# A bottom-up approach to Hilbert’s Basis Theorem

Marc Maliar

###### Abstract.

In this expositional paper, we discuss commutative algebra—a study inspired by
the properties of integers, rational numbers, and real numbers. In particular,
we investigate rings and ideals, and their various properties. After, we
introduce the polynomial ring and the fundamental relationship between
polynomials and sets of points. We prove some results in algebraic geometry,
notably Hilbert’s Basis Theorem.

###### Contents

  1. 1 Introduction
  2. 2 Rings and ideals
    1. 2.1 Definitions of a ring
    2. 2.2 Definition of an ideal
    3. 2.3 Another definition of an ideal
    4. 2.4 What is in an ideal?
    5. 2.5 Finitely generated ideals
  3. 3 Algebraic varieties and the polynomial ring
    1. 3.1 Introducing plane curves
    2. 3.2 The polynomial ring
    3. 3.3 The relationship between points and polynomials
    4. 3.4 Other properties of algebraic sets
    5. 3.5 Correspondences between points and polynomials

##  1\. Introduction

Mathematics often begins with concrete problems, for example finding the
length of the hypotenuse of a triangle. But as a mathematician investigates
such problems, they begin to abstract away various amounts of details,
uncovering patterns and defining properties.

In this expositional paper, we take this approach to commutative algebra.

In Section $2$, we attempt to find the patterns and defining properties hidden
in algebra. In particular, we investigate sets of numbers such as the
integers, rational numbers and real numbers. We define a ring, the
generalization of the properties of such sets of numbers. We realize that
there exist special sets in the integers called ideals—named in this way
because they behave in an “ideal” way. Finally, we explore properties of these
rings and ideals.

But we realize that polynomials behave as integers do: we can add and multiply
polynomials, identify “prime” polynomials, and find ideals. So we begin
Section $3$ by defining the polynomial ring. Unlike the integers, though,
polynomials have an interesting relationship with sets of points in a field.
And so the rest of section $3$ is devoted to applying commutative algebra to
this relationship between sets of points and polynomials. For example, we
establish that, given a set of polynomials $S$ in two variables (or any number
of variables actually), the set of points $X=(x,y)$ such that $f(x,y)=0$ for
all $f\in S$ can be written as the intersection of finitely many sets of
points $X_{1},\dots,X_{n}$, where $X_{i}=(x,y)$ such that $f_{i}(x,y)=0$ for
some polynomial $f_{i}$. This result is a corollary of Hilbert’s Basis Theorem
(Theorem 3.11).

By the end, we are able to use abstractions to prove nontrivial theorems about
sets of points and polynomials.

##  2\. Rings and ideals

We begin this section by exploring rings. What is a ring? Consider the set of
integers, $\mathbb{Z}$. Recall that this fundamental set comes equipped with
two binary operations (the usual addition and multiplication). We generalize
the properties of this set and arrive at the definition of a ring.

We then consider special subsets of a ring, called ideals. We motivate this
study by again considering $\mathbb{Z}$. In particular, we look at sets of
integers that are divisible by a certain integer, and we note that they
possess interesting properties. We then define ideals by these interesting
properties. In the rest of the section, we discover other interesting
properties of ideals and investigate related definitions.

###  2.1. Definitions of a ring

######  Definition 2.1.

A _ring_ $R$ is a set with two binary operations addition and multiplication,
denoted by $+$ and $*$ respectively. These operations have the following
properties:

  1. (1)

Addition is associative and commutative; there exists an additive identity
(denoted by $0$); each element in the ring has a (unique) additive inverse.

  2. (2)

Multiplication is associative; there exists a multiplicative identity (denoted
by $1$).

  3. (3)

Multiplication is distributive over addition. That is,
$r_{1}*(r_{2}+r_{3})=r_{1}*r_{2}+r_{1}*r_{3}$ and
$(r_{1}+r_{2})*r_{3}=r_{1}*r_{3}+r_{2}*r_{3}$ hold for $r_{1},r_{2},r_{3}\in
R$.

Why such a definition? The ring definition comes from trying to generalize
numbers (integers, rational numbers, and real numbers) and their various
properties. Notice that the aforementioned sets of numbers with their usual
operations satisfy these properties.

We consider commutative rings: rings in which multiplication is commutative.
From now on, we consider only commutative rings; any ring mentioned is
commutative. Notice that the integers, rational numbers, and real numbers are
commutative rings.

While each element in a ring has an additive inverse, it does not necessarily
have a multiplicative inverse. In addition, $1$ can be equal to $0$. In this
case, it is easy to see that the ring has only one element, $0$ (and that this
set of one element is a ring).

Recall the classical definition of a field: a set with addition and
multiplication satisfying the usual properties. Fields turn out to be valuable
objects in algebraic geometry, so I mention them here. Also, in the next
section, we prove some results about commutative rings that are fields, so I
introduce the following (equivalent) definition:

######  Definition 2.2.

A _field_ is a commutative ring $k$ where $1\neq 0$ and every element has a
multiplicative inverse.

###  2.2. Definition of an ideal

Consider the ring of integers; now, consider the set of all elements divisible
by $3$ (a certain subset of this ring). This set has some interesting
properties.

First, if you add any two elements divisible by $3$, you end up with an
integer divisible by $3$. Second, if you multiply an element in the set with
any other integer, you end up with an integer divisible by $3$. Ideals are
defined by these two properties.

######  Definition 2.3.

An _ideal_ $I$ is a subset of a ring $R$ that, inheriting the addition and
multiplication operations from $R$, satisfies the following properties:

  1. (1)

Closure with respect to addition (if $i_{1},i_{2}\in I$ then $i_{1}+i_{2}\in
I$); associativity with respect to addition; additive identity is in the ideal
($0\in I$); each element’s additive inverse is also in the ideal (if $i\in I$,
$-i\in I$).

  2. (2)

If $r\in R$ and $i\in I$, $r*i\in I$.

