Galois resolvent is

${\tilde{P}}_{1}(t)\enskip=\enskip
t^{2}-2t+2\enskip=\enskip\bigl{(}t-(x_{2}-x_{1})\bigr{)}\bigl{(}t-(x_{3}-x_{1})\bigr{)}\enskip,$

and the Galois group reduces to

$G=\\{id,(2,3)\\}\enskip.$ $None$

Note that if we had chosen $v=x_{2}-x_{3}$ then the Galois resolvent would
have been

$t^{2}+4\enskip=\enskip\bigl{(}t-(x_{2}-x_{3})\bigr{)}\bigl{(}t+(x_{2}-x_{3})\bigr{)}$

and the Galois group would still be as in 6.13. It is easy to see from this
example that as a subgroup of $S_{3}$, $G$ does depend on our labeling of the
roots. Nevertheless we are going to show that as a group of permutations of
the set $-1,i,-i$, $G$ only depends on the equation ${\tilde{E}}_{n}(t)$ and
the given field ${\cal F}$.

To this end, we need to introduce two classes of subgroups of $S_{n}$. We
shall set

${\cal A}\enskip=\enskip\\{\hskip 2.5ptH\subseteq S_{n}\hskip 2.5pt:\hskip
2.5pt\Phi\in{\cal F}[x_{1},\ldots,x_{n}]\hskip 2.5pt\&\hskip
2.5ptS_{\Phi}\supseteq H\hskip 2.5pt\Longrightarrow\hskip
2.5pt{\tilde{\Phi}}\in{\cal F}\hskip 2.5pt\\}$ $None$

and

${\cal B}\enskip=\enskip\\{\hskip 2.5ptH\subseteq S_{n}\hskip 2.5pt:\hskip
2.5pt\Phi\in{\cal F}[x_{1},\ldots,x_{n}]\hskip 2.5pt\&\hskip
2.5pt{\tilde{\Phi}}\in{\cal F}\hskip 2.5pt\Longrightarrow\hskip
2.5ptS_{\Phi}\supseteq H\hskip 2.5pt\\}\enskip.$ $None$

In words, a subgroup $H$ of $S_{n}$ belongs to ${\cal A}$ if and only if any
polynomial $\Phi\in{\cal F}[x_{1},\ldots,x_{n}]$ whose value is invariant
under $H$ has its value ${\tilde{\Phi}}$ in ${\cal F}$. In the same vein we
can say that a subgroup $H$ of $S_{n}$ belongs to ${\cal B}$ if and only if
any polynomial $\Phi\in{\cal F}[x_{1},\ldots,x_{n}]$ with its value
${\tilde{\Phi}}$ in ${\cal F}$ must remain invariant (by value) under all
elements of $H$. It is immediate from the definitions 6.14 and 6.15 that for
any two groups $H,K\subseteq S_{n}$ we have

$H\in{\cal A}\hskip 2.5pt\&\hskip 2.5ptK\supseteq H\hskip
2.5pt\Longrightarrow\hskip 2.5ptK\in{\cal A}\enskip\enskip\enskip\enskip{\rm
and}\enskip\enskip\enskip\enskip H\in{\cal B}\hskip 2.5pt\&\hskip
2.5ptK\subseteq H\hskip 2.5pt\Longrightarrow\hskip 2.5ptK\in{\cal B}$ $None$

In words, ${\cal A}$ and ${\cal B}$ are respectively upper and lower ideals of
subgroups of $S_{n}$ (under containement). Remarkably, we have the following
basic fact

Theorem 6.2

${\cal A}$ and ${\cal B}$ are both principal ideals with $G$ as their unique
extremal element. That is

${\cal A}\hskip 2.5pt\cap\hskip 2.5pt{\cal B}\enskip=\enskip\\{G\\}$ $None$

Proof

We start by proving that

$G\in{\cal A}\hskip 2.5pt\cap\hskip 2.5pt{\cal B}\enskip.$ $None$

Given a $\Phi\in{\cal F}[x_{1},\ldots,x_{n}]$ and using Proposition 6.1 we may
write

$\gamma{\tilde{\Phi}}\enskip=\enskip\Theta_{\Phi}(\gamma{\tilde{v}})\hskip
36.135pt(\enskip\forall\gamma\in S_{n}\enskip)$ $None$

Thus if $S_{\Phi}\supseteq G$ we have

${\tilde{\Phi}}\enskip=\enskip{1\over|G|}\hskip 2.5pt\sum_{\gamma\in G}\hskip
2.5pt\Theta_{\Phi}(\gamma{\tilde{v}})$

Now the right hand side of this expression is a symmetric polynomial
(*)(*)with coefficients in ${\cal F}$ in the roots of ${\tilde{P}}_{1}(t)$.
This shows that ${\tilde{\Phi}}$ may equally be expressed as a polynomial (*)
in the coefficients of ${\tilde{P}}_{1}(t)$ which themselves are in ${\cal
F}$. This implies that ${\tilde{\Phi}}\in{\cal F}$ and that

$G\in{\cal A}\enskip.$

Conversely, if ${\tilde{\Phi}}=a\in{\cal F}$ then using the same polynomial
$\Theta_{\Phi}$ we may rewrite this as

$\Theta_{\Phi}({\tilde{v}})-a\enskip=\enskip 0\enskip.$

But this says that the polynomial

$\Theta_{\Phi}(t)-a\in{\cal F}[t]$

has a root in common with ${\tilde{P}}_{1}(t)$. Thus it must vanish at all the
other roots of ${\tilde{P}}_{1}(t)$. That is we must have

$\gamma{\tilde{\Phi}}\enskip=\enskip\Theta_{\Phi}(\gamma{\tilde{v}})\enskip=\enskip
a\enskip=\enskip{\tilde{\Phi}}\hskip
36.135pt(\enskip\forall\enskip\enskip\gamma\in G\enskip)\enskip.$

This implies that $S_{\Phi}\supseteq G$ and that

$G\in{\cal B}$ $None$

To complete our argument we must show that $G$ is contained in all the other
elements of ${\cal A}$ and that $G$ contains all the other elements of ${\cal
B}$.

Note that if $G_{1}\in{\cal A}$, the polynomial

$Q(t)\enskip=\enskip\prod_{\gamma\in G_{1}}\hskip
2.5pt(t-\gamma{\tilde{v}})\enskip.$

(whose coefficients are necessarily invariant under $G_{1}$) must belong to
${\cal F}[t]$. Since it has the root ${\tilde{v}}$ in common with
${\tilde{P}}_{1}(t)$ and ${\tilde{P}}_{1}(t)$ is irreducible $Q(t)$ must be
divisible by ${\tilde{P}}_{1}(t)$. This gives that

$G\subseteq G_{1}\enskip.$

Conversely, let $G_{1}\in{\cal B}$. Consider the polynomial

$\Phi(x)\enskip=\enskip{\tilde{P}}_{1}\bigl{(}v(x)\bigr{)}\in{\cal
F}[x_{1},\ldots,x_{n}]\enskip.$

Since its value ${\tilde{\Phi}}={\tilde{P}}_{1}({\tilde{v}})=0$ is clearly in
${\cal F}$ it must remain invariant under every element of $G_{1}$. That is we
must have

${\tilde{P}}_{1}(\gamma{\tilde{v}})=0$

for all $\gamma\in G_{1}$. This shows that

$G_{1}\subseteq G$

and completes our proof.

Remark 6.2

Since the definitions 6.14 and 6.15 of the classes ${\cal A}$ and ${\cal B}$
only involve the given field ${\cal F}$ and the roots of the equation
${\tilde{E}}_{n}(t)$, we see that one of the consequences of Theorem 6.2 is
that also $G$ itself only depends on ${\cal F}$ and ${\tilde{E}}_{n}(t)$. When
in our developments we keep the given equation fixed and only vary the field,
for simplicity, we shall use the notation ${\cal A}({\cal F})$, ${\cal
B}({\cal F})$, $G({\cal F})$ and leave the dependence on
${{\tilde{E}}_{n}(t)}$ implicit. In all other cases we will indicate this
dependence with a subscript. We should also keep in mind that one of the
immediate consequence of our definition of a Galois group is that if ${\cal
F}\subseteq{\cal F}_{1}$ are two fields then we necessarily must have
$G_{{\tilde{E}}_{n}(t)}({\cal F}_{1})\subseteq G_{{\tilde{E}}_{n}(t)}({\cal
F})$.

