Tsemo Aristide

College Boreal

1, Yonge Street, Toronto, ON

<EMAIL_ADDRESS>

Some properties of schemes in groups theory and Top couples.

Abstract.

Let $G$ be a group, and $H$ a $G$-group defined by an imbedding map
$G\rightarrow H$; in [12] we have defined a topology on a subset of normal
subgroups of $H$, the so-called prime ideals. In this work, we generalize this
topology to other classes of groups. We hope that such topologies define new
effective tools to tackle well-known problems in groups theory. We study
properties of the topology defined in [12], for example we define the notions
of functor of points, dimension and outline a Galois theory.

1\. Introduction.

Classical algebraic geometry is the study of solutions of polynomial equations
defined on commutative rings. The nature of the objects involved here can be
very different when one allows the base ring to vary; for example varieties
defined over the fields of complex numbers are complex manifolds with
singularities, thus, inherit a separated topology from their complex
structure, this is not the case of a variety defined over a finite field. The
first tools developed in classical algebraic geometry where related to
topology or complex geometry on the purpose to study complex schemes. In fact,
Serre has shown in the G.A.G.A [8] deep relations between the objects of these
theories.

To study the variety defined on more general rings, some tools issued from
complex algebraic geometry have been adapted. The most efficient work in that
direction has been done by A. Grothendieck with the theory of schemes; he has
introduced topological concepts to objects which does not look a priori
topologically by generalizing the Zariski topology. This approach has allowed
to give new interpretations of many conjectures and often led to their
solutions. We can for examples quote the Weil conjectures and more recently
the Milnor conjecture.

Many generalizations of the theory of schemes have been achieved for non
commutative algebras. In group theory, many classical sets are defined by
algebraic-like equations. Thus formally, they are similar to the objects
studied in classical algebraic geometry. Recently, Baumslag, Myasnikov and
Remeslennikov [2] have initiated the study of varieties of groups. They
consider a group $G$, and $G[X_{1},...,X_{n}]$ the free product of $G$ and the
free group generated by a set of cardinality $n$. For any group $H$ endowed
with an injective morphism $G\rightarrow H$, the elements of
$G[X_{1},...,X_{n}]$ define polynomial functions on $H^{n}$ which took their
values in $H$. The solutions of these equations define a Zariski-like
topology. These authors have adapted tools from the theory of algebraic
varieties like Zariski topology and irreducible subsets in groups theory.

In our paper [12], we have generalized the construction of Baumslag, Myasnikov
and Remeslennikov by defining a theory of schemes in this context, The purpose
of this work is to continue to study this notion; we also generalize it by
extending it to others subcategories of the category of groups. We hope that
these new concepts will also lead to the solutions of some problems in groups
theory by using the topological intuition.

Plan.

1\. Introduction

2\. Some properties of groups

3\. Topology in groups theory

4\. The geometry of $G$-groups

5\. The functor of points

6\. Noetherian $G$-schemes

7\. Dimension of affine schemes

8\. Galois theory.

2\. Some properties of groups.

Let $G$ be a group, the comma category over $G$ is the category whose objects
$(H,\phi_{H})$ are morphisms of groups $\phi_{H}:G\rightarrow H$. A morphism
$f:(H,\phi_{H})\rightarrow(H^{\prime},\phi_{H^{\prime}})$ of the comma
category is a morphism of groups $f:H\rightarrow H^{\prime}$ such that
$f\circ\phi_{H}=\phi_{H^{\prime}}$.

We denote by $C(G)$ the subcategory of the comma category such that each
object $(H,\phi_{H})$ of $C(G)$ is an object of the comma category such that
$\phi_{H}$ is injective.

Let $(H,\phi_{H})$ be an object of $C(G)$ and $x$ an element of $H$. For every
element $g\in G$, we denote $\phi_{H}(g)x\phi_{H}(g^{-1})$ by $x^{g}$. We
denote by $G(x)$ the subgroup of $H$ generated by $\\{x^{g},g\in G\\}$. We say
that the element $x\in H$ is invertible if and only if
$G(x)\cap\phi_{H}(G)\neq 1$.

Let $H$ be a group, and $g$, $g^{\prime}$ elements of $H$; the commutator
$[g,g^{\prime}]$ of $g$ and $g^{\prime}$ is the element
$gg^{\prime}g^{-1}{g^{\prime}}^{-1}$. If $U$ and $V$ are subgroups of $H$,
$[U,V]$ is the normal subgroup generated by $[u,v],u\in U,v\in V$.

Let $H$ be a group, we denote by $D^{0}H$ the group $H$, suppose defined
$D^{n}H$, for an integer $n$, then $D^{n+1}H=[D^{n}H,D^{n}H]$.

We say that an element $x$ of $H$ distinct of the neutral element is a divisor
of zero if and only if there exists an element $y$ of $H$ distinct of the
neutral element such that the group of commutators $[G(x),G(y)]=1$. Remark
that the element $y$ in this situation is also divisor of zero.

Proposition 2.1.

The element $x$ of $H$ is a divisor of zero if and only if there exists a non
trivial element $y$ of $H$ such that for every element $g\in G$,
$[x,y^{g}]=1$.

Proof.

Let $x$ be an element of $H$ which is a divisor of zero, and $y$ an element of
$H$ such that $[G(x),G(y)]=1$. The set $U=\\{[x,y^{g}],g\in G\\}$ is contained
in $[G(x),G(y)]$; we deduce that all elements of $U$ are equal to the neutral
element.

Conversely, suppose that all elements of $U$ are equal to the neutral element.
Let $u,v\in G$,

$[x^{u},y^{v}]=uxu^{-1}vyv^{-1}ux^{-1}u^{-1}vy^{-1}v^{-1}.$

We deduce that:

$u^{-1}[x^{u},y^{v}]u=xu^{-1}vyv^{-1}ux^{-1}u^{-1}vy^{-1}v^{-1}u=[x,y^{u^{-1}v}]$

Since $[x,y^{u^{-1}v}]=1$,we deduce that $[x^{u},y^{v}]=1$.

Recall that if $a,b,c$ are elements of a group, if $a$ commutes with $b$ and
$c$, then $a$ commutes with $b^{-1}$ and $bc$, this implies that $G(x)$
commutes with $G(y)$.

Remark.

Let $N_{G}(x)$ be the normal subgroup generated by $G(x)$, we can find an
example of a $G$-group $H$ such that $[G(x),G(y)]=1$ and
$[N_{G}(x),N_{G}(y)]\neq 1$:

Consider $H=SL(4,C)$ the group of linear automorphisms of the $4$-dimensional
complex vector space $C^{4}$ whose determinant is $1$. Let
$e_{1},e_{2},e_{3},e_{4}$ be a basis of $C^{4}$. Let $G=SL(2,C)$ be the
subgroup of $H$ such that for every $f\in G$, $f(e_{3})=e_{3},f(e_{4})=e_{4}$
and $f(e_{1})$ and $f(e_{2})$ are elements of the vector subspace
$Vect(e_{1},e_{2})$ generated by $e_{1}$ and $e_{2}$. Let $g$ be an element of
$H$ such that $g(e_{1})=e_{1},g(e_{2})=e_{2}$ and $g(Vect(e_{3},e_{4}))\subset
Vect(e_{3},e_{4})$, $[g,G]=1$, but $[N_{G}(g),G]$ is not the trivial subgroup:
if $f:C^{4}\rightarrow C^{4}$ is defined by
$f(e_{1})=e_{3},f(e_{2})=e_{4},f(e_{3})=e_{1}$ and $f(e_{4})=e_{2}$,
$fgf^{-1}\in G$, and we can choose $g$ such that $fgf^{-1}$ is not in the
center of $G$.

Proposition 2.2.

Suppose that $G$ is a solvable $G$-group different of $\\{1\\}$, then any
$G$-group $H$ has a divisor of zero. If $H$ is a finite dimensional simply
connected Lie group which does not have zero divisors, then $H$ is simple.

Proof.

Let $G$ be a solvable group different of $\\{1\\}$; there exists $n$ such that
$D^{n}G$ is not trivial and $D^{n+1}G$ is trivial. For every element $x\in
D^{n}G$ different of its neutral element we have $[G(x),G(x)]=1$, thus $x$ is
a divisor of zero; if $H$ is a $G$-group, $x$ is also a zero divisor of $H$.

