Back to Algebraic Geometry Page

On this page:

• Introduction
• Affine varieties
• Radical
• Hilbert’s Nullstellensatz
• TODO: polynomial functions and coordinate rings

[example] [extitle]Conventions Assumed[/extitle]

On this page:

• $K$ will always be assumed to be an algebraically closed field.
• All rings are commutative unital rings,

[/example]

Introduction

This page will largely be focused on a classical algebraic geometry perspective: affine varieties. We begin by defining the ambient space we are working with.

Affine varieties

[definition] [deftitle]Definition %counter% (Affine $n$-space)[/deftitle]

We define affine $n$-space over $K$ to be the set

\(\AA^n \coloneqq \AA_K^n \coloneqq \{(a_1, \ldots, a_n) \mid a_i \in K \text{ for } i \in [n]\}.\) [/definition]

It is important that we define affine $n$-space as a set. We ignore the canonical $K$-vector space structure on $K^n$ when it comes to $\AA_K^n$. Affine $n$-space will serve as the ambient space of the zero loci of polynomials that we will consider on these pages.

[definition] [deftitle]Definition %counter% (Affine variety)[/deftitle]

Let $S \subseteq K[x_1, \ldots, x_n]$. We call

an affine variety. [/definition]

[example] [extitle]Remark %counter%[/extitle]

In our definition, we will follow the convention of Gathmann of dropping the requirement that affine varieties are irreducible and we will work with this as a separate requirement later on. [/example]

[example] [extitle]Example %counter% ($V(x^2 + y^2 - 1)$)[/extitle]

A classic and easy-to-understand example is the zero locus of the polynomial $x^2 + y^2 - 1$ in $\RR^2$. Obviously $\RR$ isn’t algebraically closed, but working in $\RR^2$ allows us to easily draw some pictures. The set of points cut out by $x^2 + y^2 - 1$ are the solutions to the polynomial equation

which, of course, we know to be the circle of unit radius centered at the origin.

[/example]

[example] [extitle]Example %counter% ($V(y^2 - x^3)$)[/extitle]

Another very classic example arises from the polynomial equation $y^2 = x^3$.

[/example]

The definition of an affine variety admits some easy-to-see results.

[theorem] [thmtitle]Proposition %counter% ($V(-)$ is inclusion-reversing)[/thmtitle]

Given $S \subseteq T \subseteq K[\mathbf{x}]$, then $V(S) \supseteq V(T)$. [/theorem]

[proof] Let $x \in V(T)$. Since $S \subseteq T$, every $f\in S$ satisfies $f(x) = 0$. Thus, $x \in V(S)$ and we are done. [/proof]

[theorem] [thmtitle]Proposition [counter]union_of_affine_varieties[/counter] ($V(S) \cup V(T) = V(ST)$)[/thmtitle]

Let $S, T \subseteq K[\mathbf{x}]$. Then $V(S) \cup V(T) = V(ST)$ where

[/theorem]

[proof] ($\subseteq$) Let $x \in V(S) \cup V(T)$ and, WLOG, assume $x \in V(S)$. Let $fg \in ST$. Since $f$ vanishes on $x$, it follows that

Thus, $x \in V(ST)$ as desired.

($\supseteq$) If $V(ST) \subseteq V(S)$, then we are done. If it is not, then there is some point $x \in V(ST)$ for which there is a $f \in S$ such that $f(x) \neq 0$. But since $x \in V(ST)$, given any $g \in T$, we must have

Since $f(x) \neq 0$, and $K$ is an integral domain, it follows that $g(x) = 0$. Thus, $x \in V(T)$ and we are done. [/proof]

[theorem] [thmtitle]Proposition [counter]intersection_of_affine_varieties[/counter] ($V(S) \cap V(T) = V(S\cup T)$)[/thmtitle]

Let $S, T \subseteq K[\mathbf{x}]$. Then $V(S) \cap V(T) = V(S\cup T)$. [/theorem]

[proof] ($\subseteq$) Let $x \in V(S) \cap V(T)$. Then every $f\in S$ and $g\in T$ vanishes on $x$. But this just means that $x \in V(S \cup T)$ as desired.