In trying to extend Lagrange’s Theorems 4.1 $\&$ 4.2 to the Galois setting we
should be tempted to let the Galois group $G$ play the role of $S_{n}$ in the
arguments. However, given a polynomial $\Psi\in{\cal
F}[x_{1},x_{2},\ldots,x_{n}]$ we may not be in a position to write down the
coset decomposition in 4.5 with $S_{n}$ replaced by $G$ for the simple reason
that we may not have $G_{\Psi}\subseteq G$. Using $G_{\Psi}\cap G$ in place of
$G_{\Psi}$ doesn’t get us anywhere for the simple reason that the intersection
$G_{\Psi}\cap G$ may only consist of the identity permutation. Nor we can use
$S_{\Psi}$ instead of $G_{\Psi}$ for it may be too big and as we have seen it
may not even be a group. It develops that the optimal choice turns out to be
the intersection $S_{\Psi}\cap G$ which here and after will be denoted by
${\tilde{G}}_{\Psi}$ and referred to as the Galois stabilizer of $\Psi$. In
fact, ${\tilde{G}}_{\Psi}$ is neither too small nor to big and remarkably it
can be easily shown that

Theorem 6.3

For any polynomial $\Psi\in{\cal F}[x_{1},x_{2},\ldots,x_{n}]$ the Galois
stabilizer

${\tilde{G}}_{\Psi}\enskip=\enskip S_{\Psi}\cap G$

is a group

Proof

Since by construction ${\tilde{G}}_{\Psi}$ contains the identity, we need only
show that

$\eta,\xi\in{\tilde{G}}_{\Psi}\hskip 2.5pt\Longrightarrow\hskip
2.5pt\eta^{-1}\xi\in{\tilde{G}}_{\Psi}\enskip.$ $None$

However, this is immediate. In fact, $\eta,\xi\in{\tilde{G}}_{\Psi}$ implies
that

$\xi{\tilde{\Psi}}-\eta{\tilde{\Psi}}=0\hskip 2.5pt\in{\cal F}\enskip.$

But since $G\in{\cal B}$ we must necessarily have

$S_{\xi\Psi-\eta\Psi}\supseteq G$

In other words

$\gamma\xi{\tilde{\Psi}}-\gamma\eta{\tilde{\Psi}}\enskip=\enskip
0\enskip.\hskip 36.135pt(\enskip\forall\enskip\enskip\gamma\in G\enskip)$

In particular this must hold true for $\gamma=\eta^{-1}$. That is

$\eta^{-1}\xi{\tilde{\Psi}}\enskip=\enskip{\tilde{\Psi}}\enskip,$

which gives 6.22 as desired.

We are now in a position to state and prove the Galois version of Theorems 4.1
and 4.2.

Theorem 6.4

If $\Psi\&\Phi\in{\cal F}[x_{1},x_{2},\ldots,x_{n}]$ and

${\tilde{G}}_{\Psi}\subseteq{\tilde{G}}_{\Phi}$ $None$

then we can construct a polynomial $\Theta(t)\in{\cal F}[t]$ which gives

$\gamma{\tilde{\Phi}}\enskip=\enskip\Theta(\gamma{\tilde{\Psi}})\hskip
36.135pt(\enskip\forall\enskip\enskip\gamma\in G\enskip)$ $None$

Proof

Let

$G\enskip=\enskip\tau_{1}\hskip
2.5pt{\tilde{G}}_{\Psi}\enskip+\enskip\tau_{2}\hskip
2.5pt{\tilde{G}}_{\Psi}\enskip+\enskip\cdots\enskip+\enskip\tau_{k}\hskip
2.5pt{\tilde{G}}_{\Psi}\hskip 36.135pt(\enskip\tau_{1}=identity\enskip)$
$None$

and set

$A(t)\enskip=\enskip\sum_{i=1}^{k}\hskip 2.5pt\tau_{i}{\tilde{\Phi}}\hskip
2.5pt\prod_{j=1}^{k}\hskip-0.72229pt\ {}^{(i)}\hskip
2.5pt(t-\tau_{j}{\tilde{\Psi}})\enskip\enskip\hskip 2.5pt,\hskip
2.5pt\enskip\enskip B(t)\enskip=\enskip\prod_{i=1}^{k}\hskip
2.5pt(t-\tau_{i}\hskip 2.5pt{\tilde{\Psi}})\enskip.$ $None$

Now from 6.23 and 6.25 we get that $A(t)$ and $B(t)$ are $G$-invariant, and
$G\in{\cal A}$ gives that

$A(t)\hskip 2.5pt\&\hskip 2.5ptB(t)\in{\cal F}[t]\enskip.$

Since by construction $B(t)$ has distinct roots we can find two polynomials
$p(t),q(t)\in{\cal F}[t]$ such that

$p(t)B^{\prime}(t)+q(t)B(t)\enskip=\enskip 1\enskip.$

Setting $t=\gamma{\tilde{\Psi}}$ gives

$p(\gamma{\tilde{\Psi}})B^{\prime}(\gamma{\tilde{\Psi}})\enskip=\enskip 1$
$None$

On the other hand we see from 6.26 that if $\gamma=\tau_{i}h$ with
$h\in{\tilde{G}}_{\Psi}$ then

$A(\gamma{\tilde{\Psi}})\enskip=\enskip
A(\tau_{i}{\tilde{\Psi}})\enskip=\enskip\tau_{i}{\tilde{\Phi}}\hskip
2.5ptB^{\prime}(\tau_{i}{\tilde{\Psi}})\enskip.$

Using 6.23 we may rewrite this as

$A(\gamma{\tilde{\Psi}})\enskip=\enskip\gamma{\tilde{\Phi}}\hskip
2.5ptB^{\prime}(\gamma{\tilde{\Psi}})\enskip.$

Multiplying both sides by $p(\gamma{\tilde{\Psi}})$ and using 6.27 we finally
get

$\gamma{\tilde{\Phi}}\enskip=\enskip
p(\gamma{\tilde{\Psi}})A(\gamma{\tilde{\Psi}})$

which is 6.24 with

$\Theta(t)\enskip=\enskip p(t)A(t)\enskip.$

This Theorem has an immediate consequence which can be helpful in the
construction of the Galois group of an equation.

Corollary 6.1

Let $G$ be the Galois group of ${\tilde{E}}_{n}(t)\in{\cal F}[t]$ and suppose
that for some $\Psi\in{\cal F}[x_{1},x_{2},\ldots,x_{n}]$ we have

$\eqalign{&(1)\enskip\enskip
S_{\Psi}=G_{\Psi}\enskip,\cr&(2)\enskip\enskip{\tilde{\Psi}}\in{\cal
F}\enskip.\cr}$

Then we must necessarily have

$G\hskip 2.5pt\subseteq\hskip 2.5ptG_{\Psi}\enskip.$

Proof

In view of Theorem 6.2 we need only show that $G_{\Psi}\in{\cal A}({\cal F})$.
To this end let $\Phi\in{\cal F}[x_{1},x_{2},\ldots,x_{n}]$ and let
$S_{\Phi}\supseteq G_{\Psi}$. Condition $(1)$ then assures that
${\tilde{G}}_{\Phi}\supseteq{\tilde{G}}_{\Psi}$. We can thus use Theorem 6.4
and derive that for some $\theta(t)\in{\cal F}[t]$ we have
${\tilde{\Phi}}=\theta({\tilde{\Psi}})$. But then condition $(2)$ yields us
that ${\tilde{\Phi}}\in{\cal F}$ as desired.

Before we can proceed any further we need to establish the following basic
fact

Proposition 6.3

Let $B(t)\in{\cal F}[t]$ be a polynomial of degree $k$ which is irreducible in
${\cal F}[t]$, and let

$B({\tilde{\Psi}})\enskip=\enskip 0$

for some $\Psi\in{\cal F}[x_{1},x_{2},\ldots,x_{n}]$. Then the values

$1\hskip 2.5pt,\hskip 2.5pt{\tilde{\Psi}}\hskip 2.5pt,\hskip
2.5pt{\tilde{\Psi}}^{2}\hskip 2.5pt,\hskip 2.5pt\ldots,{\tilde{\Psi}}^{k-1}$
$None$

form a basis of a vector space $V$ over ${\cal F}$ which is also a field

Proof

Suppose that for some $c_{o},c_{1},c_{2},\ldots,c_{k-1}\in{\cal F}$ not all
vanishing we had

$c_{o}+c_{1}{\tilde{\Psi}}+c_{2}{\tilde{\Psi}}^{2}+\cdots+c_{k-1}{\tilde{\Psi}}^{k-1}\enskip=\enskip
0$

then the polynomial $R(t)=c_{o}+c_{1}t+c_{2}t^{2}+\cdots+c_{k-1}t^{k-1}$ would
have a root in common with $B(t)$ and the greatest common divisor of $R(t)$
and $B(t)$ would yield a non trivial factorization of $B(t)$ in ${\cal F}[T]$
contradicting the irreducibility of $B(t)$. This shows that the elements in
6.28 are independent over ${\cal F}$.