Suppose that $H$ is a finite dimensional simply connected Lie group which does
not have divisors of zero and its radical is not trivial. let $R$ be the
radical of $H$, recall that $R$ is a maximal solvable normal subgroup of $H$.
Let $n$ be an integer such that $D^{n}R$ is not trivial, and $D^{n+1}R$ is
trivial. Since $R$ is a normal subgroup of $H$, for every element $x\in
D^{n}R$ and every element $g\in G$, $x^{g}\in D^{n}R$, this implies that
$[G(x),G(x)]=1$ for every element $x\in D^{n}R$. In particular, if $x$ is not
the neutral element it is a divisor of zero. This is in contradiction with the
fact that we have supposed that $G$ does not have divisors of zero. Thus the
radical of $H$ is trivial and $H$ is semi-simple. If $H$ is not simple, then
it is the product of two non trivial groups $H_{1}$ and $H_{2}$. Let $h_{1}$
and $h_{2}$ be respectively elements of $H_{1}$ and $H_{2}$ distinct of the
neutral element, we have $[G((h_{1},1));G((1,h_{2}))]=1$. Thus $H$ has
divisors of zero.

Remark.

The previous proposition shows that a non commutative nilpotent group $G$ has
divisors of zero. In fact, every element in the center of $G$ is a divisor of
zero.

Proposition 2.3.

Let $k$ be a field, and $G$ a subgroup of $PSL(n,k)$ suppose that the action
of $G$ on the $n$-dimensional $k$-projective space $Pk^{n}$ is reducible, then
$G$ has divisors of zero. If $G$ is a connected semi-simple Lie group defined
over the complex numbers imbedded in $PSL(n,C)$ or if $G$ is a finite group
imbedded in $PSL(n,k)$ where $k$ is a field whose characteristic does not
divide the order of $G$, then $G$ does not preserves the image of a proper
linear vector space in $P^{n}C$ (resp. $P^{n}k)$ if $G$ does not have zero
divisors.

Proof.

Suppose that the action of $G$ is reducible. Let $G^{\prime}$ be the inverse
image of the group $G$ by the projection map $p:Sl(n,k)\rightarrow PSL(n,k)$.
We can write $k^{n}=U\oplus V$, where $U$ and $V$ are non trivial vector
subspaces of $k^{n}$ and are stable by $G^{\prime}$. Let $u^{\prime}$ be a non
trivial invertible linear map of $U$ and $v^{\prime}$ be a non trivial
invertible linear map of $V$. Consider the linear morphism $u$ of $k^{n}$
defined by $u(x)=u^{\prime}(x)$ if $x\in U$, $u(x)=x$ if $x\in V$. Let $v$ be
the linear morphism of $k^{n}$ defined by $v(x)=x$ if $x\in U$,
$v(x)=v^{\prime}(x)$ if $x\in V$. We have $[G^{\prime}(u),G^{\prime}(v)]=1$.
It results that the image of $u$ and $v$ in $PSL(n,k)$ are divisors of zero.

Suppose that $G$ is a connected semi-simple subgroup of $PSL(n,C)$, (resp. $G$
is a finite subgroup of $PSl(n,k)$ such that the order of $G$ and the
characteristic of $k$ are prime each other), $p^{-1}(G)$ is semi-simple (resp.
finite subgroup) and $C^{n}$ (resp $k^{n}$) under the action of $G$ can be
decomposed in the direct summand of irreducible vector spaces. Thus $PSL(n,C)$
(resp. $PSL(n,k)$ has zero divisors if $p^{-1}(G)$ preserves a proper linear
subspace.

Definitions 2.1.

A $G$-domain $(H,\phi_{H})$ of $C(G)$ is an object of $C(G)$ such that $H$
does not have divisors of zero.

A normal subgroup $I$ of $H$ is an ideal of $H$ if and only if $I\cap
G=\\{1\\}$.

A normal subgroup $P$ of $H$ is a prime ideal if and only if:

\- $P$ is an ideal distinct of $H$ and

\- $G/P$ is a $G$-domain, this fact is equivalent to saying that for every
$x,y\in H$, $[G(x),G(y)]\in P$ if and only if $x$ is in $P$ or $y$ is in $P$.

\- The $G$-group $H$ is $G$-simple if $\\{1\\}$ is the the only ideal of $H$.

Proposition 2.4.

Suppose that $G$ is a connected simple Lie group over a field of
characteristic zero, and $H$ a connected Lie group in $C(G)$, if $H$ is
$G$-simple then $H$ is simple Lie group.

Proof.

We can write $H$ has a semi-product of a semi-simple group $U$ and a solvable
group $L$ by using the Levi decomposition. Without restricting the generality,
we can suppose that $G$ is contained in $U$ since $G$ is simple. Since $G\cap
L=\\{1\\}$, $L$ is an ideal. We deduce that $L$ is trivial since the only
prime of $H$ is $1$. If the subgroup $U$ is not simple, there will exist a
simple component of $U$ which intersection with $G$ is trivial since the
simple Lie group $G$ is contained in a simple component of $H$. This fact is
impossible since all ideals of $H$ are trivial.

Remark.

The previous result is not true if we don’t assume that the groups $G$ and $H$
are connected as shows the following example: Let $S_{n}$ be the group of
permutations of the set of $n$ elements, $n>4$. The subgroup $A_{n}$ of
permutations of even signature is a simple group. The group $S_{n}$ is endowed
with an $A_{n}$-structure. It is well-known that the only normal subgroups of
$S_{n}$ are the trivial subgroup, $A_{n}$ and $S_{n}$. Thus $S_{n}$ is an
$A_{n}$-simple group, and $S_{n}$ is not simple.

Proposition 2.5.

Let $k$ be a field and $G$ a Zariski dense subgroup of $PSl(n,k)$, the
$G$-group $PSl(n,k)$ does not have divisors of zero.

Proof.

Let $x$ be a divisor of zero and $y$ a non trivial element of $PSl(n,k)$ such
that for every element $g\in G$, $[x,y^{g}]=1$. The equation $[x,y^{u}]=1$ is
an algebraic equation in $PSl(n,k)$. This implies that the Zariski adherence
of $G$ satisfies this equation. Since $G$ is Zariski dense in $PSL(n,k)$, this
implies that for every $g\in PSl(n,k)$, $[x,y^{g}]=1$. Let $x^{\prime}$ (resp.
$y^{\prime}$) an element of $Sl(n,k)$ above $x$ (resp. above $y$). Thus
$[x^{\prime},{y^{\prime}}^{g}]=cI_{n}$ is in the center of $SL(n,k)$ for every
$g\in SL(n,k)$. Recall that the elements of the center of $SL(n,k)$ are
$c_{1}I_{1},...,c_{n}I_{n}$ where $c_{1},...,c_{n}$ are the $n$-roots of
unity. The subset $G_{i}$ of $SL(n,k)$ such that
$[x^{\prime},{y^{\prime}}^{g}]=c_{i}^{-1}I_{n}$ is closed, and $SL(n,k)$ is a
finite union of the $G_{i}$. Since $SL(n,k)$ is irreducible, there exists a
$i$ such that $G_{i}=SL(n,k)$. It results that
$[{x^{\prime}}^{n},{y^{\prime}}^{g}]=I_{n}$ for every $g\in SL(n,k)$. Let
$E_{h}$ be an eigenvector space of $y^{\prime}$ associated to the eigenvalue
$h$. Let $p$ be the dimension of $E_{h}$, for every vector subspace $E$ of
dimension $p$, there exists an element of $SL(n,k)$ such that $g(E_{h})=E$,
this implies that $E$ is the eigenvector space of ${y^{\prime}}^{g}$ and
${x^{\prime}}^{n}$ preserves $E$. We deduce that ${x^{\prime}}^{n}=cI$, $c\in
k$. The same argument shows also that ${y^{\prime}}^{n}=dI_{n}$, $d\in k$.
Thus $x^{\prime}$ and $y^{\prime}$ are diagonalizable maps. If $g$ is an
element of $Sl(n,k)$, $g(E_{h})$ is an eigenvector space of ${y^{\prime}}^{g}$
associated to $h$. This implies that ${x^{\prime}}(g(E))$ is the eigenvector
space of ${y^{\prime}}^{g}$ associated to $c_{i}^{-1}h$ if
$[x^{\prime},{y^{\prime}}^{g}]=c_{i}I_{n}$. Thus $x^{\prime}$ permutes the
eigenvector spaces of $y^{\prime}$. Since $x^{\prime}$ and $y^{\prime}$ are
not homothetic maps since $x$ and $y$ are not trivial, $y^{\prime}$ has two
distinct eigenvector spaces $E_{1}$ and $E_{2}$ associated respectively to the
eigenvalue $c_{1}$ and $c_{2}$; without restricting the generality, since
$x^{\prime}$ permutes the eigenvector space of ${y^{\prime}}^{g}$ for every
$g\in SL(n,k)$, we can suppose that $x^{\prime}(g(E_{1}))=g(E_{2})$. This last
equality is not possible for every element of $SL(n,k)$.