($\supseteq$) Let $x \in V(S \cup T)$. Then, for any $f\in S$ and $g\in T$, both have $x$ as a common vanishing point. Therefore, $x \in V(S) \cap V(T)$. [/proof]

Notice that the later two propositions tell us that (a) the union/intersection of affine varieties is, again, an affine variety; and (b) a formula for each.

A much more interesting observation about affine varieties is how they interplay with polynomials. Notice that if $f$ and $g$ vanish on $X \subseteq \AA^n$, then so does their sum $f + g$. Additionally, it is also clear that given any $h$, the product $hf$ vanishes as well. Accordingly, it suffices to consider stuff only for ideals.

[theorem] [thmtitle]Proposition [counter]promotion_to_ideal[/counter] ($V(S) = V(\generator{S})$)[/thmtitle]

Let $S \subseteq K[\mathbf{x}]$. Then $V(S) = V(\generator{S})$. [/theorem]

[proof] The $\supseteq$ direction follows immediately from the inclusion-reversing property of $V(-)$. To show $\subseteq$, let $x \in V(S)$. Now, given an arbitrary element $f \coloneqq \sum_{i} c_i f_i$ of $\generator{S}$, each $f_i$ vanishes on $x$ so $f(x) = 0$. Thus, $x \in V(S)$ as desired. [/proof]

[theorem] [thmtitle]Corollary [counter]varieties_are_zero_loci_of_finitely_many_polynomials[/counter] ($V(S) = V(f_1, \ldots, f_k)$)[/thmtitle]

Let $S \subseteq K[x_1, \ldots, x_n]$. Then there exist finitely many polynomials $f_1, \ldots, f_k \in K[x_1, \ldots, x_n]$ such that $V(S) = V(f_1, \ldots, f_k)$. [/theorem]

[proof] By the Hilbert Basis Theorem, $\generator{S}$ is finitely generated (because fields are trivially Noetherian). Therefore, there are $f_1, \ldots, f_k \in \generator{S}$ such that

Applying the previous proposition then tells us that

\(V(S) = V(\generator{S}) = V(f_1, \ldots, f_k).\) [/proof]

Notice that this result says a lot about the structure of affine varieties — they can always be realized as the zero locus of finitely many polynomials. We will finish off this section by metnioning some more straightforward results about $V(-)$.

[theorem] [thmtitle]Proposition %counter% ($V(I) \cup V(J) = V(IJ) = V(I \cap J)$)[/thmtitle]

If $I, J \unlhd K[\mathbf{x}]$, then $V(I) \cup V(J) = V(IJ) = V(I \cap J)$. [/theorem]

The first equality is immediate by Proposition [countref]union_of_affine_varieties[/countref] above.

Radical

In algebra, one learns about the radical of a subset of a ring. A classic example is the radical of the zero ideal, which gives the nilpotent elements. For instance, in linear algebra, nilpotence of matrices is used to give us information about Jordan blocks. Here, we explore radicals in the context of affine varieties.

[definition] [deftitle]Definition %counter% (Radical)[/deftitle]

Let $R$ be a commutative ring and $S \subseteq R$. The radical of $S$ is the set

[/definition]

Of great interest is the radical of an ideal. For starters, the radical of an ideal is actually an ideal itself.

[theorem] [thmtitle]Proposition %counter% ($\sqrt{I}$ is an ideal)[/thmtitle]

Let $R$ be a commutative ring and $I \unlhd R$. Then $\sqrt{I}$ is an ideal of $R$. [/theorem]

[proof] Clearly, $0 \in \sqrt{I}$ since $0 \in I$. Absorption of products follows immediately by the commutativity of $R$. Now let $x, y\in R$ and say $x^k = y^\ell = 0$. Then

by the Binomial Theorem. Since we have either a factor of $x^k$ or $y^\ell$ in each term, it follows that $(x + y)^{k + \ell} = 0$. [/proof]

When I first saw the radical of an ideal, I thought it was a very strange concept. I personally find it more comfortable to think about with a couple of examples on hand:

[example] [extitle]Example %counter% (Radicals in $\ZZ$)[/extitle]