To complete the proof we need to show that every non vanishing element $v$ of
$V$ has an inverse in $V$. Now such an element would be given by a linear
combination

$v\enskip=\enskip
c_{o}+c_{1}{\tilde{\Psi}}+c_{2}{\tilde{\Psi}}^{2}+\cdots+c_{k-1}{\tilde{\Psi}}^{k-1}\hskip
36.135pt(\enskip c_{i}\in{\cal F}\enskip)$

with some $c_{i}\neq 0$. For the same reasons as above, the polynomial
$R(t)=c_{o}+c_{1}t+c_{2}t^{2}+\cdots+c_{k-1}t^{k-1}$ cannot have any root in
common with $B(t)$. Thus we can use the Berlekamp algorithm and construct two
polynomials $p(t),q(t)\in{\cal F}[t]$ giving

$p(t)R(t)+q(t)B(t)\enskip=\enskip 1\enskip.$

Setting $t={\tilde{\Psi}}$ we get

$p({\tilde{\Psi}})\hskip 2.5ptR({\tilde{\Psi}})\enskip=\enskip 1\enskip,$

which shows that $R({\tilde{\Psi}})$ is invertible and that its inverse in $V$
is given by $p({\tilde{\Psi}})$.

Q.E.D.

The vector space $V$ will here and after be denoted by ${\cal
F}[{\tilde{\Psi}}]$ and referred to as the Extension of ${\cal F}$ by
${\tilde{\Psi}}$. The integer $k$ giving the dimension of ${\cal
F}[{\tilde{\Psi}}]$ will be called the degree of the extension. We shall also
say that ${\cal F}[{\tilde{\Psi}}]$ is obtained by adjoining ${\tilde{\Psi}}$
to ${\cal F}$. The following theorem provides the crucial tools needed in the
applications of Galois theory to the theory of equations.

Theorem 6.5

Let $\Psi\in{\cal F}[x_{1},x_{2},\ldots,x_{n}]$, set

$G\enskip=\enskip\tau_{1}{\tilde{G}}_{\Psi}\enskip+\enskip\tau_{2}{\tilde{G}}_{\Psi}+\cdots+\tau_{k}{\tilde{G}}_{\Psi}\hskip
36.135pt(\enskip\tau_{1}=id\hskip 2.5pt)$ $None$

and let
${\tilde{\Psi}}_{1}=\tau_{1}{\tilde{\Psi}},{\tilde{\Psi}}_{2}=\tau_{2}{\tilde{\Psi}},\ldots,{\tilde{\Psi}}_{k}=\tau_{k}{\tilde{\Psi}}$
denote the conjugates of ${\tilde{\Psi}}$ in $G$. Then

(i) The polynomial $B(t)=\prod_{i=1}^{k}\hskip 2.5pt(t-{\tilde{\Psi}}_{i})$ is
irreducible in ${\cal F}[t]$.

(ii) By adjoining ${\tilde{\Psi}}$ to ${\cal F}$ the Galois group of the
equation ${\tilde{E}}_{n}(t)=0$ is reduced to ${\tilde{G}}_{\Psi}$.

(iii) The Galois group of $B(t)$ is the subgroup $\Gamma_{\tilde{\Psi}}$ of
$S_{k}$ corresponding to the action of $G$ on the left cosets of
${\tilde{G}}_{\Psi}$. In particular we have the isomorphism

$\Gamma_{\tilde{\Psi}}\hskip 2.5pt\cong G\hskip 2.5pt/\hskip
2.5pt\bigcap_{i=1}^{k}\hskip
2.5pt\tau_{i}{\tilde{G}}_{\Psi}\tau_{i}^{-1}\enskip.$ $None$

(iv) Set ${\cal F}_{1}={\cal F}[{\tilde{\Psi}}]$ and let $Aut_{\cal F}({\cal
F}_{1})$ denote the group of automorphisms of ${\cal F}_{1}$ that leave ${\cal
F}$ elementwise fixed. This given, $Aut_{\cal F}({\cal F}_{1})$ can be
identified with the set
$\\{\tau_{i}{\tilde{G}}_{\Psi}:\tau_{i}{\tilde{G}}_{\Psi}={\tilde{G}}_{\Psi}\tau_{i}\hskip
2.5pt\\}$. In particular we have the isomorphism

$Aut_{\cal F}({\cal F}_{1})\hskip
2.5pt\cong\Bigl{(}\sum_{\tau_{i}{\tilde{G}}_{\Psi}={\tilde{G}}_{\Psi}\tau_{i}}\enskip\tau_{i}{\tilde{G}}_{\Psi}\Bigr{)}/{\tilde{G}}_{\Psi}$
$None$

Proof

Proof of (i)

Suppose that a polynomial $B_{1}(t)\in{\cal F}[t]$ divides $B(t)$ and shares
the root $\tau_{i}{\tilde{\Psi}}$ with $B(t)$. Then the polynomial
$B_{1}(\tau_{i}\Psi)\in{\cal F}[x_{1},x_{2},\ldots,x_{n}]$ has the value
$B_{1}(\tau_{i}{\tilde{\Psi}})=0\in{\cal F}$ and $G\in{\cal B}$ gives that
$S_{B_{1}(\tau_{i}\Psi)}\supseteq G$. In other words we must have
$B_{1}(\gamma\tau_{i}{\tilde{\Psi}})=0$ for all $\gamma\in G$. Since the
action of $G$ on
${\tilde{\Psi}}_{1},{\tilde{\Psi}}_{2},\ldots,{\tilde{\Psi}}_{k}$ is
transitive, we see that $B_{1}(t)$ has to vanish at all the roots of $B(t)$,
which forces $B_{1}(t)=B(t)$. Thus $B(t)$ can’t have a proper factor in ${\cal
F}[t]$.

Proof of (ii)

Since $B({\tilde{\Psi}})=0$ we can use Proposition 6.3 to construct the field
${\cal F}_{1}={\cal F}[{\tilde{\Psi}}]$. This given we need only verify that

${\cal A}({\cal F}[{\tilde{\Psi}}])\cap{\cal B}({\cal
F}[{\tilde{\Psi}}])\enskip=\enskip\\{{\tilde{G}}_{\Psi}\\}\enskip.$

Now this is immediate. In fact, if for some $\Phi\in{\cal
F}[x_{1},x_{2},\ldots,x_{n}]$ we have $S_{\Phi}\supseteq{\tilde{G}}_{\Psi}$
then we must have ${\tilde{G}}_{\Phi}\supseteq{\tilde{G}}_{\Psi}$ as well and
Theorem 6.4 gives that for some $\Theta(t)\in{\cal F}[t]$

${\tilde{\Phi}}=\Theta({\tilde{\Psi}})\hskip 2.5pt\in{\cal F}[\Psi]\enskip.$

Thus ${\tilde{G}}_{\Psi}$ is in ${\cal A}({\cal F}[{\tilde{\Psi}}])$.

Conversely let $\Phi\in{\cal F}[x_{1},x_{2},\ldots,x_{n}]$ and
${\tilde{\Phi}}\in{\cal F}[{\tilde{\Psi}}]$. This means that for some
$\Theta(t)\in{\cal F}[t]$ we have

${\tilde{\Phi}}=\Theta({\tilde{\Psi}})\enskip.$

In other words for the polynomial $\Xi(x)=\Phi(x)-\Theta(\Psi(x))\in{\cal
F}[x_{1},x_{2},\ldots,x_{n}]$ we have

${\tilde{\Xi}}=0\hskip 2.5pt\in{\cal F}$

so from $G\in{\cal B}({\cal F})$ we get that we must have

$\gamma{\tilde{\Phi}}-\Theta(\gamma{\tilde{\Psi}})=\gamma{\tilde{\Xi}}\enskip=\enskip
0\hskip 36.135pt(\enskip\forall\enskip\enskip\gamma\in G\enskip)$

However, for $\gamma\in{\tilde{G}}_{\Psi}$ this yields that

$\gamma{\tilde{\Phi}}\enskip=\enskip\Theta({\tilde{\Psi}})\enskip=\enskip{\tilde{\Phi}}\enskip.$

In other words ${\tilde{\Phi}}\in{\cal F}[\Psi]$ implies that
$S_{\Phi}\supseteq{\tilde{G}}_{\Psi}$, which gives ${\tilde{G}}_{\Psi}\in{\cal
B}({\cal F}[{\tilde{\Psi}}])$ as desired. This completes the proof of $(ii)$.