Proposition 2.6.

Let $H$ be an element of $C(G)$, suppose that $G$ is a normal subgroup of $H$,
then for every ideal ideal $I$, $[I,G]=1$.

Proof.

Let $g$ be an element of $G$, and $i$ and element of $I$, we have
$[u,i]=uiu^{-1}i^{-1}$ is an element of $G\cap I=\\{1\\}$.

Definitions 2.2.

Let $(H,\phi_{H})$ be an object of $C(G)$, an element $x$ of $H$ is nilpotent
of length $n$ if and only if the group $G(n)(x)=[G(x),[G(x)[,...,G(x)]=1$
where in the formula $G(x)$ appears $n$ times.

Let $H$ be an element of $C(G)$, the radical $rad_{G}(H)$ of $H$ is the
intersection of all prime ideals of $H$. We denote by $Com_{H}(G)$ the
subgroup of $H$ such that for every element $h\in Com_{H}(G)$ and $g\in G$, we
have $[g,h]=1$. The group $Com_{H}(G)$ is included in every element $P\in
Spec_{G}(H)$ and so is its normalizer $N(Com_{H}(G))$. This implies that
$N(Com_{H}(G))$ is a subset of $Rad_{G}(H)$.

Proposition 2.7.

Let $P$ be a prime and $x$ an element of $H$, suppose that $G(n)(x)$ is
contained in $P$, then $x$ is an element of $P$. In particular The normal
subgroup $Nil(H)$ generated by nilpotent elements of $H$ is contained in $P$.

Proof.

We show the result recursively. Let $P$ be a prime, suppose that every element
$u$ such that $G(n-1)(u)\in P$ is in $P$. Let $x$ be an element of $H$ such
that, $G(n)(x)$ is contained in $P$. Suppose that $x$ is not in $P$. Let $u$
be an element of $G(n-1)(x)$, we have $[G(x),u]\subset G(n)(x)\subset P$.
Since $P$ is prime, $u$ is an element of $P$, we deduce that $G(n-1)(x)$ is
contained in $P$, the iteration hypothesis implies that $x$ is an element of
$P$.

Proposition 2.8.

Suppose that $G$ is a simple group, and $H$ is an element of $C(G)$, then
every normal subgroup of $H$ is an ideal or contain $G$.

Proof.

Let $I$ be a normal subgroup of $H$, $I\cap G$ is a normal subgroup of $G$;
since $G$ is simple we deduce that $I\cap G=G$ or $I\cap G=\\{1\\}$.

3\. Topologies in group theory.

In this part, we are going to generalize the topology introduced in [12]. Let
$C$ be a subcategory of the category of groups, we denote by $D$ a subclass of
the class of objects of $C$ which satisfies the following properties:

T1 If $G$ is an object of $C$, $G^{\prime}$ is an object of $D$ such that
there exists an injective $C$-morphism $i:G\rightarrow G^{\prime}$, then $G\in
D$.

T2 Let $G$ be an element of $D$, and $I$, $J$ normal subgroups of $G$,
$[I,J]=1$ implies that $I=1$ or $J=1$.

Definitions 3.1.

Let $H$ be an element of $C$, an ideal of $H$ is a normal subgroup $I$ of $H$
such that $H/I\in C$ and the projection $H\rightarrow H/I$ is a morphism of
$C$.

An ideal $P$ of $H$ is a prime if and only if $H/P$ is an object of $D$ and
$P$ is distinct of $H$.

T3 We suppose that the inverse image of an ideal by a morphism of $C$ is an
ideal.

A couple $(C,D)$ which satisfies the properties T1, T2 and T3 will be called a
Top couple.

Let $H$ be an element of $C$ and $I$ a normal subgroup of $H$, we denote by
$V(I)$ the set of prime ideals of $H$ which contain $I$.

Proposition 3.1.

Let $H$ be an element of $C$, and $I$, $J$ normal subgroups of $H$, we have:

$V([I,J])=V(I)\bigcup V(J)$

Let $(I_{a})_{a\in A}$ a family of normal subgroups of $H$, and $I_{A}$ the
normal subgroup generated by $(I_{a})_{a\in A}$ we have:

$V(I_{A})=\cap_{a\in A}V(I_{a})$

Proof.

Let $P$ be an element of $V(I)\bigcup V(J)$, $P$ contains $I$ or $J$. This
implies that $P$ contains $[I,J]$.

Let $P$ be an element of $V(I,J])$ and $p:H\rightarrow H/P$ the natural
projection map, we have $[p(I),p(J)]=1$. Since $H/P$ is an object of $D$, we
deduce that $p(I)=1$ or $p(J)=1$. This is equivalent to saying that $I\subset
P$ of $J\subset P$.

Let $P$ be an element of $V(I_{A})$, for every $a\in A$, we have $I_{a}\subset
I_{A}\subset P$ this implies $P\in\cap_{a\in A}V(I_{a})$.

Let $P$ be an element of $\cap_{a\in A}V(I_{a})$, for every $a\in
A,I_{a}\subset P$ this implies that $I_{A}\subset P$ and $P\in V(I_{A})$, we
deduce that $V(I_{A})=\cap_{a\in A}V(I_{a})$.

Remarks.

Let $(C,D)$ be a Top couple and $H$ be an object of $C$, and $Spec(H)$ the set
of prime ideals of $H$. The set $V(I)$ are the closed subsets of a topology
defined on $Spec(H)$; $V(H)$ is the empty subset, and $V(\\{1\\})=Spec(H)$.

Proposition 3.2.

Let $(C,D)$ be a Top couple and $H$, $H^{\prime}$ two objects of $C$. A
$C$-morphism $f:H\rightarrow H^{\prime}$ induces a continuous morphism
$c(f):Spec(H^{\prime})\rightarrow Spec(H)$ by setting $c(f)(P)=f^{-1}(P)$.

Proof.

Let $P$ be an ideal of $H^{\prime}$, the property T3 implies that $f^{-1}(P)$
is an ideal subgroup of $H$ since the morphism $H/f^{-1}(P)\rightarrow
H^{\prime}/P$ is injective, the property T1 implies that $H/f^{-1}(P)\in D$.
This implies that $f^{-1}(P)$ is a prime ideal.

The morphism $c(f)$ is continuous since $c(f)^{-1}(V(I))=V(f(I))$.

Examples of Top couples.

We are going to present examples of Top couple, we start by the following
example which is already presented in Tsemo [12]:

Proposition 3.3.

Let $G$ be a group, and $D$ the class of $G$-groups which do not have divisors
of zero, the couple $(C(G),D)$ is a Top couple.

proof.

We have to show that the couple $(C(G),D)$ satisfies the properties $T1$, $T2$
and $T3$.

If $H$ is a $G$-group which does not have zero divisors, any $G$-subgroup of
$H$ does not have zero divisors of zero, thus the property $T1$ is satisfied.

We show now that the couple $(C(G),D)$ satisfies the property T2. Let $H$ be a
group which does not have divisors of zero, and $I,J$ two normal subgroups of
$H$ such that $[I,J]=1$. Suppose that $I\neq 1$ and $J\neq 1$. This implies
the existence of elements $x\in I$ and $y\in J$ different of the neutral
elements. We have $[G(x),G(y)]\subset[I,J]=1$. This is equivalent to saying
that $x$ and $y$ are divisors of zero. This fact is a contradiction to the
hypothesis; thus $I=1$ or $J=1$.

It remains to show that the couple $(C,D)$ satisfies the property $T3$. Recall
that an ideal $I$ of the element $H$ of $C(G)$ is a normal subgroup $I$ of $H$
such that $I\cap G=1$, if $f:H\rightarrow H^{\prime}$ is a $G$-morphism, and
$I$ an ideal of $H^{\prime}$, $f^{-1}(I)$ is an ideal of $H$ since the
restriction of $f$ on $G$ is injective.

We present now another example of a Top couple. One of the most interesting
problem in groups theory is to determine if a group has a finite subgroup
which is isomorphic to the free group generated by two elements. Tits Jacques
has shown that a subgroup of a linear group is virtually solvable or he
contains a free subgroup isomorphic to $F_{2}$. The Top couple that we are
presenting now is related to this subject.

Here $C$ is $C(F_{2})$ the category whose objects are injective morphisms
$F_{2}\rightarrow H$. An object $F_{2}\rightarrow H$ of $C$ is an object of
$D$ if and only if $H$ is a free group. Remark that the rank of an object of
$D$ is greater or equal to $2$. Let $(H,\phi_{H})$ be an object of $C(F_{2})$,
an ideal $I$ of $H$ is a normal subgroup $I$ such that
$I\cap\phi_{H}(F_{2})=\\{1\\}$. From this definition, we deduce that the ideal
$P$ of $H$ is a prime if and only if $H/P$ is a free group.