  3. (3)

If $r\in R$ and $i\in I$, $i*r\in I$.

######  Remark 2.4.

Since we work with commutative rings, it is only necessary to prove either
$(2)$ or $(3)$. In non-commutative rings, an ideal that satisfies $(2)$ but
not necessarily $(3)$ is a left-ideal, and an ideal that satisfies $(3)$ but
not necessarily $(2)$ is a right-ideal.

Note that we do not need to explicitly prove all of these statements. Indeed,
proving closure with respect to addition, existence of additive inverse, and
$(2)$ (or $(3)$ as discussed) is sufficient.

  1. (a)

Associativity is inherited from the ring definition.

  2. (b)

For $i\in I$, $i*(-1)=-i\in I$.

An ideal cannot be empty because it must contain $0$.

Let us find other ideals of the rings we have mentioned (integers, rational
numbers, and real numbers). To begin, note that, for any ring, both $\\{0\\}$
and $R$ are ideals.

But are these the only ideals? Consider first the integers.

######  Proposition 2.5.

For all $z\in\mathbb{Z}$, define $I_{z}=\\{z*r\mid r\in\mathbb{Z}\\}$. $I_{z}$
is an ideal. These are the only ideals of $\mathbb{Z}$.

###### Proof.

The proof that $I_{z}$ is an ideal follows from my previous remark. So then
suppose there exists an ideal $I$ that cannot be written as an $I_{z}$.
$I\neq\\{0\\}$ because then $I=I_{0}$, so there exists at least one nonzero
integer in $I$. Consider the set of nonzero positive integers in $I$. This set
must be nonempty; an empty set would imply that the nonzero integer $z$ we
found earlier is negative, but $z*(-1)\in I$—a contradiction. By the Well-
Ordering Principle, find the smallest positive nonzero integer $z\in I$.

Because $I\neq I_{z}$, there must exist an element $z^{\prime}\in I$ that is
not in the form $z*r$ for any $r\in\mathbb{Z}$. Suppose it is positive (if it
is negative multiply it by $-1$ and it will still be in the ideal). We know
that $z^{\prime}>z$ because $z$ is the smallest positive integer. Now, observe
that $z^{\prime}+(-z)$ is still in the ideal and also will not be in the form
$z*r$. Repeat until the result is less than $z$. The result will be in $I$,
positive and nonzero (because it is not in the form $z*r$). Thus, we have
found a smaller positive nonzero integer in $I$ that is smaller than $z$—a
contradiction. ∎

In other words, the only ideals of the ring of integers are multiples of an
integer. Now, consider rational numbers and real numbers. It turns out that we
can make a statement about fields in general:

######  Proposition 2.6.

The only ideals of a field $k$ are the field itself and the zero ideal. The
converse is also true: if a commutative ring has exactly two ideals, the zero
ideal and the entire ring, then the commutative ring is a field.

###### Proof.

First, we prove that that the only ideals of $k$ are the field itself and the
zero ideal. Let $I$ be an arbitrary ideal.

  1. Case 1:

$I$ contains no non-zero elements. Then, $I=\\{0\\}$.

  2. Case 2:

$I$ contains a non-zero element. Let $i$ be this element. Then, because $k$ is
a field, there exists a multiplicative inverse of $i$, denoted by $i^{-1}$. By
the property of ideals, $i*i^{-1}=1\in I$. Any element $j\in k$ is in the
ideal now because $1*j=j\in I$, so then $I=k$.

Second, we prove that if the only ideals of a ring $R$ are the ring and the
zero ideal, the ring is a field. Let $r$ be a non-zero element in our
commutative ring $R$. Construct a set $I=\\{rq\mid q\in R\\}$. It is easy to
check that this is an ideal. Because this ideal contains a nonzero element
$r$, it cannot be the zero ideal, so by our given, the ideal must be the
entire ring. Then, there exists an element $r^{\prime}$ such that
$r(r^{\prime})=1$ since $1$ is in our ideal. Thus, $r$ has an inverse. ∎

From this proof, we can also see the following statement:

######  Corollary 2.7.

Let $I$ be an ideal of $R$. Then, $1\in I\iff I=R$.

###  2.3. Another definition of an ideal

Consider the following definition:

######  Definition 2.8.

A _ring homomorphism_ is a map $\phi$ from one ring $R$ to another ring $S$
with the following properties:

  1. (1)

$\phi(r_{1}+r_{2})=\phi(r_{1})+\phi(r_{2})$,

  2. (2)

$\phi(r_{1}*r_{2})=\phi(r_{1})*\phi(r_{2})$, and

  3. (3)

$\phi(1_{R})=1_{S}$, where $1_{R},1_{S}$ are the multiplicative identities of
$R$ and $S$ respectively.

We introduce an equivalent definition of the ideal using ring homomorphisms:

######  Definition 2.9.

An _ideal_ $I$ is a subset of a ring $R$ that is the kernel of a ring
homomorphism. That is, a set $I\subset R$ is an ideal if there exists a ring
homomorphism $\phi\colon R\to S$, where $S$ is some other ring, such that
$\phi^{-1}(0)=I$.

######  Proposition 2.10.

Definition 2.3 is equivalent to Definition 2.9.

###### Proof.

First, we show Definition 2.9 from Definition 2.3. We need to find a ring $S$
and a function $\phi$ as described. We need $\phi$ such that for all $i\in I$,
$\phi(i)=0$. Furthermore, we need $\phi$ such that adding $i\in I$ to any
$r\in R$ will not change the computed result (i.e., $\phi(r)=\phi(r+i)$).

Let $r$ be an element in $R$. Define $\phi\colon r\mapsto X_{r}:=\\{r+i\ :\
i\in I\\}$. Define $R/I=\\{X_{r}\mid r\in R\\}$.