Proof of (iii)

Let $\Gamma_{\tilde{\Psi}}$ be the image of $G$ in the symmetric group $S_{k}$
given by the permutation action of $G$ on the left cosets of
${\tilde{G}}_{\Psi}$. We want to show that

${\cal A}_{B}({\cal F})\cap{\cal B}_{B}({\cal
F})\enskip=\enskip\\{\Gamma_{\tilde{\Psi}}\\}\enskip.$ $None$

To this end let $\Phi\in{\cal F}[y_{1},y_{2},\ldots,y_{k}]$ and suppose that
for any $\gamma\in\Gamma_{\tilde{\Psi}}$ we have

$\Phi({\tilde{\Psi}}_{\gamma_{1}},{\tilde{\Psi}}_{\gamma_{2}},\cdots,{\tilde{\Psi}}_{\gamma_{k}})\enskip=\enskip\Phi({\tilde{\Psi}}_{1},{\tilde{\Psi}}_{2},\ldots,{\tilde{\Psi}}_{k})\enskip.$
$None$

Since each $g\in G$ induces a permutation of
${\tilde{\Psi}}_{1},{\tilde{\Psi}}_{2},\ldots,{\tilde{\Psi}}_{k}$ by an
element $\gamma\in\Gamma_{\tilde{\Psi}}$, 6.32 implies that

$S_{\Phi(\Psi_{1},\Psi_{2},\ldots,\Psi_{k})}\hskip 2.5pt\supseteq\hskip
2.5ptG\enskip.$

But then $G\in{\cal A}_{{\tilde{E}}_{n}}({\cal F})$ gives that
$\Phi({\tilde{\Psi}}_{1},{\tilde{\Psi}}_{2},\ldots,{\tilde{\Psi}}_{k})\in{\cal
F}$. Thus $\Gamma_{\tilde{\Psi}}\in{\cal A}_{B}({\cal F})$.

Conversely, say
$\Phi({\tilde{\Psi}}_{1},{\tilde{\Psi}}_{2},\ldots,{\tilde{\Psi}}_{k})=a\in{\cal
F}$. Then $G\in{\cal B}_{{\cal E}_{n}}({\cal F})$ gives that
$g\Phi({\tilde{\Psi}}_{1},{\tilde{\Psi}}_{2},\ldots,{\tilde{\Psi}}_{k})=\Phi({\tilde{\Psi}}_{1},{\tilde{\Psi}}_{2},\ldots,{\tilde{\Psi}}_{k})$
for all $g\in G$. But if $\gamma=\gamma(g)$ is the image of $g$ in $S_{k}$,
this simply says that

$\Phi({\tilde{\Psi}}_{\gamma_{1}},{\tilde{\Psi}}_{\gamma_{2}},\cdots,{\tilde{\Psi}}_{\gamma_{1}})\enskip=\enskip\Phi({\tilde{\Psi}}_{1},{\tilde{\Psi}}_{2},\ldots,{\tilde{\Psi}}_{k})\enskip.$

This gives $\Gamma_{\tilde{\Psi}}\in{\cal B}_{B}({\cal F})$. To complete the
proof of $(iii)$ we need only observe that the Galois stabilizer of a
conjugate ${\tilde{\Psi}}_{i}$ is simply the conjugate subgroup
$\tau_{i}{\tilde{G}}_{\Psi}\tau_{i}^{-1}$, thus the only elements of $g$ that
leave invariant all the conjugates of ${\tilde{\Psi}}$ are those that belong
to the intersection

$\cap_{i=1}^{k}\hskip 2.5pt\tau_{i}{\tilde{G}}_{\Psi}\tau_{i}^{-1}\enskip.$

This yields 6.30.

Proof of (iv)

We assume here that ${\tilde{E}}_{n}(t)\in{\cal F}[t]$ as before and that
${\cal E}={\cal F}[\alpha_{1},\alpha_{2},\ldots,\alpha_{n}]$. Note that since
every element of ${\cal F}_{1}$ is of the form

$\theta({\tilde{\Psi}})\enskip=\enskip
c_{o}+c_{1}{\tilde{\Psi}}+\cdots+c_{k-1}{\tilde{\Psi}}^{k-1}\hskip
36.135pt(\enskip c_{i}\in{\cal F}\enskip)$ $None$

with $k$ the index of ${\tilde{G}}_{\Psi}$ in $G$, to find the image by a
$g\in Aut_{\cal F}({\cal F}_{1})$ of any element of ${\cal F}_{1}$ we need
only know $g{\tilde{\Psi}}$. This is because, every $c_{i}$ remaining
unchanged by $g$ we must necessarily have
$g\theta({\tilde{\Psi}})=\theta(g{\tilde{\Psi}})$. Moreover, since
$B(t)=\prod_{i=1}^{k}(t-{\tilde{\Psi}}_{i})\in{\cal F}[t]$, the identity
$B({\tilde{\Psi}})=0$ forces $B(g{\tilde{\Psi}})=0$ as well. In particular, we
deduce that $g{\tilde{\Psi}}={\tilde{\Psi}}_{i}=\tau_{i}{\tilde{\Psi}}$ for
some $i=1,2,\ldots,k$. In addition $g{\tilde{\Psi}}\in{\cal F}_{1}$ yields
that we must have
${\tilde{\Psi}}_{i}=\tau_{i}{\tilde{\Psi}}=\theta({\tilde{\Psi}})$ with
$\theta(t)\in{\cal F}$. But this gives that
${\tilde{G}}_{\Psi}\subseteq{\tilde{G}}_{\Psi_{i}}$ and since
${\tilde{G}}_{\Psi_{i}}=\tau_{i}{\tilde{G}}_{\Psi}\tau_{i}^{-1}$ the latter
inclusion can hold true if and only if

${\tilde{G}}_{\Psi_{i}}={\tilde{G}}_{\Psi}\enskip.$

Thus we see that the elements of $Aut_{\cal F}({\cal F}_{1})$ can be simply
identified with the left cosets $\tau_{i}{\tilde{G}}_{\Psi}$ such that
$\tau_{i}{\tilde{G}}_{\Psi}={\tilde{G}}_{\Psi}\tau_{i}$. This gives 6.31 and
completes the proof of the Theorem.

We should note that the argument used in the proof of part $(i)$ of this
Theorem leads to the following basic property of the Galois group of an
equation.

Proposition 6.4

The polynomial $B(t)=\prod_{i=1}^{n}(t-\alpha_{i})\in{\cal F}[t]$ is
irreducible in ${\cal F}[t]$ if and only if its Galois group $G=G_{B}({\cal
F})$ acts transitively on $\alpha_{1},\alpha_{2},\ldots,\alpha_{n}$.

Proof

Let $B_{1}(t)\in{\cal F}[t]$ be an irreducible factor of $B(t)$. If
$\alpha_{i}$ is any of the roots of $B_{1}(t)$ then
$B_{1}(\alpha_{i})=0\in{\cal F}$ and the fact that $G\in{\cal A}_{B}({\cal
F})$ gives that $B_{1}(\alpha_{\gamma_{i}})=0$ for all $\gamma\in G$. But then
the transitivity of $G$ yields that $B_{1}(t)$ can’t be a proper factor of
$B(t)$. Conversely suppose that $G$ is intransitive. This means that the orbit
of any of the roots, say $\alpha_{1}$ can’t consist of all the roots of
$B(t)$. Denote this orbit by $Orb(\alpha_{1})$ and set

$B_{1}(t)=\prod_{\alpha_{i}\in Orb(\alpha_{1})}(t-\alpha_{i})\enskip.$

Since $B_{1}(t)$, by construction is invariant under $G$ then $G\in{\cal
A}_{B}({\cal F})$ gives that $B_{1}(t)\in{\cal F}[t]$. Moreover, also by
construction, $B_{1}(t)$ does not contain all then roots of $B(t)$. Thus
$B_{1}(t)$ is necessarily a proper factor of $B(t)$ and the latter must
therefore be reducible in ${\cal F}[t]$.