Proposition 3.4.

The couple $(C(F_{2}),D)$ is a Top couple.

Proof.

Let show that $(C(F_{2}),D)$ satisfies $T1$. Let $i:H\rightarrow H^{\prime}$
be an injective $F_{2}$-morphism, the fact that $H^{\prime}$ is an object of
$D$ is equivalent to saying that $H^{\prime}$ is free, this implies that $H$
is free since a subgroup of a free group is free.

Property T2. Let $H$ be an object of $D$ and $I,J$ ideal of $H$ such that
$[I,J]=1$. Since $H$ is a free group, we deduce that $I=1$ or $J=1$.

Property T3 Let $f:H\rightarrow H^{\prime}$ be an $F_{2}$-morphism, and $P$ an
ideal of $H^{\prime}$, $f^{-1}(P)$ is an ideal of $H$ since the restriction of
$f$ on $F_{2}$ is injective, the canonical map $H/f^{-1}(P)\rightarrow
H^{\prime}/P$ is injective and a subgroup of a free subgroup is free.

4\. The geometry of $G$-groups.

In the sequel we are going to consider only the topology mentioned at
proposition 3.3.

Proposition 4.1.

Let $H$ be an element of $C(H)$, for every normal subgroups $I$ and $J$ of
$H$, we have $V([I,J])=V(I\cap J)$.

Proof.

We know that $V([I,J])=V(I)\bigcup V(J)$. Since $[I,J]\subset I\cap J$, we
have $V(I\cap J)\subset V([I,J])$. Let $P$ be an element of $V([I,J])$,
suppose that $P$ is not an element of $V(I\cap J)$, this implies that there
exists an element $x$ of $I\cap J$ which is not an element of $P$. For every
element $g\in G$, $[x,x^{g}]\in[I,J]$ since $x\in I\cap J$. Since $P$ is a
$G$-domain, we deduce that $x$ is an element of $P$. This is a contradiction,
thus $P$ contains $I\cap J$.

Proposition 4.2.

Suppose that $G$ is a solvable group distinct of $\\{1\\}$, for every object
$H$ of $C(G)$, $Spec_{G}(H)$ is empty.

Proof.

Let $P$ be a prime ideal of an object $H$ of $V(I)$. Since $G$ is a solvable
group distinct of $1$, the proof of the proposition 2.2 shows that there
exists an element $x\in G$ which is a divisor of zero. This implies that $x\in
P$. This is impossible since $P\cap G=\\{1\\}$.

Remark.

Let $G$ be the trivial subgroup $\\{1\\}$, any group $H$ is an element of
$C(G)$. Let $P$ be a prime ideal of $H$, for any element $x\in H$, $x\in P$
since $[x,x]=1$. This implies that $Spec_{G}(H)$ is empty since a prime of $H$
is distinct of $H$.

Let $H$ be an element of $C(G)$, the proposition 3.3 shows that the subsets
$V(I)$ where $I$ is a normal subgroup of $H$ are the closed subsets of a
topology on the set $Spec_{G}(H)$ of prime ideals of $H$. The empty closed
subset is $V(H)$, and the $Spec_{G}(H)=V(\\{1\\})$.

We will say that $Spec_{G}(H)$ is an affine scheme.

The structural sheaf.

Let $(H,\phi_{H})$ be an element of $C(G)$, and $U$ an open subset of
$Spec_{G}(H)$, we denote by $Rad(U)$ the intersection of the elements of $U$.
We define on $Spec_{G}(U)$ the structural presheaf:

$U\longrightarrow P_{H}(U)=H/Rad(U)$

Let $U$ and $V$ be two open subsets of $Spec_{G}(H)$ such that $U\subset V$;
$Rad(V)\subset Rad(U)$, this implies the existence of a map
$r_{U,V}:P_{H}(V)=H/Rad(V)\rightarrow P_{H}(U)=H/Rad(U)$ which is the
restriction map of the presheaf $P_{H}$.

The structural sheaf of $O_{H}$ of $Spec_{G}(H)$ is the sheaf associated to
the presheaf $P_{H}$.

Recall that if $x$ is an element of $Spec_{G}(H)$ and $U(x)$ the set of open
subsets of $Spec_{G}(H)$ which contains $x$, the stalk ${P_{H}}_{x}$ of $x$ is
the inductive limit ${P_{H}}_{x}=lim_{V\in U(x)}P_{H}(V)$. The etale space
$E_{P_{H}}$ of $P_{H}$ is the topological space which is the disjoint union
$\\{{P_{H}}_{x},x\in Spec_{G}(H)\\}$ endowed with the topology generated by
$\\{(s_{x})_{x\in U},s\in P_{H}(U)\\}$ where $U$ is any open subset of
$Spec_{G}(H)$. The sheaf $O_{H}$ is a the sheaf of continuous sections of the
canonical morphism $p:E_{P_{H}}\rightarrow Spec_{G}(H)$ which sends an element
of ${P_{H}}_{x}$ to $x$.

Definition 4.1.

Let $(X,O_{X})$ and $(Y,O_{Y})$ be two topological spaces $X$ and $Y$
respectively endowed with the sheaves $O_{X}$ and $O_{Y}$. A morphism
$(f,f^{\prime}):(X,O_{X})\rightarrow(Y,O_{Y})$ is defined by a continuous map
$f:X\rightarrow Y$ and a morphism of sheaves $f^{\prime}:O_{Y}\rightarrow
f^{-1}O_{X}$ which commutes with the restriction maps.

Proposition 4.3.

Let $(H,\phi_{H})$ and $(L,\phi_{L})$ elements of $C(G)$ with trivial
radicals, there exists a natural bijection between morphisms:
$(f,f^{\prime}):(Spec_{G}(H),O_{H})\rightarrow(Spec_{G}(L),O_{L})$ and
$Hom_{G}(L,H)$.

Proof.

Let $(f,f^{\prime}):(Spec_{G}(H),O_{H})\rightarrow Spec_{G}(L),O_{L})$ be a
morphism. We associate to $(f,f^{\prime})$ the $G$-morphism
$f^{\prime}(Spec_{G}(L)):O_{L}(Spec_{G}(L))=L\rightarrow
O_{H}(Spec_{G}(H))=H$.

Let $u:L\rightarrow H$ be a $G$-morphism of $C(G)$, then $u$ induces a
continuous map
$(f_{u},f^{\prime}_{u}):(Spec_{G}(H),\phi_{H})\rightarrow(Spec_{G}(L),\phi_{L})$
defined as follows:

Let $P$ be an element of $Spec_{G}(H)$, then $f_{u}(P)=u^{-1}(P)$. The
proposition 3.3 shows that $f_{u}$ is well defined. Let $V$ be an open subset
of $Spec_{G}(L)$, the composition by $u$ induces a morphism
$f^{\prime}_{u}(V):O_{L}(V)\rightarrow f^{-1}(O_{H})(V)$.

The correspondences defined above between $Hom_{G}(L,H)$ and the set of
morphisms between $(Spec(H),O_{H})$ and $(Spec(L),O_{L})$ are inverse each
others.

Definition 4.2.

A topological manifold $X$ is an $G$-scheme if and only if there exists a
covering $(U_{i})_{i\in I}$ of $X$ and a sheaf $O_{X}$ on $X$, such that for
every $i\in I$, there exists an affine $G$-scheme
$(Spec_{G}(H_{i}),O_{H_{i}})$ and an homeomorphism of sheafed spaces
$f_{i}:(Spec_{G}(H_{i}),O_{H_{i}})\rightarrow(U_{i},O_{U_{i}})$, where
$O_{U_{i}}$ is the restriction of $O_{X}$ to $U_{i}$.

Proposition 4.4.

Let $(H,\phi_{H})$ be an element of $C(H)$, and $Nil(H)$ the normal subgroup
of $H$ generated by its nilpotent elements. The projection morphism
$p^{\prime}:H\rightarrow H/Nil(H)$ induces an homeomorphism between
$p:Spec_{G}(H/Nil(H))\rightarrow Spec_{G}(H)$.

Proof.

Let $P$ be a prime ideal of $H/Nil(H)$, $p^{\prime}({p^{\prime}}^{-1})(P)=P$.
This implies that $p$ is injective. Let $Q$ be a prime ideal of $H$,
${p^{\prime}}^{-1}(p^{\prime}(Q))=Q$ since $Q$ contains $Nil(H)$. This implies
that $p$ is surjective thus bijective. Let $I$ be an ideal of $H/Nil(H)$,
$p(V(I))=V({p^{\prime}}^{-1}(I))$, this implies that $p$ is open. We conclude
that $p$ is an homeomorphism.