Consider the ideal $\generator{12}$ in $\ZZ$. Taking the radical gives

Here, the radical annihilates the exponents on the prime factors. This holds much more generally in $\ZZ$ because we can exploit the UFD property of $\ZZ$. [/example]

[example] [extitle]Example %counter% (Radicals in $K[x]$)[/extitle]

The ideal generated by $(x - 2)^4(x - 3)^5$ has radical $\generator{(x-2)(x-3)}$ by the classical UFD theorems applied to $K[x]$. Again, we’re just annihilating the powers of the irreducible factors. [/example]

[example] [extitle]Example %counter% (Nilradical)[/extitle]

One case where we need to be careful with the naive intuition in the previous two examples is $\sqrt{0} \unlhd R$. The radical of zero is called the nilradical of the ring and its elements are the nilpotent elements of $R$. [/example]

Another way I also like to think about radicals is in a geometric context. Consider the variety $V((y^2 - x^3)^3)$:

The variety looks just like $V(y^2 - x^3)$! So, in a certain sense, it suffices to just drop the powers of the irreducible factors to obtain the zero locus of a particular polynomial (or a set of polynomials) and the radical is the tool that precisely allows that. This is justified by the following proposition:

[theorem] [thmtitle]Proposition [counter]variety_of_sqrt[/counter] ($V(\sqrt{J}) = V(J)$)[/thmtitle]

Let $J \unlhd K[\mathbf{x}]$. Then $V(\sqrt{J}) = V(J)$. [/theorem]

[proof] ($\subseteq$) Note that $J \subseteq \sqrt{J}$. Thus, $V(J) \supseteq V(\sqrt{J})$ by the inclusion-reversing property of $V(-)$.

($\supseteq$) Let $x \in V(J)$ and let $f \in \sqrt{J}$. There is some $k \in \ZZ_{> 0}$ for which $f^k \in J$. So $f^k(x) = 0$. Since $K$ is a field, it follows that $f(x) = 0$. Thus, $x \in V(\sqrt{J})$ and we are done. [/proof]

The (extremly basic!) theory of radicals above is precisely what lets us prove the following property that should seem “obvious” from a glance.

[theorem] [thmtitle]Proposition %counter% ($V(I) \cup V(J) = V(IJ) = V(I\cap J)$)[/thmtitle]

Let $I, J \unlhd K[\mathbf{x}]$. Then $V(I) \cup V(J) = V(IJ) = V(I\cap J)$. [/theorem]

[proof] The first equality is immediate by Proposition [countref]union_of_affine_varieties[/countref] so we only show the second equality. A classic result of commutative algebra is that $\sqrt{IJ} = \sqrt{I \cap J}$. Given this result, it is immediate that $V(IJ) = V(I \cap J)$ by Proposition [countref]variety_of_sqrt[/countref]. Let us now prove the aforementioned result.

($\subseteq$) Let $f \in \sqrt{IJ}$. Then there is some $k \in \ZZ_{> 0}$ for which $f^k \in IJ$. Since $K[\mathbf{x}]$ is a UFD, it follows that we can find $k_1$ and $k_2$ such that

with $f^{k_1} \in I$ and $f^{k_2} \in J$. Setting $k’ = \max\{k_1, k_2\}$, we see that $f^{k’} \in I \cap J$. Thus, $f \in \sqrt{I \cap J}$.

($\supseteq$) Let $f \in \sqrt{I\cap J}$. Then there is some $k\in \ZZ_{> 0}$ for which $f^k \in I\cap J$. But this means that $f^k \cdot f^k = f^{2k} \in IJ$ so $f \in \sqrt{IJ}$ as desired. [/proof]

And then, for completion, we include the following property:

[theorem] [thmtitle]Proposition %counter% ($V(I) \cap V(J) = V(I+J)$)[/thmtitle]

Let $I, J \unlhd K[\mathbf{x}]$. Then $V(I) \cap V(J) = V(I+J)$. [/theorem]

[proof] Proposition [countref]intersection_of_affine_varieties[/countref] implies that

But, from commutative ring theory, we know that $I + J$ is the ideal generated by $I\cup J$ so Proposition [countref]promotion_to_ideal[/countref] implies the desired result. [/proof]

Hilbert’s Nullstellensatz

There are at least four results that I’m aware that people call the classical Nullstellensatz and I’ve never been able to figure out which one is actually the classical Nullstellensatz. In any case, we will give one version of it as the other versions are not too hard to see from the version we will be given here.