Remark 6.3

We should note that parts $(i)$, $(ii)$ and $(iii)$ of Theorem 6.5 are Galois’
fundamental breakthroughs in the theory of equations. In the original Galois
context groups came first and fields were only accessories. In later
interpretations and additions to Galois’ work started by Kronecker [], brought
to completion by Dedekind [] and Weber and popularized by E. Artin [],[], this
viewpoint has been reversed and Galois theory was made to become part of the
theory of fields. In particular, part $(iv)$ of Theorem 6.5 in its present
interpretation is a later addition.

Since our presentation of Galois theory differs from most available textbooks
on the subject, perhaps a few words might be needed to connect this writing to
present day literature. For instance in Artin’s monograph [] an extension
field ${\cal F}_{1}$ of a field ${\cal F}$ is called a normal extension if the
group of automorphisms of ${\cal F}_{1}$ that leave ${\cal F}$ fixed has
${\cal F}$ for its fixed field. We should note that in [] a field ${\cal F}$
is said to be fixed  by an automorphism $g$ if $g$ fixes every element of
${\cal F}$. We see then that one of our fields ${\cal F}_{1}={\cal F}[\Psi]$
is a normal extension of ${\cal F}$ if and only if the only elements of ${\cal
F}_{1}$ that remain fixed under every element of $Aut_{{\cal F}}({\cal
F}_{1})$ (as given by 6.31) are the elements of ${\cal F}$ itself.

This given, we can easily convert Theorem 6.5 into the collection of results
that in [] is referred to as the Fundamental Theorem of Galois Theory.

Theorem 6.6

Let $G$ be the Galois group of
${\tilde{E}}_{n}(t)=\prod_{i=1}^{n}(t-\alpha_{i})$ in ${\cal F}$ and let
${\cal E}={\cal F}[\alpha_{1},\alpha_{2},\ldots,\alpha_{n}]$. Then

(i) Each subgroup $G_{1}\subseteq G$ is the Galois group of $E_{n}(t)$ with
respect to an intermediate field ${\cal F}_{1}={\cal F}[{\tilde{\Psi}}]$.
Different groups $G_{1},G_{2}$ corresponding to different fields ${\cal
F}_{1},{\cal F}_{2}$.

(ii) The subgroup $G_{1}$ is a normal subgroup of $G$ if and only if the
corresponding field ${\cal F}_{1}$ is a normal extension of ${\cal F}$. In
that case the group of automorphisms of ${\cal F}_{1}$ that leave ${\cal F}$
fixed is isomorphic to the quotient group $G/G_{1}$.

(iii) For each $G_{1}\subseteq G$ the dimension of ${\cal F}_{1}$ over ${\cal
F}$ is $|G|/|G_{1}|$ and the dimension of ${\cal E}$ over ${\cal F}_{1}$ is
$|G_{1}|$.

Proof

proof of (i)

We have seen (Theorem 6.1) that given any subgroup $G_{1}\subseteq G$ we can
find $\Psi\in{\cal F}[x_{1},x_{2},\ldots,x_{n}]$ such that
${\tilde{G}}_{\Psi}=G_{1}$. From Theorem 6.5 we get that the Galois group of
${\tilde{E}}_{n}$ in ${\cal F}_{1}={\cal F}[\Psi]$ is $G_{1}$. If for
$G_{1},G_{2}\subseteq G$ we have ${\tilde{G}}_{\Psi_{1}}=G_{1}$,
${\tilde{G}}_{\Psi_{2}}=G_{2}$ then ${\cal
F}[{\tilde{\Psi}}_{1}]\subseteq{\cal F}[{\tilde{\Psi}}_{2}]$ gives
${\tilde{\Psi}}_{1}\theta({\tilde{\Psi}}_{2})$ with $\theta(t)\in{\cal F}[t]$
thus also ${\tilde{G}}_{\Psi_{2}}\subseteq{\tilde{G}}_{\Psi_{1}}$ so the
equality ${\cal F}[\Psi_{1}]={\cal F}[\Psi_{2}]$ forces the equality
$G_{1}={\tilde{G}}_{\Psi_{1}}={\tilde{G}}_{\Psi_{2}}=G_{2}$. This proves
$(i)$.

proof of (ii)

Suppose that $G_{1}={\tilde{G}}_{\Psi}$ is not a normal subgroup of $G$. Then
by relabeling the elements $\tau_{i}$ appearing in 6.29 so that 6.34 may be
rewritten as

$Aut_{\cal F}({\cal F}_{1})\enskip=\enskip\sum_{i=1}^{s}\hskip
2.5pt\tau_{i}{\tilde{G}}_{\Psi}\enskip\hskip 36.135pt(\enskip
for\enskip\enskip s<k\enskip)$

Now note that the polynomial

$B_{1}(t)\enskip=\enskip\prod_{i=1}^{s}(t-{\tilde{\Psi}}_{i})\in{\cal
F}_{1}[t]$

cannot be in ${\cal F}[t]$. This is clear since otherwise the irreducibility
of $B(t)=\prod_{i=1}^{k}(t-{\tilde{\Psi}}_{i})$ (Theorem 6.5 $(i)$) would be
contraddicted. In particular one of the elementary symmetric functions
$e_{i}(y_{1},y_{2},\ldots,y_{s})$ must take a value
$e_{i}({\tilde{\Psi}}_{1},{\tilde{\Psi}}_{2},\ldots,{\tilde{\Psi}}_{s})$ not
in ${\cal F}$. However, since the latter is invariant under permutations of
${\tilde{\Psi}}_{1},{\tilde{\Psi}}_{2},\ldots,{\tilde{\Psi}}_{s}$ we see (from
6.31) that we have a element of ${\cal F}_{1}$ that remains invariant under
all elements of $Aut_{\cal F}({\cal F}_{1})$. So ${\cal F}_{1}$ in this case
is not a normal extension of ${\cal F}$. Conversely suppose that
$G_{1}={\tilde{G}}_{\Psi}$ is a normal subgroup of $G$ then 6.31 gives that

$Aut_{\cal F}({\cal
F}_{1})\enskip=\enskip\Bigl{(}\sum_{i=1}^{k}\enskip\tau_{i}{\tilde{G}}_{\Psi}\Bigr{)}/{\tilde{G}}_{\Psi}\enskip=\enskip
G/G_{1}\enskip.$ $None$

Since every element of ${\cal F}_{1}$ is already fixed by
${\tilde{G}}_{\Psi}$, we see from 6.34 that an element $f\in{\cal F}_{1}$ is
fixed by $Aut_{\cal F}({\cal F}_{1})$ if and only if it is fixed by every
element of $G$, but then $G\in{\cal A}_{{\tilde{E}}_{n}}({\cal F})$ gives that
$f$ must lie in ${\cal F}$. This gives that ${\cal F}$ is the fixed field of
$Aut_{\cal F}({\cal F}_{1})$ and completes the proof of $(ii)$.

proof of (iii)

Given that $G_{1}={\tilde{G}}_{\Psi}$ and given that we have 6.29, Proposition
6.3 combined with $(i)$ of Theorem 6.5 yields that the dimension of ${\cal
F}_{1}$ as a vector space over ${\cal F}$ is precisely given by
$k=|G|/|G_{1}$. It develops that the last assertion of the Theorem is an
immediate consequence of part $(ii)$ of Theorem 6.5. In fact, to construct the
Galois group of ${\tilde{E}}_{n}$ relative to ${\cal F}_{1}$ we can also
resort to the original definition. That is we break up the polynomial
${\tilde{P}}(t)$ given in 6.11 into its irreducible factors in ${\cal
F}_{1}[t]$ and take the collection of permutations which give the irreducible
factor $P_{1,1}(t)$ that has ${\tilde{v}}$ as a root. (*)(*)Note that
$P_{1,1}(t)$ must also be an irreducible factor of the polynomial
${\tilde{P}}_{1}(t)$ which gave us the Galois group $G$ of ${\tilde{E}}_{n}$
in ${\cal F}$.