Definition 4.3.

Let $G$ and $L$ be two elements of $C(G)$ with trivial radical. A morphism
$(f,f^{\prime}):(Spec_{G}(H),O_{H})\rightarrow(Spec_{G}(L),O_{L})$ is of
finite type if and only if the $L$-group $H$ is the quotient of a free group
$L[x_{1},...,x_{n}]$ by an ideal generated by a finite subset for the
$L$-structure of $H$ induced by $(f,f^{\prime})$.

The smallest $n$ such that $H$ is the quotient of $L[x_{1},...,x_{n}]$ by an
ideal of $L[x_{1},...,x_{n}]$ generated by a finite subset is called the
$L$-rank of $G$.

Suppose that $G$ is a domain and the radical of $H$ is trivial, we say that
$(Spec_{G}(H),)O_{H})$ is of finite type if and only if the canonical morphism
$(Spec_{G}(H),O_{H})\rightarrow(Spec_{G}(G),O_{G})$ is a morphism of finite
type.

We endow now the category of affine schemes with a product:

Proposition 4.5.

Let $H$ and $K$ be two elements of $C(G)$, the affine scheme
$Spec_{G}(H*_{G}K)$ is the categorical product of $Spec_{G}(H)$ and
$Spec_{G}(K)$.

Proof.

A couple of morphisms $f:Spec_{G}(L)\rightarrow Spec_{G}(H)$ and
$g:Spec_{G}(L)\rightarrow Spec_{G}(K)$ is defined by a couple of morphisms of
$G$-groups $f^{\prime}:H\rightarrow L$ and $g^{\prime}:K\rightarrow L$ which
induces a unique morphism $(f^{\prime},g^{\prime}):H*_{G}K\rightarrow L$ such
that $f^{\prime}=(f^{\prime},g^{\prime})\circ i_{H}$ and
$g^{\prime}=(f^{\prime},g^{\prime})\circ i_{K}$ where $i_{H}:H\rightarrow
H*_{G}K$ is the map defined by $i_{H}(h)=h*_{G}1$. This is equivalent to
saying that $Spec_{G}(H*_{G}K)$ is the product of $Spec_{G}(H)$ and
$Spec_{G}(K)$.

Proposition 4.6.

The category of affine $G$-schemes has sums.

Proof.

Let $H$ and $K$ be two elements of $C(G)$, we can endow the product of these
groups $H\times K$ with the diagonal action of $G$; Let
$f:Spec_{G}(H)\rightarrow Spec_{G}(L)$ and $g:Spec_{G}(K)\rightarrow
Spec_{G}(L)$ be morphisms of $G$-schemes. These morphisms defined respectively
by $G$-morphisms: $f^{\prime}:L\rightarrow H$ and $g^{\prime}:L\rightarrow K$
which induce a $G$-morphism $(f^{\prime},g^{\prime}):L\rightarrow H\times K$,
which defines the sum $(f,g):Spec_{G}(H\times K)\rightarrow Spec_{G}(L)$.

Definitions 4.4.

Let $H$ be an element of $C(G)$, we define a functor $c_{H}:C(G)\rightarrow
C(H)$ which associates to $L$ the amalgamated sum $H*_{G}L$. We say that the
$H$-scheme $Spec_{H}(H*_{G}L)$ is obtained from $L$ by a change of the basis.
In particular if $H$ is an algebraic closure $G_{al}$ of the group $G$,
$Spec_{G_{al}}(G_{al}*_{G}L)$ will be called the geometric scheme associated
to $H$.

Definitions 4.5.

Recall that an irreducible topological set is a set which cannot be the union
of two proper distinct closed subsets.

Let $(H,\phi_{H})$ be an element of $C(G)$. The normal subgroup $I$ of $H$ is
a radical, if $I$ is the intersection of all the primes ideals which contains
$I$.

Proposition 4.7,

Let $(H,\phi_{H})$ be an element of $C(H)$, if $I$ a prime ideal of $H$, then
$V(I)$ is irreducible. Conversely, suppose that $I$ is a radical ideal, then
if $V(I)$ is irreducible $I$ is a prime ideal.

Proof.

Suppose that $I$ is a prime ideal and $V(I)$ is the union of the proper closed
subsets $V(J)$ and $V(K)$. Since $I$ is a prime ideal, $I$ is an element of
$V(I)$; thus $I$ is an element of $V(J)$ or $I$ is an element of $V(K)$. If
$I$ belongs to $V(J)$; this implies that $I$ contains $J$, thus $V(I)\subset
V(J)$. This is a contradiction with the fact that $V(J)$ is a proper closed
subset of $V(I)$. The same argument shows that $I$ cannot belong to $V(K)$.

Suppose now that $V(I)$ is an irreducible closed subset and $I$ is a radical
ideal. Let $x$ and $y$ be elements of $H$ such that $[G(x),G(y)]\subset I$.
For every element $P\in V(I)$, $[G(x),G(y)]\subset P$, this implies that $x\in
P$ or $y\in P$. Let $V_{x}$ the subset of elements of $V(I)$ which contains
$x$ and $V_{y}$ the subset of elements of $V(I)$ which contains $y$, $V(I)$ is
the union of $V(x)$ and $V(y)$. Let $N(x)$ be the minimal subgroup of $H$
which contains $x$, if $P$ is an ideal which contains $x$, $N(x)\subset P$.
This implies that $V(x)=V(I)\cap V(N(x))$ and $V(y)=V(I)\cap V(N(y))$. Since
$V(I)$ is irreducible, we deduce that $V(I)\cap V(N(x))=V(I)$ or $V(I)\cap
V(N(y))=V(I)$. Suppose that $V(I)\cap N(x)=V(I)$, this implies that
$V(I)\subset V(N(x))$ and for every prime ideal $P\in V(I)$, $x\in P$, since
$I$ is a radical ideal, $x\in I$. By the same argument, we deduce that if
$V(I)\cap V(N(y))=V(I)$, $y\in I$.

Corollary 4.1.

Suppose that $G$ does not have divisors of zero, $Spec_{G}(H)$ is irreducible
if and only if the radical of $H$ is a prime ideal.

Proof.

Recall that the radical $Rad_{G}(H)$ is the intersection of all the prime
ideals of $Spec_{G}(H)$. Thus by definition, $Rad_{G}(H)$ is a radical ideal,
we can apply proposition 4.7.

Remark.

If $Rad_{G}(H)$ is a prime, it is a generic point of $Spec_{G}(H)$, that is,
its adherence is $Spec_{G}(H)$.

5\. Functor of points.

Let $H$ be an element of $C(G)$. Consider the $n$-uple $[h]=(h_{1},...,h_{n})$
of $H^{n}$; it induces a $G$-morphism:

$u_{[h]}:G[X_{1},...,X_{n}]\rightarrow H$

defined by $u_{[h]}(X_{i})=h_{i}$, for every element $f\in
G[X_{1},...,X_{n}]$, we set $f_{H}([h])=u_{[h]}(f)$, we have thus defined a
morphism $f_{H}:H^{n}\rightarrow H$.

Let $S$ be a subset of $G[X_{1},...,X_{n}]$ in their paper [2], Baumslag and
al defined the subset:

$V_{H}(S)=\\{[h]=(h_{1},...,h_{n})\in H^{n}:\forall f\in S,f_{H}([h])=1.\\}$

We are going to present this notion with the concept of functor of points.

Definition 5.1.

We denote by $Spec_{G}^{0}$ opposite to the category $Spec_{G}$ of affine
$G$-schemes. For every object $H$ of $C(G)$, the functor of points
$h_{Spec_{G}(H)}:Spec_{G}^{0}\rightarrow Sets$ is the functor defined by:
$h_{Spec_{G}(H)}(Spec_{G}(K))=mor_{G}(Spec_{G}(K),Spec_{G}(H))=Mor_{G}(H,K)$.

In particular, if $H$ is a scheme of finite type, $H$ is the quotient of
$G[X_{1},...,X_{n}]$ by an ideal $I$, an element $f\in Mor_{G}(H,K)$ is
defined by a $n$-uple $(k_{1},...,k_{n})\in K^{n}$ such that for every $f\in
I$, $f(k_{1},...,k_{n})=1$. Thus the elements of $Mor_{G}(H,K)$ in this
situation correspond to the elements of the closed subset $V_{G}(K)$ defined
in [2], where $S$ is replaced by $I$.

6.1 Noetherian $G$-schemes.

The elements of $G[X_{1},...,X_{n}]$ are often called equations in
$n$-variables over $G$. Every set of equations $E$ generates a normal subgroup
$I_{E}$ of $G[X_{1},...,X_{n}]$.

Definition 6.1.