We now define an equivalence relation between elements in $R$. We say that
$r\sim q$ for $r,q\in R$ if and only if $\phi(r)=\phi(q)$. It follows from
properties of an ideal that $\sim$ is an equivalence relation. Notice that,
under this definition, $\phi(i)=\phi(0)\;\forall\;i\in I$.

Define addition and multiplication as follows: for $\phi(r),\phi(q)\in R/I$,
let $\phi(r)+\phi(q)=\phi(r+q)$ and $\phi(r)*\phi(q)=\phi(r*q)$. Both of these
are well-defined. Consider, for example, addition. Let
$r_{1},r_{2},q_{1},q_{2}$ be elements in $R$ with $r_{1}\sim r_{2}$ and
$q_{1}\sim q_{2}$. Let $i,j$ be elements in $I$ such that $r_{1}=r_{2}+i$ and
$q_{1}=q_{2}+j$. Then,

$\phi(r_{1})+\phi(q_{1})=\phi(r_{1}+q_{1})=\phi(r_{2}+q_{2}+i+j)=\phi(r_{2}+q_{2})=\phi(r_{2})+\phi(q_{2}).$

We conclude that equivalent inputs implies equivalent outputs under our
addition operation. Multiplication follows similarly as well.

Under these operations, $R/I$ forms a ring. To finish the proof, define
$S=R/I$ and $\phi$ as defined above. By our definition of $\phi$, we already
know that $\phi(r_{1}+r_{2})=\phi(r_{1})+\phi(r_{2})$ and
$\phi(r_{1}*r_{2})=\phi(r_{1})*\phi(r_{2})$. It suffices to show that
$\phi(1_{R})=1_{S}$ where $1_{S}$ is the multiplicative identity of $R/I$. The
proof follows from the definition.

Second, we show Definition 2.3 from Definition 2.9. As discussed, we must
prove only three statements:

  1. (a)

Closure with respect to addition: let $i_{1},i_{2}\in I$. We know
$\phi(i_{1}+i_{2})=\phi(i_{1})+\phi(i_{2})=0+0=0$. Then, $i_{1}+i_{2}\in I$
because $\phi(i_{1}+i_{2})=0$.

  2. (b)

Existence of additive inverse: $\phi(0)=0$ so $0\in I$.

  3. (c)

If $i\in I$ and $r\in R$ then $i*r\in I$: let $i\in I,r\in R$.
$\phi(i*r)=\phi(i)*\phi(r)=0*\phi(r)=0$. Then, $i*r\in I$ because
$\phi(i*r)=0$.

∎

Consider the ideals of $\mathbb{Z}$ we have found. What is the ring $S$ and
ring homomorphism $\phi$ that we should be able to find, according to our new
definition of an ideal?

  1. (1)

$I_{z}=\\{nz\mid z\in\mathbb{Z}\\}$: we use $S=\mathbb{Z}/n\mathbb{Z}$ ($S$ is
the set of integers modulo $n$—a ring) and $\phi$ where $\phi(r)=r\mod n$. The
set of elements that map to $0$ is $\\{nz\mid z\in\mathbb{Z}\\}$.

  2. (2)

$I=\\{0\\}$: we use $S=R$ and $\phi$ where $\phi(r)=r$. The only element that
maps to $0$ is $0$.

  3. (3)

$I=\mathbb{Z}=R$: we use $S=\\{0\\}$ and $\phi$ where $\phi(r)=0$. All
elements map to $0$.

###  2.4. What is in an ideal?

Consider an ideal $I$ of ring $R$. How can we imagine what is in this ideal?

Let us try to come up with an expression that can represent any element in the
ideal. Let $I$ be an ideal and let $\\{i_{\mu}\\}$ be a subset of $I$ indexed
by $\mu$. Consider the expression $\sum_{\mu}i_{\mu}*r_{\mu}$ where $i_{\mu}$
is an element of the ideal and $r_{\mu}$ is an element in $R$. Can we write
any element in $I$ in this form for some $r_{\mu}$’s? If so, we denote the
ideal $I$ by

$I=(\\{i_{\mu}\\})$

and say that it is generated by $\\{i_{\mu}\\}$. For example, $I=R=(1)$ (as we
saw in Corollary 2.7). So, now we have come up with a way to imagine elements
in our ideal.

Some things to emphasize: in rings that extend outwards infinitely (like the
integers and the ring of polynomials introduced later), non-zero ideals grow
large up to arbitrary finite size. For example, with the ideal of integers
divisible by $3$, we can keep multiplying and finding larger and larger
elements.

In addition, note that just because we can write an element $i\in I$ as a sum
of $i_{\mu}*r_{\mu}$’s doesn’t mean that for any two elements $a,b\in R$ such
that $a*b=i$, either $a$ or $b$ in $I$. An ideal in which this is true for
every $i\in I$ is called a prime ideal:

######  Definition 2.11.

A _prime ideal_ $I$ of ring $R$ is an ideal that satisfies the following
properties:

  1. (1)

If $k\in I$ and $i*j=k$, then either $i\in I$ or $j\in I$.

  2. (2)

$I$ is not equal to the ring $R$.

Consider the ideal of integers divisible by $3$ in the ring of integers: this
ideal is a prime ideal because we can write every element as $3$ multiplied by
an integer, and $3$ is in our ideal. However, the ideal of integers divisible
by $6$ does not have this property; for example $6=2*3$ but neither $2$ nor
$3$ are divisible by $6$. The generalization of these statements corresponds
to the following statement:

######  Proposition 2.12.

In $\mathbb{Z}$, ideals generated by a prime number are prime. The zero ideal
is also prime. All other ideals are not prime.

###### Proof.

Let $I$ be an ideal generated by a prime natural number $p$. Suppose $I$ is
not prime. Then, there exists an element $k\in I$ and $i,j\in I$ such that
$i*j=k$ but $i\not\in I$ and $j\not\in I$. This means that $k$ is divisible by
$p$ but neither $i$ nor $j$ are divisible by $p$, which is impossible.

The zero ideal is prime because no integer divides $0$.