However, part $(ii)$ of Theorem 6.5 gives that this procedure must deliver the
polynomial

$P_{1,1}(t)\enskip=\enskip\prod_{\tau\in{\tilde{G}}_{\Psi}}\hskip
2.5pt(t-\tau{\tilde{v}})\enskip.$ $None$

Since $P_{1,1}(t)$ is in ${\cal F}_{1}[t]$ we can use Proposition 6.3 and
deduce that the extension ${\cal F}_{1}[{\tilde{v}}]$ must be of dimension
$|{\tilde{G}}_{\Psi}|$ as a vector space over ${\cal F}_{1}$. But then $(iii)$
follows from the fact that ${\cal F}_{1}[{\tilde{v}}]$ and ${\cal
F}[\alpha_{1},\alpha_{2},\ldots,\alpha_{n}]$ are one and the same. In fact, we
trivially have ${\cal F}_{1}[{\tilde{v}}]\subseteq{\cal
F}[\alpha_{1},\alpha_{2},\ldots,\alpha_{n}]$ and the reverse containement
${\cal F}_{1}[{\tilde{v}}]\supseteq{\cal
F}[\alpha_{1},\alpha_{2},\ldots,\alpha_{n}]$ follows from Theorem 6.1. This
completes our proof.

Remark 6.4

We should note that under the definition of some texts (see [] ,[]) one of our
extensions ${\cal F}_{1}={\cal F}[{\tilde{\Psi}}]$ would be called $normal$ if
and only if any irreducible polynomial $Q(t)\in{\cal F}[t]$ that has a root in
${\cal F}_{1}$ has all the other roots in ${\cal F}_{1}$. Now we can easily
show that, again this happens if and only if ${\tilde{G}}_{\Psi}$ is a normal
subgroup of the Galois group $G=G_{{\tilde{E}}_{n}(t)}({\cal F})$. Let us use
the same notation as in the proof of Theorem 6.5 and let us set
$Q(t)=\prod_{i=1}^{m}(t-\beta_{i})$. Note first that if the “normality”
condition in [] and [] is satisfied then one of the polynomials all whose
roots, by this condition, would have to lie in ${\cal F}_{1}$ is the
polynomial $B(t)$ itself. But as we have seen this is equivalent to
${\tilde{G}}_{\Psi}$ being a normal subgroup of $G$.

Conversely suppose that ${\tilde{G}}_{\Psi}$ is a normal subgroup of
$G=G_{{\tilde{E}}_{n}(t)}({\cal F})$. Then $\beta_{1}\in{\cal F}_{1}$, simply
means that $\beta_{1}=\theta({\tilde{\Psi}})$ with $\theta(t)\in{\cal F}[t]$.
In other words we have

$\beta_{1}=\Phi(\alpha_{1},\alpha_{2},\ldots,\alpha_{n})=\theta(\Psi(\alpha_{1},\alpha_{2},\ldots,\alpha_{n}))\enskip.$

But then we must also have

$\gamma\beta_{1}=\gamma{\tilde{\Phi}}=\theta(\gamma{\tilde{\Psi}}).$

Letting $\Delta$ denote a set of representatives for the left cosets of
${\tilde{G}}_{\Phi}$ in $G$ construct the polynomial

$Q_{1}(t)\enskip=\enskip\prod_{\tau\in\Delta}(t-\tau{\tilde{\Phi}})\in{\cal
F}[t]$

Since $Q_{1}(t)$t shares a root with $Q(t)$, the irreducibility of the latter
forces all the roots of $Q(t)$ to be roots of $Q_{1}(t)$. In other words,
every root $\beta_{i}$ has an expression of the form

$\beta_{i}=\tau_{j_{i}}{\tilde{\Phi}}=\theta(\tau_{j_{i}}{\tilde{\Psi}})\enskip.$

Now, since the normality of ${\tilde{G}}_{\Psi}$ forces all the conjugates of
${\tilde{\Psi}}$ in $G$ to belong to ${\cal F}[{\tilde{\Psi}}]$ we must
conclude that each $\beta_{i}\in{\cal F}[{\tilde{\Psi}}]$ as desired.

We should note that $Q(t)$ being a factor of $Q_{1}(t)$ yields that $m\leq
k=degree\hskip 2.5ptB(t)$.

7\. Solving cyclic equations

In these notes we shall say that an $n^{th}$ degree polynomial
${\tilde{E}}_{n}(t)=\prod_{i=1}^{n}(t-\alpha_{i})\in{\cal F}[t]$ and the
corresponding equation ${\tilde{E}}_{n}(t)=0$ is $cyclic$ in ${\cal F}[t]$ if
its Galois group $G=G_{{\tilde{E}}_{n}(t)}({\cal F})$ is the cyclic group on
$n$ letters. More precisely, ${\tilde{E}}_{n}(t)$ is cyclic with respect to
${\cal F}[t]$ if and only if by a suitable labeling of the roots
$\alpha_{1},\alpha_{2},\ldots,\alpha_{n}$ we have

$G=G({\cal F})=\\{id,\gamma,\gamma^{2},\ldots,\gamma^{n-1}\\}$ $None$

with

$\gamma\alpha_{i}\enskip=\enskip\alpha_{i+1}\hskip 36.135ptfor\enskip\hskip
2.5pti=1,2,\ldots,n\enskip\enskip\enskip(\alpha_{n+1}=\alpha_{1})$ $None$

We should note that in our definition of cyclicity we implicitely assume that
$E_{n}(t)$ has distinct roots. This given, the following basic fact is helpful
in establishing cyclicity.

Theorem 7.1

${\tilde{E}}_{n}(t)$ is cyclic with respect to ${\cal F}[t]$ if and only if

(1) It is irreducible in ${\cal F}[t]$,

(2) We have a polynomial $\theta(t)\in{\cal F}[t]$ and a labeling
$\alpha_{1},\alpha_{2},\ldots,\alpha_{n}$ of the roots of ${\tilde{E}}_{n}(t)$
such that

$\alpha_{i+1}\enskip=\enskip\theta(\alpha_{i})\hskip 36.135ptfor\enskip\hskip
2.5pti=1,2,\ldots,n\enskip\enskip\enskip(\alpha_{n+1}=\alpha_{1})$ $None$

Proof

Suppose ${{\tilde{E}}_{n}(t)}$ is cyclic. Then under the labeling which gives
7.1 and 7.2 set

$Q(t)\enskip=\enskip\sum_{s=1}^{n}\hskip 2.5pt\alpha_{s+1}\hskip
2.5pt\prod_{j=1}^{(s)}(t-\alpha_{j})\enskip.\hskip 36.135pt(\hskip
2.5pt\alpha_{n+1}=\alpha_{1}\hskip 2.5pt)$

Note that for $i=1,2,\ldots,n$ we have

$Q(\alpha_{i})\enskip=\enskip\alpha_{i+1}\hskip
2.5pt{\tilde{E}}_{n}^{\prime}(\alpha_{i})$ $None$

Since the roots $\alpha_{1},\alpha_{2},\ldots,\alpha_{n}$ are supposed to be
distinct ${\tilde{E}}_{n}(t)$ and its derivative ${\tilde{E}}_{n}^{\prime}(t)$
have no roots in common. Thus using the Berlekamp algorithm we can construct
two polynomials $A(t),B(t)\in{\cal F}[t]$ such that
$A(t){\tilde{E}}_{n}(t)+B(t){\tilde{E}}_{n}^{\prime}(t)=1$. Since setting
$t=\alpha_{i}$ yields $B(\alpha_{i}){\tilde{E}}_{n}^{\prime}(\alpha_{i})=1$,
we see that we can rewrite 7.4 in the form

$\theta(\alpha_{i})\enskip=\enskip\alpha_{i+1}$

with

$\theta(t)\enskip=\enskip B(t)Q(t)\in{\cal F}[t]\enskip.$

This proves property $(2)$. To show $(1)$ we need only observe that, in view
of the transitivity of the action in 7.2, the irreducibility of
${{\tilde{E}}_{n}(t)}$ follows from Proposition 6.4.