[definition] [deftitle]Definition %counter% (Ideal of a subset of $\AA_K^n$)[/deftitle]

Let $X \subseteq \AA_K^n$. Then the set

is called the ideal of $X$. [/definition]

There is an induced operation $I(-)$ which sends subsets $X$ of affine $n$-space to an ideal $I(X) \unlhd K[\mathbf{x}]$. The rather remarkable (and, fairly intuitive!) fact that will arise is that $I(-)$ and $V(-)$, in a certain sense, are inverse maps of each other. This fact is known as the Nullstellensatz which we state now.

[theorem] [thmtitle]Theorem %counter% (Hilbert’s Nullstellensatz)[/thmtitle]

Let $K$ be an algebraically closed field and define

to

the maps

and

are inverse to each other. [/theorem]

As it turns out, the only hard part of this theorem is showing that $I(V(J)) \subseteq J$ for some radical ideal $J \unlhd K[\mathbf{x}]$. If $K = \RR$ (i.e. not algebraically closed), then we have ideals like $J = \generator{x^2 + 1}$ from which it follows that $V(J) = \emptyset$ and every polynomial in $\RR[x]$ vanishes on the empty set. The assumption that $K$ is algebraically closed is crucial. The proofs that I am aware of this to show the Nullstellensatz result are based on Noether Normalization which I am most definitely not going to replicate here. That being said, there are some consequences of the Nullstellensatz that are straightforward.

[theorem] [thmtitle]Corollary %counter% (Weak Nullstellensatz)[/thmtitle]

Let $K$ be an algebraically closed field and $J \unlhd K[\mathbf{x}]$. If $J \neq K[\mathbf{x}]$, then $V(J) \neq \emptyset$. [/theorem]

[proof] If $V(J) \neq \emptyset$, then the Nullstellensatz implies $

Since $1 \in \sqrt{J}$, it follows that $1 \in J$ which is a contradiction. [/proof]

[example] [extitle]Remark %counter%[/extitle]

Notice that this result is in, some sense, a generalization of the algebraic closure property in univariate polynomial rings. But, then, it is also totally natural to ask the question given a set of points, does there exist a set of polynomials whose zero locus is said set. [/example]

[theorem] [thmtitle]Corollary %counter%[/thmtitle]

Let $K$ be an algebraically closed field. Then the maximal ideals of $K[x_1, \ldots, x_n]$ are of the form

[/theorem]

[proof] It is an easy induction exercise to show that these kinds of ideals are maximal. To show that any maximal ideal is of this form, on the other hand, is a bit more involved.

The maximal ideals of the form in the statement are in bijection with the points of $\AA_K^n$ by the Nullstellensatz (and noting that maximal ideals are radical). On the other hand, we note that if $\mathfrak{m}$ is a maximal ideal, then $V(\mathfrak{m})$ consists of at least one point. If it is one point, then we reduce to the statement of the result by Nullstellensatz. If there is more than one point, then we can set

for some $a \coloneqq \{a_1, \ldots, a_n\} \in V(\mathfrak{m})$. Then, applying $I(-)$ gives

It is straightforward to see that $I(-)$ is inclusion-reversing, so we have

But $I(a) = \generator{x_i - a_i \mid i\in [n]}$ which contradicts the maximality of $\mathfrak{m}$. Thus, $V(\mathfrak{m})$ consists of exactly one point and we are done. [/proof]

[example] [extitle]Remark %counter%[/extitle]

Notice that this result says that there is bijection between the points of $\AA_K^n$ and the maximal ideals of $K[x_1, \ldots, x_n]$. This allows us to start to build up a dictionary of “equivalent” geometric terms and algebraic terms. [/example]

TODO: polynomial functions and coordinate rings