We already showed that all ideals of $\mathbb{Z}$ are generated by one
element, and so it is only necessary to look at ideals generated by nonprime
elements. Let $I$ be an ideal generated by a nonprime natural number $n$.
Because $n$ is not prime, $n=\ell*m$ for some $\ell,m$ not divisible by $n$,
so $\ell$ and $m$ are not in $I$; then, $I$ is not prime. The ideal $I$
generated by $-n$ is the same as the one generated by $n$ so the same logic
applies. ∎

We have an alternate definition of prime ideal as well. This definition
relates to Definition 2.11 similarly to how Definition 2.9 relates to
Definition 2.3.

######  Definition 2.13.

A _prime ideal_ $I$ of a ring $R$ is the kernel of a ring homomorphism
$\phi\colon R\rightarrow S$, where $S$ is a ring that is not equal to
$\\{X_{0}\\}$ and has no zero-divisors (no elements $s$ such that for some
$t\in S$, $s*t=0$).

######  Proposition 2.14.

Definition 2.11 is equivalent to Definition 2.13.

###### Proof.

First, we prove that Definition 2.11 is equivalent to Definition 2.13.

Let $I$ be a prime ideal by Definition 2.11. Suppose that Definition 2.13 does
not hold. By Definition 2.9, there must exist a set $S$ and $\phi$ such that
$I$ is the kernel of $\phi$. Then, take $S$ and $\phi$ that Definition 2.13
refers to. Consider two cases:

  1. Case 1:

$S=\\{X_{0}\\}$. $I$ is not equal to $R$. Then, there exists an element $r$ in
$R$ that is not in $I$, and $\phi(r)\neq X_{0}$.

  2. Case 2:

$S$ has a zero-divisor. Let $s$ be a zero-divisor that is not equal to $0$
with a corresponding $t$ that is not equal to $0$ such that $s*t$ is equal to
$0$. By construction, $\phi$ is surjective, so there exist $r,q\in R$ such
that $\phi(r)=s$ and $\phi(q)=t$. Then,

$0=s*t=\phi(r)*\phi(q)=\phi(r*q),$

so $r*q\in I$. However, $r\not\in I$—otherwise $\phi(r)=0$ (and similar
argument for $s$). Then, $I$ is not prime according to Definition 2.11, so we
have arrived at a contradiction.

Second, we prove that Definition 2.13 is equivalent to Definition 2.11.

For sake of contradiction, let $I$ be an ideal that is not prime by Definition
2.11. By Definition 2.9, there exists a set $S$ and $\phi$ such that $I$ is
the kernel of $\phi$. Consider two cases:

  1. Case 1:

$I=R$. Then, every element $i\in I$ maps to $0$, so $S=\\{X_{0}\\}$.

  2. Case 2:

There exist elements $r,q\in R$ such that $r*q$ is in $I$ but neither $r$ nor
$q$ is in $I$. Then,

$0=\phi(r*q)=\phi(r)*\phi(q),$

but neither $\phi(r)$ nor $\phi(q)$ are equal to $0$ because neither $r$ nor
$q$ are in $I$. Then, both $\phi(r)$ and $\phi(q)$ are zero-divisors of $S$ so
we have arrived at a contradiction.

∎

There are many other interesting ways to characterize rings and ideals. See
[4, Chapter 1].

###  2.5. Finitely generated ideals

Let us go back to our equation for elements in an ideal. Can we make some
claim about the sets $i_{\mu}$ that generate $I$?

Note that such a set must exist. If we choose $\\{i_{\mu}\\}=I$ then $I$ is
generated by $\\{i_{\mu}\\}$. Let $i_{\lambda}$ be an element in $I$. Set
$r_{\mu}=1$ for $\mu=\lambda$ and $r_{\mu}=0$ for $\mu\neq\lambda$.

But can we find other sets? Specifically, can we find a finite set
$\\{i_{\mu}\\}$ that generates $I$? Such an ideal is said to be finitely
generated. Furthermore, for a ring $R$, is every ideal $I\subset R$ finitely
generated? Such a ring is said to be a _Noetherian_ ring.

Consider the ring of integers. Since every ideal of $\mathbb{Z}$ consists of
all multiples of some element, every ideal $I$ is finitely generated and
$\mathbb{Z}$ is Noetherian.

Later, I will give an example of a ring that is not Noetherian.

There are two other alternate definitions of Noetherian rings:

######  Proposition 2.15.

The following are equivalent:

  1. i

For every ideal $I\subset R$, there exist $x_{1},\dots,x_{n}\in I$ such that
$I=(x_{1},\dots,x_{n})$.

  2. ii

Every chain of ideals $I_{1}\subset\dots\subset I_{k}\dots$ eventually
terminates such that $I_{\ell}=I_{\ell+1}=\dots$.

  3. iii

Every nonempty set of ideals of $R$ has an ideal that is not contained in any
other.

###### Proof.

(i) $\Rightarrow$ (ii): Take a chain of ideals, and let $I$ be the union of
them all. For $I$, there exist $x_{1},\dots,x_{n}$ that generate $I$. All of
these $x_{i}$ should appear at some point in the chain of ideals. When they
appear, they stay in the chain because each ideal contains all previous
ideals. Because there are finitely many $x_{i}$, we can find an $I_{\ell}$ for
some $\ell$ that finally contains all of them. It is clear that $I_{\ell}=I$
because they are generated by the same elements. To prove the chain
terminates, let $k$ be a natural number bigger than $l$. $I_{l}$ is a subset
of $I$ and therefore is a subset of $I_{k}$. $I_{k}$ is a subset of $I_{l}$
because we have a chain of ideals. Then, $I_{l}=I_{k}$, so the chain
terminates.
(ii) $\Rightarrow$ (iii): At the ideal at which the chain terminates, that
ideal is not contained in any other.
(iii) $\Rightarrow$ (i): Let $I$ be an ideal and $\Sigma=\\{J\subset I\mid
J\text{ is a finitely generated ideal}\\}$. By our given, $J$ has an element
$J_{0}$ that is not contained in any other. Suppose that $J_{0}\neq I$. Then,
there exists $x\in I,x\not\in J_{0}$. Let $J_{0}^{\prime}$ be the ideal
generated by all the finitely many generators in $J_{0}$ and also $x$.
$J_{0}^{\prime}$ is still finitely generated and contains $J_{0}$,
contradicting our assumption that $J_{0}$ is not contained in any other. Then,
$I=J_{0}$ is finitely generated. ∎