Suppose now that ${{\tilde{E}}_{n}(t)}$ satisfies $(1)$ and $(2)$. Then, using
the labeling that gives 7.3, we may define $\gamma$ as the circular
permutation that gives $\gamma\hskip 2.5pt\alpha_{i}=\alpha_{i+1}$. Let us
also recursively define the polynomials $\theta_{i}(t)\in{\cal F}[t]$ by
setting $\theta_{o}=t$ and

$\theta_{i+1}(t)\enskip=\enskip\theta(\theta_{i}(t))\hskip 36.135pt(\enskip
i=1,2,\ldots,n-1\enskip)\enskip.$

Let as also set for any $\Phi(x_{1},x_{2},\ldots,x_{n})\in{\cal
F}[x_{1},x_{2},\ldots,x_{n}]$

$R_{\Phi}(t)\enskip=\enskip\Phi\bigl{(}\theta_{o}(t),\theta_{1}(t),\theta_{2}(t),\ldots,\theta_{n-1}(t)\bigr{)}\enskip.$
$None$

Now let $G=\\{id,\gamma,\gamma^{2},\cdots,\gamma^{n-1}\\}$ and suppose that

$S_{\Phi}\supseteq G$ $None$

Using 7.2 and 7.5 we may rewrite this as

$R_{\Phi}(\alpha_{1})=R_{\Phi}(\alpha_{2})=\cdots=R_{\Phi}(\alpha_{n})\enskip.$
$None$

In particular, we must have

$\Phi(\alpha_{1},\alpha_{2},\ldots,\alpha_{n})=R_{\Phi}(\alpha_{1})\enskip=\enskip{1\over
n}\bigl{(}R_{\Phi}(\alpha_{1})+R_{\Phi}(\alpha_{2})+\cdots+R_{\Phi}(\alpha_{n})\bigr{)}$

However since

${1\over
n}\bigl{(}R_{\Phi}(x_{1})+R_{\Phi}(x_{2})+\cdots+R_{\Phi}(x_{n})\bigr{)}\in
Sym[{\cal F};x_{1},x_{2},\ldots,x_{n}]$

its value at $\alpha_{1},\alpha_{2},\ldots,\alpha_{n}$ must be expressible as
a polynomial in the coefficients of ${{\tilde{E}}_{n}(t)}$, This gives that

$\Phi(\alpha_{1},\alpha_{2},\ldots,\alpha_{n})=a\in{\cal F}$ $None$

and establishes that $G\in{\cal A}_{{{\tilde{E}}_{n}(t)}}({\cal F})$. To
finish the proof we need only show that we also have

$G\in{\cal B}_{{{\tilde{E}}_{n}(t)}}({\cal F})\enskip.$ $None$

To this end suppose that 7.8 hold true for some $\Phi\in{\cal
F}[x_{1},x_{2},\ldots,x_{n}]$. Using 7.5 we may translate this property into
the statement that the polynomial

$R_{\Phi}(t)-a\in{\cal F}[t]$

vanishes for $t=\alpha_{1}$. Now under (1) ${{\tilde{E}}_{n}(t)}$ is
irreducible and this forces it to be a factor of $R_{\Phi}(t)-a$. In other
words we must have

$R_{\Phi}(\alpha_{i})=a\hskip 36.135pt(\hskip 2.5ptfor\enskip\enskip
i=1,2,\ldots,n\enskip)\enskip.$

This shows that $S_{\Phi}\supseteq G$ and that 7.9 holds true as desired.

Note that although cyclicity is a field dependent property, here and in the
following we shall drop the appendage “$in\enskip{\cal F}[t]$” or “with
respect to ${\cal F}[t]$” in all cases in which the identity of the base field
${\cal F}$ is clear from the context.

Theorem 7.1 has two immediate important applications:

Corollary 7.1

If a binomial equation $t^{p}-a=0$ with $0\neq a\in{\cal F}$ and $p$ a prime
is irreducible in ${\cal F}[t]$, then it is cyclic if and only if ${\cal F}$
contains a primitive $p^{th}$-root of unity.

Proof

Let $w\in{\cal F}$ be a primitive $p^{th}$-root of unity and let $\alpha$ be a
root of $t^{p}-a$. Then since for no $s<p$ we may have $w^{s}=1$ (*)(*)the
smallest such $s$ would have to divide $p$ the powers $w^{i}$ (for
$i=1,..,p-1$) are all distinct. This gives that the roots of $\enskip
t^{p}-a\enskip$ are simply

$\alpha_{1}=\alpha\hskip 2.5pt,\hskip 2.5pt\alpha_{2}=w\alpha\hskip
2.5pt,\hskip 2.5pt\alpha_{3}=w^{2}\alpha\hskip 2.5pt,\hskip 2.5pt\ldots,\hskip
2.5pt,\hskip 2.5pt\alpha_{n}=w^{n-1}\alpha$

and so we have 7.3 with $\theta(t)=wt$. Thus the cyclicity of $t^{p}-a$
follows from Theorem 7.1.

Conversely suppose that $t^{p}-a$ is cyclic. Let
$\alpha_{1},\alpha_{2},\ldots,\alpha_{p}$ be the labeling of its roots that
gives 7.2 so that its Galois group is $G=\\{id,\gamma,\ldots,\gamma^{p-1}\\}$.
Set $w=\alpha_{2}/\alpha_{1}=\alpha_{2}\alpha_{1}^{p-1}/a$. Clearly we must
have $w^{p}=1$ and we can’t have $w=1$ since that would contradict the
irreducibility of $t^{p}-a$. This given, since the the elements
$w^{i-1}\alpha_{1}$ are distinct and all satisfy $t^{p}-a=0$ they must be a
permutation of $\alpha_{1},\alpha_{2},\ldots,\alpha_{p}$. Thus

$t^{p}-a\enskip=\enskip\prod_{i=1}^{p}\hskip
2.5pt(t-w^{i-1}\alpha_{1})\enskip.$

We thus deduce that

$\alpha_{i}=w^{h_{i}}\alpha\hskip 36.135pt(\enskip i=1,\ldots p\enskip)$

where $h_{1},h_{2},\ldots,h_{p}$ is a permutation of $0,1,\ldots,p-1$. From
this we derive that $\gamma w=\alpha_{3}/\alpha_{2}=w^{h_{3}-1}$. Since
$\alpha_{3}\neq\alpha_{2}$ we see that $h_{3}\neq 1\enskip mod\enskip p$.
Setting $h_{3}-1=e$ we get $\gamma^{k}w=w^{e^{k}}$. But then a theorem of
Euler gives that $s^{p-1}\cong 1$ mod $p$. In particular we get that
$\gamma^{p-1}w=w$. Now, $\gamma^{p-1}=\gamma^{-1}$ and $\gamma^{-1}$ generates
$G$ as well as $\gamma$. Thus $w$ is invariant under $G$ and must necessarily
belong to ${\cal F}$ as we wanted to show.

Corollary 7.2

The cyclotomic polynomial

$\Phi_{p}(t)=1+t+t^{2}+\cdots+t^{p-1}\hskip 36.135pt(for\enskip p\enskip
a\enskip prime\enskip)$ $None$

is cyclic with respect to the field ${\cal Q}$ of rational numbers

Proof

We start by showing that $\Phi_{p}$ is irreducible in ${\cal Q}[t]$. Suppose
not. Then by Gauss theorem we will have a non trivial factorization

$\Phi_{p}(t)\enskip=\enskip A(t)B(t)$ $None$

where $A(t)$ and $B(t)$ are both monic polynomials with integer coefficients.
Setting $t=1$ gives

$p=\Phi_{p}(1)\enskip=\enskip A(1)B(1)\enskip.$

So one of $A(1),B(1)$ must be $\pm 1$. Say it is $A(1)$. Now let $\alpha$ be a
root of $\Phi_{p}$. Since $\alpha^{p}=1$ we can’t have $\alpha^{s}=1$ for any
$1\leq s\leq p-1$. In particular, for such an $s$, the elements

$\alpha^{s}\hskip 2.5pt,\hskip 2.5pt\alpha^{2s}\hskip 2.5pt,\hskip
2.5pt\ldots\hskip 2.5pt,\hskip 2.5pt\alpha^{(p-1)s}$

are just a rearrangement of the roots of $\Phi_{p}$. Since at least one of
them is a root of $A(t)$, we are forced to conclude that

$A(\alpha^{s})A(\alpha^{2s})\cdots A(\alpha^{(p-1)s})\enskip=\enskip
0\enskip.\hskip 36.135pt(\enskip\forall\enskip 1\leq s\leq p-1\enskip)$

This yields that the polynomial

$R(t)=A(t)A(t^{2})\cdots A(t^{(p-1)})$

must be divisible by $\Phi_{p}$. Since $R(t)$ is monic with integer
coefficients, again by Gauss theorem, we shall have the factorization

$R(t)=(1+t+t^{2}+\cdots+t^{p-1})R_{1}(t)$

with $R_{1}(t)$ also monic with integer coefficients. Now setting $t=1$ we are
forced to the impossible conclusion that

$\pm 1\enskip=\enskip p\hskip 2.5ptR_{1}(1)\enskip.$

Thus $\Phi_{p}$ must be irreducible as asserted.