We say that the group $G$ is Noetherian if and only if for every set of
equations $E$, there exists a finite subset $E_{0}$ of $E$ such that
$V_{G}(I_{E})=V_{G}(I_{E_{0}})$.

The problem of the existence of a solution of a finite set of equations
defined on a group $G$ has been studied by many authors. This situation is
analog the study of the roots of polynomial equations in field theory. In
fact, Scott [7] has defined a notion of algebraically closed group that we
recall.

Definition 6.2.

A set $E$ of equations and inequations defined on a group $G$ is consistent,
if there exists an embedding $G\rightarrow H$ such that $E$ has a solution in
$H$.

An example of a set of equations and inequations which is not consistent is:
$x^{2}=1,x^{3}=1,x\neq 1$.

Definitions 6.3.

A group is algebraically closed if and only if every finite set of equations
consistent defined on $G$ has a solution in $G$.

Scott has shown that every group can be imbedded in an algebraically closed
group. We show now the following proposition (See also Baumslag and al [2] p.
62)

Proposition 6.1.

Let $G$ be an algebraically closed group, then for every finitely generated
maximal $M$ ideal of $G[X_{1},...,X_{n}]$, there exists elements
$a_{1},...,a_{n}$ in $G$ such that $M$ is generated by
$\\{X_{1}a_{1}^{-1},...,X_{n}a_{n}^{-1}\\}$.

Proof.

Let $H$ be the quotient $H=G[X_{1},...,X_{n}]/M$, and $x_{1},...,x_{n}$, the
image of $X_{1},...,X_{n}$ by the natural projection
$G[X_{1},...,X_{n}]\rightarrow H$. Then $(x_{1},...x_{n})$ is a solution of
every equation defined by elements of $M$. Since $G$ is algebraically closed
and $M$ is finitely generated, there exists an element $(a_{1},...,a_{n})$ in
$G^{n}$ which is a solution of the equations defined by elements of $M$. This
implies that the ideal generated by $X_{1}a_{1}^{-1},...,X_{n}a_{n}^{-1}$
contains $M$ henceforth this ideal is equal to $M$ since $M$ is maximal.

Definition 6.4.

A topological set is Noetherian if and only if every descending chain of
closed subspaces $Y_{1}\supseteq Y_{2}....\supseteq Y_{i}....$ stabilizes.
This means that there exists $n$ such that $Y_{i}=Y_{n}$ for every $n\geq i$.

Let $H$ be an object of $C(G)$, $H$ is a Noetherian group if and only if every
family of ideals $I_{1}\subset I_{2}\subset...\subset I_{n}...$ stabilizes,
this is equivalent to saying that there exists an $i$ such that for every
$n\geq i$, $I_{n}=I_{i}$.

Remark that $Spec_{G}(H)$ is Noetherian if and only if $H$ is a Noetherian
group.

Proposition 6.2.

The object $H$ of $C(G)$ is Noetherian if and only if every ideal of $H$ is
finitely generated.

Proof.

Suppose that $H$ is Noetherian, let $I$ be an ideal of $H$, consider the set
$A$ whose objects are finitely generated ideals contained in $I$ ordered by
the inclusion. Every totally ordered family $(I_{l})_{l\in L}$ of $A$ has a
maximal element. If not we can construct a subfamily $(I_{n})_{n\in N}$ of
$(I_{l})_{l\in L}$ such that $I_{n}$ is strictly contained in $I_{n+1}$. This
is in contradiction with the fact that $H$ is Noetherian. Let $M$ be a maximal
element of $A$, $M=I$, if not there exists an element $x$ of $I$ which is not
in $M$ and the ideal generated by $M$ and $x$ contains strictly $M$.

Suppose that every ideal of $H$ is finitely generated. Let $I_{0}\subset
I_{2}...\subset I_{n}...$ be an ascending sequence of ideal, $\bigcup_{n\in
N}I_{n}=I$ is a finitely generated ideal by $x_{1},...,x_{p}$. The element
$x_{i}$ belongs to $I_{m(i)}$. Let $m$ be the maximum of $m(i)$, $I=I_{m}$.

Proposition 6.3.

Suppose that $Spec_{G}(H)$ is a Noetherian topological space, then the radical
of $H$ is the intersection of a finite number of primes ideals. In particular
if the intersection of all the elements of $Spec_{G}(H)$ is the neutral
element, then there exists a finite number of elements of $Spec_{G}(H)$ whose
intersection is the neutral element.

Proof.

We know that a Noetherian topological space is the union of a finite number of
irreducible closed subsets. This implies that there exists a finite number of
primes ideals $P_{1},...,P_{n}$ such that
$Spec_{G}(H)=\bigcup_{i=1,...,n}V(P_{i})$. The proposition 4.1 implies that
$Spec_{G}(H)=V(\bigcap_{i=1,...,n}P_{i})$. This implies that every prime ideal
$P$ of $H$ contains $\bigcap_{i=1,...,n}P_{i}$, thus
$\bigcap_{i=1,...,n}\subset Rad_{G}(H)$. It results that
$Rad_{G}(H)=\bigcap_{i=1,...,n}P_{i}$ since $Rad_{G}(H)$ is the intersection
of all the elements of $Spec_{G}(H)$.

7\. Dimension of affine schemes.

We want to introduce the notion of dimension of an affine schemes. In
commutative algebra, the dimension of a ring can be defined to be its height
that is, the maximal length of chains $0\subset P_{1}...\subset P_{n}$ of
prime ideals such that $P_{i}$ is distinct of $P_{i+1}$. We can adopt the same
definition here for the prime ideals of elements of $H$ for every object $H$
of $C(G)$. But the following proposition shows that the height of $G*Z$ can be
infinite.

Proposition 7.1.

Let $G$ be a group which does not have divisors of zero, then the height of
$G*Z$ is infinite.

Proof.

Let $I_{n}$ be the ideal of $G*Z$ generated by $2^{n}$,
$(G*Z)/I_{n}=G*(Z/2^{n})$. Suppose that $G*Z/2^{n}$ have divisors of zero
$x,y$. For every $g\in G$, we have $[x,y^{g}]=1$, $x$ and $y$ cannot be in a
subgroup conjugated to $G*1$ since $G$ does not have zero divisors; $x,y$
cannot be in a subgroup conjugated to $1*Z$ in this case, $[x,y^{g}]$ is not
trivial for every non trivial element of $g\in G$. If $x$ is not in a
conjugated of $G*1$ and $1*Z$, then [5] page 187 shows that for every $g\in
G$, $y^{g}$ and $x$ are the power of the same element; this is impossible. We
conclude that $I_{n}$ is a prime, thus the sequence $I_{n}\subset
I_{n-1}...\subset I_{0}$ is an infinite sequence of distinct prime ideals.

In classical algebraic geometry, the dimension of a scheme is also the the
transcendence dimension of its field of fraction. The corresponding notion
here is the $G$-rank:

Definition 7.1.

Let $H$ be an object of $C(G)$, a family of $G$-generators of $H$ is a family
of elements $(x_{i})_{i\in I}$ of $H$ such that $G$ and $(x_{i})_{i\in I}$
generate $H$. The $G$-rank of $H$ is the minimal cardinal of the set $I$ such
that there exists a family of generators $(x_{i})_{i\in I}$ of $H$.

We have the following result:

Proposition 7.2.

Let $G$ be a simple group, and $G^{n}$ the product of $n$ copies of $G$
endowed with the $G$-structure defined by the embedding of $G$ as the first
factor, $(G,1,...,1)$ then $Spec_{G}(H)$ contains only one element the prime
$(1,G,...,G)$ and the $rank_{G}(G^{n})=n-1$.

Proof.

Let $I$ be an ideal of $G^{n}$, suppose that $I$ contains an element
$u=(g_{1},...,g_{n})$ such that $g_{1}$ is different of the neutral element of
$G$. For every $g\in G$, let $i_{g}=(g,1,...,1)$, we have:
$i_{g}ui_{g}^{-1}=(gg_{1}g^{-1},g_{2},...,g_{n})\in I$. Since $G$ is simple,
there exists $g\in G$ such that $gg_{1}g^{-1}\neq g_{1}$. This implies that
$u^{-1}i_{g}ui_{g}^{-1}$ is a non trivial element of $G$. This is a
contradiction since $I$ is an ideal, thus $I$ is contained in $\\{1\\}\times
G^{n-1}$. If $I$ is distinct of $\\{1\\}\times G^{n-1}$, there exists an
element $v\in\\{1\\}\times G^{n-1}$ who does not belong to $I$. Let
$p:G^{n}\rightarrow G^{n}/I$ be the natural projection, for every element
$g\in G$, we have $[g,v]=1$ thus $[p(g),p(v)]=1$. This implies that $G^{n}/I$
is not a domain, thus $I$ is not a prime.