##  3\. Algebraic varieties and the polynomial ring

We begin our study of a polynomial ring, points, and the relationship between
them. We motivate our study first by introducing the plane curve, a
mathematical object studied heavily in the $19$th century. Next, we formally
define polynomials and the fundamental functions “taking the variety of” and
“taking the ideal of” that map polynomials to points and vice-versa,
respectively. After, we investigate how these two functions serve as inverses
to each other. We then attempt to understand the nature of plane curves (and
algebraic sets in general). In particular, we prove Hilbert’s Basis Theorem.
Finally, we investigate bijective correspondences between sets of polynomials
and points.

###  3.1. Introducing plane curves

######  Definition 3.1.

Let $k$ be a field and $f\colon k^{2}\rightarrow k$ be a polynomial. A _plane
curve_ $X$ is defined as the set of all points $\\{(x,y)\\}\in k^{2}$ that
satisfy $f(x,y)=0$ (such a function is said to “vanish” on $X$).

Figure 1. The plane curve defined by $k=\mathbb{R}$ and
$f(x,y)=y^{2}-x^{2}(x+1)$.

We are usually familiar with finding the zeros of a polynomial $f(x)$ by
setting $f(x)=0$, and we know that there are finite number of such $x$ (over a
field). Now, we have $f(x,y)$, and there can be an infinite number of $(x,y)$
that satisfy $f(x,y)=0$ (as seen in Figure 1). Algebraic geometry is concerned
with studying the relationship between these polynomial functions and sets of
points.

###  3.2. The polynomial ring

We introduce machinery for understanding how to work with these polynomial
equations.

######  Definition 3.2.

Let $k$ be a field, $n$ be a natural number, and $A$ be $\mathbb{N}^{n}$. The
_polynomial ring_ in $n$ variables $k[X_{1},\dots,X_{n}]$ is the set of all
expressions of the form $\\{\sum_{\alpha}^{A}a_{\alpha}X^{\alpha}\\}$ where
$a_{\alpha}\in k$ for all $\alpha\in A$ and
$X^{\alpha}=X_{1}^{\alpha_{1}}\dots X_{n}^{\alpha_{n}}$ where $\alpha_{i}$ is
the $i$th component in $\alpha$. A _polynomial_ in $n$ variables is an element
of the polynomial ring. Polynomials are finite sums so we impose the
additional restriction that $a_{\alpha}=0$ for all but finitely many
$\alpha\in A$. Let $\\{\sum_{\alpha}^{A}a_{\alpha}X^{\alpha}\\}$ and
$\\{\sum_{\alpha}^{A}b_{\alpha}X^{\alpha}\\}$ be two polynomials. Addition of
two polynomials is defined as
$\\{\sum_{\alpha}^{A}(a_{\alpha}+b_{\alpha})X^{\alpha}\\}$. Multiplication is
defined as
$\\{\sum_{\alpha}^{A}\Pi_{\beta}^{A}(a_{\alpha}*b_{\alpha})X^{\alpha*\beta}\\}$
where $X^{\alpha*\beta}=X_{1}^{\alpha_{1}*\beta_{1}}\dots
X_{n}^{\alpha_{n}*\beta_{n}}$. One can prove that the polynomial ring is a
ring by Definition 2.1.

Note that all concepts we defined about rings and ideals in general apply now
to polynomial rings. For example, the polynomial rings has ideals, ring
homomorphisms, generating sets, and prime ideals.

###  3.3. The relationship between points and polynomials

In the beginning of this section, we defined a plane curve. In the last
subsection, we generalized the algebraic aspect of plane curves by defining
polynomials in $n$ variables. Now, we generalize the “set of points” aspect of
plane curves.

######  Definition 3.3.

Let $V\colon k[X_{1},\dots,X_{n}]\to k^{n}$ be a function where $V(S)=\\{x\in
k^{n}\mid f(x)=0\;\forall\;f\in S\\}$. $V(S)$ is known as the _variety_ of
$S$. $X\subset k^{n}$ is an _algebraic set_ if $X=V(S)$ for some $S\subset
k[X_{1},\dots,X_{n}]$.

######  Remark 3.4.

Note that we are evaluating $f\in k[X_{1},\ldots,X_{n}]$ on a point $x\in
k^{n}$. This is done using the field operations of $k$.

Under this definition, a plane curve is just an algebraic set created from
only one ($f$) two-dimensional (variables $x,y$) polynomial.

We introduce the following proposition that follows from the definition:

######  Proposition 3.5.

$\emptyset=V(\\{k[X_{1},\dots,X_{n}]\\})$ and $k^{n}=V(\\{0\\})$ are algebraic
sets.

When one takes the variety, one maps polynomials to points. Now, we consider a
map which takes points to polynomials:

######  Definition 3.6.

Let $X$ be a subset of $k^{n}$ (not necessarily an algebraic set). The _ideal
of_ $X$, denoted by $I(X)$, is the set of polynomials in $n$ variables that
vanish on $X$.

######  Proposition 3.7.

$I(X)$ is an ideal.

###### Proof.

Follows from field operations. ∎

It is clear that the variety of a set of polynomials and the ideal of set of
points are somehow fundamentally linked. In which way are these two functions
“inverses” of each other?