We know from number theory that for any prime $p$ we can find a primitive
exponent $e\in[1,p-1]$ which has the propery that the integers

$e\hskip 2.5pt,\hskip 2.5pte^{2}\hskip 2.5pt,\hskip 2.5pt\ldots e^{p-2}$

are distinct and, in fact, give (modulo $p$) a permutation of

$1\hskip 2.5pt,\hskip 2.5pt2\hskip 2.5pt,\hskip 2.5pt\ldots,p-1$

Choosing one such exponent, we derive that if $\alpha$ is any of the roots of
$\Phi_{p}$ then the powers

$\alpha\hskip 2.5pt,\hskip 2.5pt\alpha^{e}\hskip 2.5pt,\hskip
2.5pt\ldots\alpha^{e^{p-2}}$

give back again all the roots of $\Phi_{p}$. But this means that if we label
the roots of $\Phi_{p}$ by setting $\alpha_{i}=\alpha^{e^{i-1}}$ we shall have
7.3 with

$\theta(t)\enskip=\enskip t^{e}\in{\cal Q}[t]\enskip.$

This completes our proof.

There is a further property of the binomial equation with prime exponent we
need to know here.

Proposition 7.1

The binomial $t^{p}-a$ ($a\in{\cal F}$) is reducible in ${\cal F}[t]$ if and
only if $a$ is the $p^{th}$ power of an element of ${\cal F}$.

proof

Suppose that $t^{p}-a$ is reducible in ${\cal F}[t]$ and let

$t^{p}-a\enskip=\enskip A(t)B(t)\hskip 36.135pt(\enskip A(t),B(t)\in{\cal
F}[t]\enskip)$

be a non trivial factorization. We may then write

$A(t)=\prod_{i=1}^{s}(t-w^{h_{i}}\alpha)$

where $s<p$, $\alpha$ is any root of $t^{p}-a$, and $w$ can be chosen to be
the ratio of any two roots. However, $A(t)\in{\cal F}[t]$ implies in
particular that its constant term $c$ is in ${\cal F}$. In other words

$\alpha^{s}w^{({h_{1}}+{h_{2}}+\cdots+{h_{s}})}=c\enskip\in{\cal F}$ $None$

Since $1\leq s<p$ we can find integers $h,k$ such that $hs=1+kp$. Raising both
sides of 7.12 to the power $h$ yields

$\alpha\hskip 2.5pta^{k}w^{h^{\prime}}=c^{h}\enskip,$

for a suitable integer $0\leq h^{\prime}\leq p-1$. Thus one of the roots of
$t^{p}-a$, namely $b=\alpha w^{h^{\prime}}$ lies in ${\cal F}$ and $a$ must
necessarily the $p^{th}$ power of an element of ${\cal F}$ as we asserted.

The converse is entirely trivial since when $a=b^{p}$ with $b\in{\cal F}$ we
have the factorization

$t^{p}-a=(t-b)(b^{p-1}+b^{p-2}t+\cdots+bt^{p-2}+t^{p-1})\enskip.$

To proceed any further in this section we need to update the meaning of
solving by radicals in the Galois setting. It is natural to assume that in
this setting root extraction should simply mean extending a given field ${\cal
P}$, by the adjunction of a root of a binomial equation

$t^{n}-a=0\enskip.\hskip 36.135pt(\enskip a\in{\cal P}\enskip)$

This given, solving by radicals the equation ${\tilde{E}}_{n}(t)=0$, in the
Galois setting means constructing the roots of ${\tilde{E}}_{n}(t)\in{\cal
F}[t]$ by a sequence of extensions

${\cal F}_{k-1}\hskip 2.5pt\rightarrow\hskip 2.5pt{\cal
F}_{k}\enskip=\enskip{\cal F}_{k-1}[\xi_{k}]\enskip\enskip\enskip\hskip
36.135pt(\enskip k=0,1,2,..,k_{o}\enskip)$ $None$

with

$\eqalign{&(1)\enskip\enskip\xi_{k}^{p_{k}}\in{\cal
F}_{k-1}\cr&(2)\enskip\enskip{\cal F}_{o}={\cal
F}\enskip,\cr&(3)\enskip\alpha_{1},\alpha_{2},\ldots,\alpha_{n}\in{\cal
F}_{k_{o}}\enskip.\cr}$ $None$

In particular, this will enable us to construct a formula, say for
$\alpha_{1}$, which will be of the form

$\alpha_{1}\enskip=\enskip\cdots\enskip+\enskip{\root p \of{\cdots{\root q
\of{\cdots+{\root r \of{\cdots}}+{\root s \of{\cdots}}\cdots}}}}$ $None$

Where the radicals appearing in it will be appropriately chosen solutions of
the equations

$t^{p}_{k}-a_{k}\enskip=\enskip 0\enskip\enskip\enskip\enskip{\rm
with}\enskip\enskip a_{k}\in{\cal F}_{k-1}$ $None$

If we chose to represent the element $\xi_{k}$ appearing in 7.14 $(1)$ by the
symbol $\root p_{k} \of{a}_{k}$, we are confronted with the ambiguity
resulting from the multivalued nature of the symbol “$\root p \of{.}..$”. For
instance if, at the $k^{th}$ step of the extension process, $\xi_{k}$ is to be
a primitive $6^{th}$-root of unity. It would be better to represent $\xi_{k}$
by $-{1\over 2}+{{\root 2 \of{-}3}\over 2}$ rather by $\root 6 \of{1}$. This
is because the expression $-{1\over 2}+{{\root 2 \of{-}3}\over 2}$ yields only
these two primitive roots as we specialize $\root 2 \of{-}3$ to the two
conjugate roots of the equation $x^{2}+3=0$. Clearly, it is in the nature of
the problem that any formula we may construct for $\alpha_{1}$ should have
multivalued symbols appearing in it. For $\alpha_{1}$ itself is in essence an
$n$-valued function. Indeed, labeling the roots of ${\tilde{E}}_{n}$
$\alpha_{1},\alpha_{2},\ldots,\alpha_{n}$ is an artificial device, and
“$\alpha_{1}$” should really represent only a generic  root of
${\tilde{E}}_{n}$. Now, we have seen already in the Lagrange setting, that
there is no loss in requiring that in the successive radicals $\root p_{k}
\of{a}_{k}$ the exponent $p_{k}$ should be a prime number. Now if the equation
$t^{p_{k}}-a_{k}=0$ is cyclic in ${\cal F}_{k-1}$ the adjonction of any one of
its roots to ${\cal F}_{k-1}$ will result in the same field ${\cal F}_{k}$. So
writing ${\cal F}_{k}={\cal F}_{k-1}[{\root p_{k} \of{a}_{k}}]$ does not
produce any ambiguity as far ${\cal F}_{k}$ is concerned. Moreover, we see
that if we require that each symbol $\xi_{k}={\root p_{k} \of{a}_{k}}$
appearing in 7.14 represents the same root of $t^{p_{k}}-a_{k}=0$ and if we
let each of these symbols in turn and independently describe each of the other
solutions of the corresponding equation $t^{p_{k}}-a_{k}=0$ then $\alpha_{1}$
as expressed by 7.14 will represent a $p_{1}p_{2}\cdots p_{k_{o}}$-valued
function. So that if $p_{1}p_{2}\cdots p_{k_{o}}=n$ then 7.14 by this process
will deliver each of the roots of ${\tilde{E}}_{n}$. Here and after a formula
7.15 satisfying these requirements will be called a tight  formula and the
symbols ${\root p_{k} \of{a}_{k}})$ appearing in it will likewise be called
tight radicals. More generally we shall refer to an adjunction ${\cal
P}\rightarrow{\cal P}[\xi={\root p \of{a}}]$ as a tight radical extraction if
and only if $t^{p}-a$ is cyclic in ${\cal P}$. All other formulas and root
extractions will be referred to as loose. It is not difficult to verify that
the formulas for the roots of the cubic and quartic which can be obtained by
the process given in section 5 are, in fact, tight. However as we pointed out,
the process in section 5 has the additional property that at each step the
element adjoined (roots of unity apart), is a root of a polynomial function of
the roots $x_{1},x_{2},\ldots,x_{n}$. This brings us to define 7.15 as a
natural formula if every $\xi_{k}$ is also a polynomial in
$\alpha_{1},\alpha_{2},\ldots,\alpha_{n}$ with coefficients in ${\cal F}$. In
this terminology, we can say that Lagrange showed that (as long as ${\cal F}$