Remark.

Suppose that the center of $G$ is not trivial, then for every element of
$C(G)$, $Spec_{G}(H)$ is empty and the $G$-rank of $H$ needs not to be $0$ or
$-\infty$.

Definition 7.2.

Let $H$ be an object of $C(G)$:

\- if $Spec_{G}(H)$ is empty, the dimension of $Spec_{G}(H)$ is $-\infty$.

if $Spec_{G}(H)$ is not empty, the dimension of $Spec_{G}(H)$ is the $G$-rank
of $H/Rad_{G}(H)$ the quotient of $H$ by the intersection $Rad_{G}(H)$ of all
the elements of $Spec_{G}(H)$.

Remarks.

If $H$ is a $G$-domain, then $Rad_{G}(H)=\\{1\\}$ and the dimension of
$Spec_{G}(H)$ is $rank_{G}(H)$.

Let $G$ be a group without zero divisors, and $H=G[x_{1},...,x_{n}]$, $H$ is a
$G$-domain and the dimension of $H$ is $n$.

If $H$ is a $G$-domain, then dimension of $Spec_{G}(H)=0$ is equivalent to
saying that $H$ is isomorphic to $G$.

Let $H=G^{n}$ endowed with the $G$-structure defined by the imbedding
$(G,1,...,1)$ of $G$ in $H$, then $Rad_{G}(H)=(1,G,...,G)$ and
$H/Rad_{G}(H)\simeq G$; this implies that the dimension of $Spec_{G}(H)=0$.

We know that a if $H$ is $G$-simple, then $Spec_{G}(H)$ contains only one
element, but the $G$-rank of $H$ is not necessarily $1$. But if $G$ is
algebraically closed, we have the following result:

Proposition.

Suppose that $G$ is algebraically closed and $H$ a $G$-domain of finite type
which is $G$-simple, then $H=G$, thus $dim_{G}(H)=0$.

Proof.

There exists a positive integer $n$ and an ideal $I$ of $G[x_{1},...,x_{n}]$
such that $H=G[x_{1},...,x_{n}]/I$. Since $G$ is algebraically closed, there
exists a maximal ideal $M$ (defining by a solution of the equations defined by
a finite set of generators of $I$ containing $I$) such that
$G[x_{1},...,x_{n}]/M=G$. Let $M^{\prime}$ be the image of $M$ by the
projection map $G[x_{1},...,x_{n}]\rightarrow H$, $M^{\prime}$ is an ideal of
$H$, since $H$ is $G$-simple, we conclude that $M^{\prime}=\\{1\\}$ thus
$M=I$. We deduce that $H=G$ since $H=G[x_{1},...,x_{n}]/M$.

Let $H$ be a group, the co-rank of $n$ is the maximal integer $n$ such that
there exists a surjective morphism: $H\rightarrow F[x_{1},...,x_{n}]$. We say
that the $G$-co-rank of the object $n$ in $C(G)$ is $n$ if $n$ is the maximal
integer such that there exists a $G$-morphism: $H\rightarrow
G[x_{1},...,x_{n}]$.

Contrary to the rank, a class of ideals allows to express geometrically the
idea of the co-rank,

Definition.

Let $H$ be an element of $C(G)$, a $G$-free ideal $I$ of $H$ is an ideal such
that $H/I=G[x_{1},...,x_{n}]$.

The co-dimension of $Spec_{G}(H)$ is the maximum integer $n$ such that there
exists an ideal $I$ such that $H/I=G[x_{1},...,x_{n}]$.

Proposition 7.4.

Let $H$ be an element of $C(G)$, the co-dimension of $Spec_{G}(H)$ is the
length of the maximal sequence $P_{0}\subset P_{2}\subset...\subset P_{n}$
such that for $i=1,...,n$ $P_{i}$ is $G$-free and $P_{i+1}$ contains strictly
$P_{i}$.

Proof.

Suppose that the co-dimension of $Spec_{G}(H)$ is $n$, there exists a
surjective $G$-morphism $f:H\rightarrow G[x_{1},...,x_{n}]$. We denote by
$P_{0}$ the ideal$f^{-1}(1)$ of $H$. Let $I_{l}$ be the normal subgroup of
$G[x_{1},...,x_{n}]$ generated by $x_{1},...,x_{l}$, and for $i=1,...,n$,
$P_{i}=f^{-1}(I_{i})$. The family of ideal $(P_{i})_{i\in\\{0,...,n\\}}$ is a
chain of free ideals of length $n$ such that $P_{i}$ is strictly included in
$P_{i+1}$.

Let $P_{0}\subset P_{2}\subset...\subset P_{n}$ be a maximal chain of $G$-free
ideals such that $P_{i+1}$ contains strictly $P_{i}$. Remark that the $G$-co-
rank of $H$ is inferior or equal to $n$, since if the co-rank is superior to
$n$ the first stage of the proof shows that we can construct a chain of
$G$-free prime ideals of length strictly greater than $n$ and whose $G$-co-
rank is strictly decreasing. Thus it is enough to show that $G/P_{0}\simeq
G[x_{1},...,x_{n}]$.

Suppose that the $G$-rank of $G/P_{0}$ is strictly inferior to $n$; this
implies that there exists, $i\leq n$ such that co-rank$(P_{i})$ =co-
rank$(P_{i+1})$. Let $l_{i}:H\rightarrow H/P_{i}$ be the natural projection,
$H/P_{i+1}$ is the quotient of $H/P_{i}$ by $l_{i}(P_{i+1})$. Write
$H/P_{i}=G[x_{1},...x_{n_{i}}]$ and $H/P_{i+1}=G[y_{1},...,y_{n_{i}}]$. The
quotient of $G[x_{1},...,x_{n_{i}}]$ by the normal subgroup generated by $G$
is isomorphic to the free group $F[x_{1},...,x_{n_{i}}]$. The quotient of
$G[y_{1},...,y_{n_{i}}]$ by the normal subgroup generated by $G$ is isomorphic
to the quotient of $F[x_{1},...,x_{n_{i}}]$ by the normal subgroup generated
by the image of $l_{i}(P_{i+1})$ in $F[x_{1},...,x_{n_{i}}]$. Since this group
is isomorphic to the free group of rank $n_{i}$ $F[y_{1},...,y_{n_{i}}]$, we
deduce that $l_{i}(P_{i+1})$ is included in the normal subgroup of
$G[x_{1},...,x_{n_{i}}]$ generated by $G$. This in contradiction with the fact
that the quotient of $G[x_{1},...,x_{n_{i+1}}]$ by the image of
$l_{i}(P_{i+1})$ is a free product of $G$ and a free group.

The $G$-free ideals of $H$ generate a topology as shows the following
proposition:

Proposition 7.5.

Let $G$ be a group which does no have divisors of zero, and Let $H$ be an
element of $C(G)$, for every normal subgroup $I$ of $H$, denote
$V_{F}(I)=\\{P:$ $P$ is a $G$-free prime ideal and $I\subset P\\}$, then for
every normal subgroups $I,J$ of $H$, $V_{F}(I)\bigcup V_{F}(J)=V_{F}([I,J])$.
Let $(I_{a})_{a\in A}$ be a family of normal subgroups of $H$. Denote by
$I_{A}$ the normal subgroup generated by $(I_{a})_{a\in A}$, then
$V_{F}(I_{A})=\bigcap_{a\in A}V_{F}(I_{a})$.

Proof.

Firstly we show that $V_{F}(I)\bigcup V_{F}(J)=V_{F}([I,J])$. Let $P$ be an
element of $V_{F}(I)\bigcup V_{F}(J)$, $P$ contains $I$ or $J$ this implies
that $P$ contains $[I,J]$. Let $P$ be an element of $V_{F}([I,J])$. Suppose
that $P$ is neither an element of $V_{F}(I)$ nor an element of $V_{F}(J)$.
This implies that there exists an element $x\in I$ and $y\in J$ which are not
in $P$. For every $g\in G$, $[x,y^{g}]\in P$. Let $p:H\rightarrow H/P$ be the
natural projection, we have $[p(x),p(y)^{g}]=1$. Since $H/P$ is $G$-free, we
deduce that $x$ and $y$ are elements of $G$, this is in contradiction with the
fact that $G$ does not have divisors of zeros.

Now we show that $V(I_{A})=\bigcap_{a\in A}V(I_{a})$. Let $P$ be an element of
$V(I_{A})$, since $I_{a}\subset I_{A}\forall a\in A,$ $P\subset
V(I_{a})\forall a\in A$. Let $P$ be an element of $\bigcap_{a\in A}V(I_{a})$,
$\forall a\in A,I_{a}\subset P$, this implies that $I_{A}\subset P$ and $P\in
V(I_{A})$.