######  Proposition 3.8.

The correspondences $V$ and $I$ satisfy the following properties for any field
$k$:

  1. (1)

Let $X,Y$ be subsets of $k^{n}$. If $X\subset Y$, $I(Y)\subset I(X).$

  2. (2)

Let $S,T$ be subsets of $k[X_{1},\dots,X_{n}]$. If $S\subset T$, $V(T)\subset
V(S)$.

  3. (3)

Let $X$ be a subset of $k^{n}$. Then, $X\subset V(I(X))$. $X=V(I(X))$ if and
only if $X$ is an algebraic set.

  4. (4)

Let $S$ be a subset of $k[X_{1},\dots,X_{n}]$. Then, $S\subset I(V(S))$.

###### Proof.

$(1)$ and $(2)$ follow from the definition.

For $(3)$, note that $I(X)$ is defined as the set of all polynomials that
vanish on $X$, so these polynomials will vanish on $x\in X$. For the second
part of $(3)$, we know that $X=V(I(X))$; then, $X=V(S)$ for $S=I(X)$, meaning
that $X$ is an algebraic set. Next, let us prove that $X$ is an algebraic set
implies that $X=V(I(X))$. $X=V(S)$ for some $S$. Suppose, for sake of
contradiction, that $I(X)$ does not contain $S$. Then, there exists a
polynomial $f\in S,f\not\in I(X)$. We know that $f(x)=0$ for all $x\in X$ by
definition, so then $f\in I(X)$—a contradiction. Because $S\subset I(X)$,
$V(I(X))\subset V(S)=X$ so $V(I(X))=X$ (we already know $X\subset V(I(X))$
because of the first part of $(3)$).

For $(4)$, let $S$ be such a set and let $f\in S$. By definition, we know that
$f$ vanishes at the points at which $S$ vanishes. ∎

We figured out under what conditions $X=V(I(X))$ for $(3)$ of the proposition.
But we did not figure out under what conditions $S=I(V(S))$. We discuss it
later; it is the motivation behind Hilbert’s Nullstellensatz.

Finally, we introduce some operations on elements in the polynomial ring, and
what the resulting algebraic set. We hope that these examples provide some
intuition on “taking the variety”.

######  Proposition 3.9.

Let $f_{1},f_{2}\in k[X_{1},\dots,X_{n}]$. Then,

  1. (1)

$V(f_{1}*f_{2})=V(f_{1})\cup V(f_{2})$ and

  2. (2)

$V((f_{1},f_{2}))=V(f_{1})\cap V(f_{2})$.

###### Proof.

Both follow from the definition. ∎

###  3.4. Other properties of algebraic sets

Before continuing with relationships between points and polynomials, let us
come back briefly to algebraic sets. First, consider these two statements:

######  Proposition 3.10.

Algebraic sets have the following properties:

  1. (1)

The union of finitely many algebraic sets is an algebraic set.

  2. (2)

The intersection of any collection of algebraic sets is an algebraic set.

###### Proof.

  1. (1)

Let $X=V(S)$ and $Y=V(T)$ be algebraic sets for some sets of polynomials $S$
and $T$. Indeed, $X\cup Y=V(ST)$, where $ST$ denotes the set of all products
of an element of $S$ by an element of $T$. Let $x$ be an element in $X\cup Y$.
Then, $x\in X$ or $x\in Y$. Without loss of generality, suppose $x\in X$.
Then, $f(x)=0$ for all $f\in S$ so every function in $ST$ vanishes on $x$.
Now, let $x\in V(ST)$. Without loss of generality, suppose that $x\not\in X$.
Then, there is an $f\in S$ such that $f(s)\neq 0$. All polynomials $fg\in ST$,
where $g\in T$, vanish on $x$, so then $g(t)=0$ for all $g\in T$ and $x$
vanishes on $Y\subset X\cup Y$.

By induction, the union of finitely many algebraic sets is an algebraic set as
well.

  2. (2)

Let $\\{X_{\alpha}=V(S_{\alpha})\\}$ be a collection of algebraic sets indexed
by $\alpha$. Then, $\bigcap_{\alpha}X_{\alpha}=V(\bigcup_{\alpha}X_{\alpha})$.
Let $x\in\bigcap_{\alpha}X_{\alpha}$. Then, $x\in X_{\alpha}$ for all $\alpha$
so $x\in V(S_{\alpha})$ for all $\alpha$. Now, let $x\in
V(\bigcup_{\alpha}X_{\alpha})$. By definition, $x\in V(X_{\alpha})$ for every
$\alpha$ so $x\in X_{\alpha}$ for every $\alpha$. Then,
$x\in\bigcap_{\alpha}X_{\alpha}$.

∎

These two statements work for a polynomial in any finite amount of variables,
but let us turn our attention back to plane curves. The union of finitely many
plane curves is a plane curve, and the intersection of any set of plane curves
is a plane curve.

We now return to the Noetherian property. Is the ring of polynomials in $n$
variables Noetherian? First, we prove the following statement:

######  Theorem 3.11.

Let $R$ be a ring. If $R$ is Noetherian, $R[x]$ is Noetherian.

###### Proof.

Since we work with $R[x]$, any polynomial mentioned in this proof is a
polynomial in one dimension.

First, I introduce some definitions: the degree of a polynomial is the
smallest possible natural number $i$ such that $a_{\alpha}=0$ for all
$\alpha>i$. In addition, the leading coefficient of a polynomial of degree $i$
is $a_{i}$.

Let $I$ be an ideal of $R[x]$. Define $J$ to be the set of leading
coefficients of all polynomials in $I$. $J$ is an ideal in $R$:

  1. (a)

Closure with respect to addition: let $j,k$ be elements in $J$. They
correspond to two polynomials $f,g\in I$. Since $I$ is closed under addition,
the polynomial $f+g\in I$. The polynomial $f+g$ has leading coefficient $j+k$
so $j+k\in J$.

  2. (b)

Existence of additive inverse: $0\in I$ so $0\in J$.

  3. (c)

If $i\in J_{i}$ and $r\in R$ then $i*r\in I$: let $i\in J,r\in R$. The element
$i$ corresponds to a polynomial $f\in I$. $I$ is an ideal so the polynomial
$f*r\in I$. The polynomial $f*r$ has leading coefficient $i*r$ so $i*r\in J$.

$J$ is finitely generated because $R$ is Noetherian. Let $A=\\{a_{\mu}\\}$ be
the finite set of leading coefficients that generate $J$. For each $a_{\mu}\in
A$, pick $g_{\mu}\in I$ with leading coefficient $a_{\mu}$. Let
$G=\\{g_{\mu}\\}$. By the Well-Ordering Principle, $G$ has a polynomial with
largest degree: let $k$ be this degree.

Then, for every $i\leq k$, define $J^{i}$ to be the set of leading
coefficients of all polynomials in $I$ with degree smaller than or equal to
$i$. $J^{i}$ is an ideal in $R$ (by a similar argument as given above).
$J^{i}$ is therefore also finitely generated. Let
$A^{i}=\\{a_{\mu^{i}}^{i}\\}$ be the finite set of leading coefficients that
generate $J^{i}$. For each $a_{\mu}^{i}\in A$, pick $g_{\mu}^{i}\in I$ with
leading coefficient $a_{\mu}^{i}$. Let $G^{i}=\\{g_{\mu}^{i}\\}$.

Now, let $I_{0}$ be the ideal generated by all $g^{i}$ in all $G^{i}$ and all
$g$ in $G$. There are finite number of $g^{i}$ and $g$ so $I_{0}$ is finitely
generated. I claim now that $I=I_{0}$.

Let a polynomial $f\in I_{0}$. $f$ is generated by some $g$’s and $g^{i}$’s,
which are all elements of $I$, so $f\in I$.

Now, for sake of contradiction, suppose that $I\not\subset I_{0}$. Find a
polynomial $f$ in $I$ that is not in $I_{0}$ with smallest degree. We
introduce two cases:

  1. Case 1:

The degree $\ell$ of $f$ is bigger than $k$. Let $a$ be the leading
coefficient of $f$. Then, $a\in J$. This means that
$a=\sum_{\mu}^{A}a_{\mu}*r_{\mu}$. Let
$f_{0}=\sum_{\mu}^{A}r_{\mu}X^{\text{deg}(f)-\text{deg}(g_{\mu})}g_{\mu}$. The
polynomial $f_{0}$ has the same leading coefficient as $f$ and the same degree
as $f$, so $\text{deg}(f-f_{0})<\text{deg}(f)$. Also, $f-f_{0}$ is not in
$I_{0}$ because otherwise by the definition of an ideal, $f$ is in $I_{0}$.
Then, we have found an element with smaller degree in $I$ that is not in
$I_{0}$—a contradiction.

  2. Case 2:

The degree $\ell$ of $f$ is smaller than $k$. Let $a$ be the leading
coefficient of $f$. Then, $a\in J_{\ell}$. This means that
$a=\sum_{\mu^{i}}^{A_{\ell}}a_{\mu^{i}}*r_{\mu^{i}}$. Let
$f_{0}=\sum_{\mu^{i}}^{A_{\ell}}r_{\mu^{i}}X^{\text{deg}(f)-\text{deg}(g_{\mu^{i}})}g_{\mu^{i}}$.
We get a similar contradiction as the one in Case $1$.

Note that we must separate out the proof into two cases in this way because we
do not want the degree of $X$ in $f_{0}$ to be negative. ∎

######  Corollary 3.12.

$k[X_{1},\dots,X_{n}]$ is Noetherian.

###### Proof.

$k$ is Noetherian, so $k[X_{1}]$ is Noetherian. Now, consider
$k[X_{1},X_{2}]$. Define $k[X_{1}][X_{2}]$ to be the ring of one variable
$X_{2}$ with coefficients in $k[X_{1}]$. $k[X_{1}]$ is Noetherian so
$k[X_{1}][X_{2}]$ is Noetherian. The ring $k[X_{1}][X_{2}]$ is equivalent
(isomorphic) to the ring $k[X_{1},X_{2}]$. By induction,
$k[X_{1},\dots,X_{n}]$ is Noetherian. ∎

Theorem 3.11 is known as Hilbert’s Basis Theorem. What can we use it for?

######  Proposition 3.13.

Let $X$ be an algebraic set. $X$ is the intersection of the varieties of
finitely many polynomials.

###### Proof.

$X=V(I(X))$ by Proposition 3.8. $I(X)$ is finitely generated by Corollary
3.12. So, we have $X=V((f_{1},\dots,f_{n}))$.
$V((f_{1},\dots,f_{n}))=V(f_{1})\cap\dots\cap V(f_{n})$ follows from the
definition of the variety of a set of polynomials. ∎

This fact is not only interesting when considering algebraic sets (and plane
curves), but also useful. In particular, Hilbert used it to prove an important
result in invariant theory.

We use Hilbert’s Basis Theorem again. First, we introduce a new definition:

######  Definition 3.14.

An algebraic set $X\subset k^{n}$ is irreducible if there do not exist
$X_{1},X_{2}$ such that $X_{1}\cup X_{2}=X$ with $X_{1},X_{2}\subsetneq X$.

We then prove an intermediate result:

######  Proposition 3.15.

Every descending chain of algebraic sets $X_{1}\supset X_{2}\supset\dots$
eventually terminates.

###### Proof.

Suppose that the chain does not terminate. Then, $I(X_{1})\subset
I(X_{2})\subset\dots$ does not terminate—a contradiction because
$k[X_{1},\dots,X_{n}]$ is Noetherian. ∎

Finally, we reach the proposition:

######  Proposition 3.16.

Let $X$ be an algebraic set. $X$ can be written as a finite union of
irreducible algebraic sets $X_{1},\dots,X_{n}$ where $X_{i}\not\subset X_{j}$
for all $i\neq j$ (let this be property $*$).