Remark.

Let $G$ be a group which does not have zero divisors, and $H$ an object of
$C(G)$, the previous proposition shows that the family of subspaces $V_{F}(I)$
where $I$ is any normal subgroup in $H$ defines a topology on the space
$Spec_{G}^{F}(H)$ of $G$-free prime ideals of $H$. But if $f:H\rightarrow
H^{\prime}$ is a morphism between the elements $H$ and $H^{\prime}$ of $G$ and
$P$ is an element of $Spec_{G}^{F}(H)$, $f^{-1}(P)$ is not necessarily a
$G$-free prime of $H$. For example let $G=PSL(n,Z)$ and
$H^{\prime}=PSL(n,Z)[X]$. Let $H$ be the subgroup of $H^{\prime}$ generated by
$PSL(n,Z)$ and $XNX^{-1}$ where $N$ is a nilpotent subgroup of $PSL(n,Z)$. The
canonical imbedding $i:H\rightarrow H^{\prime}$ is a $G$-morphism. The normal
trivial subgroup $I=\\{1\\}$ of $H^{\prime}$ is a prime ideal of $H$, but the
quotient of $H$ by $i^{-1}(I)$ is not $G$-free, since $H$ is not $G$-free.

The theorem of Kurosh says that a subgroup of a free product of groups
$H_{1}*H_{2}$ is the free product of a free group and groups isomorphic to
subgroups of $H_{1}$ and $H_{2}$. Thus if we choose $G=F_{n}n\geq 2$, a
$G$-morphism $f:H\rightarrow H^{\prime}$ induces a morphism
$Spec_{G}^{F}(H^{\prime})\rightarrow Spec_{G}^{F}(H)$ of $G$-free spectra. In
fact in this case the topology is defined by a Top couple (see definition
3.1).

Definition 7.4.

Let $G$ be a group without divisors of zero, and $H$ an element of $C(G)$; an
element $P$ of $Spec_{G}(H)$ is a $G$-point if and only if $H/P=G$.

Remark.

If $P$ is a $G$-point, the restriction of the projection $p:H\rightarrow H/P$
to $G$ is an isomorphism, thus this extension splits.

Proposition 7.6.

let $H$ be a be a $G$-domain if $G$ is a normal subgroup of $H$, for every
prime $P$ of $H$, we have of $[G,P]=1$. If there exists a $G$-point $P$ in
$H$, then $H=P\times G$.

Proof.

Let $P$ be a prime ideal. We have $[P,G]\subset P\cap G=\\{1\\}$ since $P$ is
an ideal and $G$ is a normal subgroup of $H$.

If $P$ is a $G$-point of $H$, then $H$ is an extension of $G$ by $P$ which
splits, since $[P,G]=1$ we deduce the result.

8\. Galois theory.

Let $G$ be a group, a polynomial of $G$ is an element of $G[X]$. The Galois
theory in classical field theory enables to study the solutions of polynomial
functions defined on a field. One of the property essential in that study is
the existence of factorization: let $P$ be a polynomial defined on a field,
and $a$ a root of $P$, we can write $P=(X-a)Q$; if $deg(P)$ is the degree of
$P$, we have: $deg(Q)=deg(P)-1$. Here we do not have factorization system, in
fact we can have a polynomial which has an infinite numbers of roots: Let
$G=Gl(2,C)$ be the group of automorphisms of the complex plan; the solutions
of the equation defined on $G$ by $X^{2}=1$ is the set of symmetries of the
complex plan. We can introduce anyway some notions related to Galois theory.

Definitions 8.1.

Let $G$ be a group, and $H$ an element of $C(G)$, an element $a\in H$ is
algebraic over $G$ if $a$ is the root of a polynomial whose coefficients are
in $G$.

An element $H$ of $C(G)$ is an algebraic extension of $G$ if and only if every
element of $H$ is algebraic over $G$.

We say that an algebraic extension of $G$ is finite if and only if the action
of $G$ on $H$ by left multiplications has a finite number of orbits

Proposition 8.1.

Suppose that $G=\\{1\\}$, then a group $H$ is an algebraic extension of $G$ if
and only if every element of $H$ has a finite order. In particular a group is
a finite extension of $\\{1\\}$ if and only if it is a finite group.

Proof.

If $G=\\{1\\}$, then $G[X]=F[X]=Z$ and the polynomials of $\\{1\\}$ are
$X^{n}$. A group $H$ is algebraic over $\\{1\\}$ if and only if for every $h$
in $H$, there exists an element $P$ of $F[X]$ such that $P(h)=1$. This is
equivalent to the existence of an integer $n$ such that $h^{n}=1$.

Suppose that $H$ is a finite group. Let $n_{H}$ be the order of $H$, for every
element $h\in H$, we have $h^{n_{H}}=1$; this implies that $H$ is algebraic
over $\\{1\\}$. The left action of $\\{1\\}$ has finite orbit since $H$ is
finite. Conversely, if the left action of $\\{1\\}$ on $H$ has finite orbits,
then $H$ is a finite group.

Remark.

An algebraic extension $H$ of $\\{1\\}$ is not necessarily finite even if $H$
is finitely generated. In 1904 Burnside asks the question wether every
finitely generated group for which every element has a finite order is finite.
There are groups which have been constructed answering negatively the problem
of Burnside.

Proposition 8.2.

Let $H$ be an element of $C(G)$, suppose that $H$ is the union of a finite
number of orbits for the action of $G$ on $H$ defined by the left
multiplication, then $H$ is algebraic over $G$. In particular if $G$ is a
normal subgroup of $H$ such that $G/H$ is finite, $H$ is algebraic over $G$.

Proof.

Let $x$ be an element of $H$; if $H$ is the union of a finite number of orbits
for the left action of $G$, there exists distinct integers $n$ and $m$ with
$n<m$ such that $x^{n}$ and $x^{m}$ are in the same orbit. This is equivalent
to saying that there exists $h\in H$, $g_{1},g_{2}\in G$ such that
$x^{n}=g_{1}h$ and $x^{m}=g_{2}h$, we deduce that
$h={g_{1}}^{-1}x^{n}={g_{2}}^{-1}x^{m}$. It results that
$g_{1}{g_{2}}^{-1}x^{m-n}=1$. Thus $x$ is algebraic over $G$.

Proposition 8.3.

Let $H$ be an algebraic extension of $G$, then the $G$-co-rank of $H$ is
strictly inferior to $1$.

Proof.

Suppose that the $G$-co-rank of $H$ is superior or equal to $1$. There exists
a $G$-surjection: $s:H\rightarrow G[x_{1},...,x_{n}]$. Let $y_{1},...,y_{n}$ a
family of algebraic elements of $H$ which generate $H$. There exists
$i\in\\{1,...,n\\}$ such that $s(y_{i})$ is not in $G$. Since $y_{i}$ is
algebraic, there exists a $G$-polynomial $P$ such that $P(y_{i})=1$. We also
have $P(s(y_{i}))=1$. This is impossible since $s(y_{i})$ is a non trivial
words written with elements of $G$ and with elements of the free group
generated by $x_{1},...,x_{n}$.

Proposition 8.4.

Let $H$ be an element of $C(G)$ and $K$ an element of $C(H)$, suppose that $H$
is a finite algebraic extension of $G$ and $K$ is a finite algebraic extension
of $H$; then $K$ is a finite algebraic extension of $G$.

Proof.

To show this proposition it is enough to show that the action of $G$ on $K$
has a finite number of orbits by applying proposition 8.2.

Since $H$ is a finite algebraic extension of $G$ and $K$ a finite algebraic
extension of $H$, we know that the orbit spaces $H/G$ and $K/H$ are finite; we
denote by $h_{1},...,h_{m}$ elements of $H$ such that the union of
$Gh_{1},...,Gh_{m}$ is $H$, and by $k_{1},...,k_{n}$ elements of $K$ such that
the union of $Hk_{1},...,Hk_{n}$ is $K$. Let $x$ be an element of $K$, there
exists $1\leq i\leq n$ such that $x\in Hk_{i}$, since $H$ is the union of
$Gh_{1},...,Gh_{m}$, we deduce that $x\in\\{Gh_{1}k_{i},...,Gh_{m}k_{i}\\}$.
This is equivalent to saying that the union of $Gh_{j}k_{i}$ $1\leq j\leq m$,
$1\leq i\leq n$ is $K$.

Definition 8.2.

Let $H$ be an algebraic extension of $G$, the Galois group $Gal_{G}(H)$ of $H$
is the group of automorphisms of $H$ whose restriction to $G$ is the identity.

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