###### Proof.

Let $\Sigma$ be the set of all algebraic sets in $k^{n}$ that do not have
property $*$. If $\Sigma$ is empty, our proof is done. Suppose that $\Sigma$
is not empty. Then, there exists an algebraic set $X$ that does not contain
any other algebraic set in $\Sigma$ (otherwise we have a descending chain of
algebraic sets that does not terminate). $X$ is not irreducible by our given.
Then, let $X_{1},X_{2}$ be algebraic subsets such that $X_{1},X_{2}\subsetneq
X$ and $X=X_{1}\cup X_{2}$. One of $X_{1},X_{2}$ cannot have property $*$;
otherwise $X$ has property $*$ (by unioning all finite irreducible polynomials
making up $X_{1}$ and $X_{2}$) so then $X\not\in\Sigma$. Thus, at least one of
$X_{1},X_{2}$ is in $\Sigma$, so there does exist an algebraic set in $\Sigma$
that is contained in $X$—a contradiction. ∎

Thus, an algebraic set can be written as the intersection of finitely many
varieties of polynomials. In addition, an algebraic set can be written as the
union of finitely many irreducible algebraic sets.

In Section $2$, we did not know of any rings that were not Noetherian. We now
give the following example:

######  Proposition 3.17.

Let $k$ be a field. $k[X_{1},X_{2},\dots]$ is not Noetherian.

###### Proof.

The chain of ideals $(X_{1})\subset(X_{1},X_{2})\subset\dots$ never
terminates. ∎

###  3.5. Correspondences between points and polynomials

We introduce a bijective correspondonce from irreducible algebraic sets to
prime ideals:

######  Proposition 3.18.

Let $X\subset k^{n}$ be an algebraic set. $X$ is irreducible $\iff$ $I(X)$ is
prime.

###### Proof.

Let us prove that if $X$ is irreducible, $I(X)$ is prime. Suppose that $I(X)$
is not prime. Then, there exist polynomials $f_{1},f_{2}\not\in I(X)$ such
that $f_{1}f_{2}\in I(X)$. Let $I_{1}$ be the ideal generated by $f$ and all
generators of $I$. $I(X)\subsetneq I_{1}$ so $V(I_{1})\subsetneq V(I(X))=X$
because $X$ is an algebraic set. The same logic applies for $f_{2}$. So we
have that $V(I_{1}),V(I_{2})\subsetneq X$. Let $x\in X$. We know that
$f_{1}f_{2}(x)=0$, so either $f_{1}(x)=0$ or $f_{2}(x)=0$, so $x\in V(I_{1})$
or $x\in V(I_{2})$. Then, $X\subset V(I_{1})\cup V(I_{2})$ so $X$ is
reducible.

Let us prove now that if $I(X)$ is prime, $X$ is irreducible. Suppose that $X$
is not irreducible. Let $X_{1},X_{2}\subsetneq X$ such that $X=X_{1}\cup
X_{2}$. Because $X_{1}\subsetneq X$, $I(X)\subsetneq I(X_{1})$, there exists
$f_{1}$ such that $f_{1}\in I(X_{1})$ but $f_{1}\not\in I(X)$ (and a similar
$f_{2}$ for $I(X_{2})$). $f_{1}f_{2}$ vanishes at all points in $X$ so
$f_{1}f_{2}\in I(X)$, so $I(X)$ is not prime. ∎

Now we return to Proposition 3.8 $(4)$ to see if we can find conditions such
that $S=I(V(S))$. If we succeed, we can define a bijective correspondence
between these sets $S$ and algebraic sets $V(S)$. Why are we not guaranteed
equality?

Instead of considering all sets of polynomials $S$, we consider ideals of the
polynomial ring $J$ only in this problem. Recall that if a set of polynomials
$S$ and an ideal $J$ both vanish on $X$ (i.e. $V(S)=V(J)$), $S\subset J$. As
we are trying to achieve $S=I(V(S))$, using $S\neq J$ would mean that
$S\subset I(V(S))$—not what we want.

First, consider the polynomial $x^{2}+1\in k[x]$, where $k=\mathbb{R}$. Let
$J$ be the ideal generated by this polynomial. $V(J)=\emptyset$, so
$I(V(J))=k^{n}$. As we can see, the solution just doesn’t exist in
$\mathbb{R}$. If we were to use $k=\mathbb{C}$, though, $V(J)$ would not be
empty (actually $|V(J)|=2$). So, in order for equality to hold, we need $k$ to
have some special property.

Second, consider the polynomial $x^{2}\in k[x]$, where $k=\mathbb{R}$. Let $J$
be the ideal generated by this polynomial. $V(J)=\\{0\\}$, and $I(V(I))=(x)$.
As we can see, the original ideal did not contain all polynomials that vanish
on its vanishing set, so we ended up with inequality. Then, if we want
$J=I(V(J))$, for every $f^{n}$ in $J$, $f$ must be in $J$.

We define two new definitions based on these two observations:

######  Definition 3.19.

A field $k$ is algebraically closed if every polynomial in $k[x]$ has a root.

######  Definition 3.20.

Let $J$ be an ideal. The _radical_ of $J$, denoted by $\sqrt{J}$, is the set
$\\{f\mid f^{n}\in J\text{ for some }n>0\\}$. $\sqrt{J}$ is an ideal.

Hilbert’s Nullstellensatz proves the bijective correspondence between radical
ideals and algebraic sets:

######  Theorem 3.21.

Let $k$ be an algebraically closed field and $J$ be an ideal with coefficients
in $k$. Then, $\sqrt{J}=I(V(J))$.

Proof of Hilbert’s Nullstellensatz requires more definitions; we do not prove
it here. We hope though that the reader has noted how useful such statements
are to understanding the polynomial ring and these sets of points.

## Acknowledgements

I am very grateful to my mentor Michael Neaton and to Dr. May for their useful
comments and suggestions. All errors are mine.

## References

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