This post contains the defintiion of affine varieties, regular functions, coordinate ring, the Nullstellensatz, and the quotient of an affine variety upon action of a finite group.

Affine varieties

Let $k$ be a field. It is clear what points in $k^n$, polynomial functions on $k^n$ etc mean.

Definition 1 An affine variety $X$ is a subset of $k^n$, that is the common zeros of a set of polynomials.

Remark

1. The set of polynomials can be taken to be finite. This is because $k[x_1,\cdots,x_n]$ is Noetherian (Hilbert basis theorem), so the ideal generated by a set of polynomials is finitely generated, and their common zero set are the same.
2. As a variety, $k^n$ is more often denoted by $\mathbb{A}^n$.

Some terminologies

1. the Zariski topology

This formalism allows us to talk about varieties more easily. Consider the topology on $\mathbb{A}^n$ with the closed sets being affine varieties. It is easy to verify that this is a topology:

• $\emptyset = \{x \in \mathbb{A}^n | \, 1 = 0 \}$, $\mathbb{A}^n = \{ x \in \mathbb{A}^n | \, 0 = 0 \}$
• Union of two closed sets: $\{x \in \mathbb{A}^n | \, f_i(x) = 0 \} \cup \{x \in \mathbb{A}^n | \, g_j(x) = 0 \} = \{ x \in \mathbb{A}^n | \, f_ig_j(x) = 0\}$
• Intersection of closed sets: just put all the equations together.

2. the ideal of X $I(X)$

This is the ideal of all the polynomials in $k[x_1,\cdots,x_n]$ that vanish on $X$.

3. the affine coordinate ring $\displaystyle A(X) = \frac{k[x_1,\cdots,x_n]}{I(X)}$

Consider $x_1,\cdots,x_n$ as the coordinates of the affine space $\mathbb{A}^n$, and restrict it on $X$. If we allow free multiplication and addition, this is what we will get.

The Nullstellensatz

It is clear that when we have a subset $S \subset \mathbb{A}^n$, we can define the corresponding ideal $I(S)$ exactly like above. On the other hand, given an ideal $\mathfrak{a} \subset k[x_1,\cdots, x_n]$, we can define $Z(\mathfrak{a})$ to be the common zeroes of all $f(x_1,\cdots,x_n) \in \mathfrak{a}$.

Some natural consequences:

• If $S \subset T \subset \mathbb{A}^n$, then $I(T) \subset I(S)$
• If $\mathfrak{a} \subset \mathfrak{b}$ are ideals of $k[x_1,\cdots,x_n]$, then $Z(\mathfrak{b}) \subset Z(\mathfrak{a})$
• If $\mathfrak{a}$ is an ideal of $k[x_1,\cdots,x_n]$, $\sqrt{\mathfrak{a}} \subset I(Z(\mathfrak{a}))$
• If $S \subset \mathbb{A}^n$, then $Z(I(S)) = \overline{S}$ (closure in Zariski topology)

Only the proof of $Z(I(S)) = \overline{S}$ is nontrivial. It is clear that $\overline{S} \subset Z(I(S))$. For the other side, suppose that $C = Z(\mathfrak{a})$ is Zariski closed and $S \subset C$. Then $\mathfrak{a} \subset I(Z(\mathfrak{a})) = I(C) \subset I(S)$, thus $Z(I(S)) \subset Z(\mathfrak{a}) = C$. This shows that $Z(I(S))$ is the smallest Zariski-closed set containing $S$ $Box$.

If $\sqrt{\mathfrak{a}} \subset I(Z(\mathfrak{a}))$ is an equality, then we can establish a one-one correspondence between the affine varieties in $\mathbb{A}^n$ and the ideals of $k[x_1,\cdots,x_n]$. This is desirable because as in the case manifolds, we want to define “varieties” without an embedding into an affine space. As we shall see, varieties are isomorphic iff their affine coordinate rings are isomorphic as $k$-algebras. Thus once we have this correspondence, we can encode the information of points into the affine coordinate ring as the first step of the definition of “varieties”.

This correspondence exists when $k$ is algebraically closed.

Theorem 1 (Hilbert’s Nullstellensatz) Let $\mathfrak{a} \subset k[x_1,\cdots,x_n]$, where $k$ is algebraically closed field. Then $I(Z(a)) = \sqrt{a}$.

In other words, if $f \in k[x_1, \cdots, x_n]$ vanishes on the common zeros of an ideal $\mathfrak{a}$, then for some $r$, $f^r \in \mathfrak{a}$.

Regular functions

Basically we want to deal with polynomial maps. However, when we restrict on affine varieties, there are some more possibilities. For example, if $X \subset \mathbb{A}^n$ is an affine variety, and $f$ is a polynomial that never vanish on $X$, then $\frac{g}{f}$ makes sense for any polynomial $g$.

However, the following proposition tells us that allowing such division does not enlarge the class of maps concerned, when $k$ is algebraically closed.

Proposition 1 Suppose that $f$ is a polynomial that never vanishes on an affine variety $X \subset \mathbb{A}^n$, where $k$ is algebraically closed. Then $f$ is a unit in $A(X)$.

Proof

Let $X = Z\left((g_1,\cdots,g_k)\right)$. As $f$ never vanishes on $X$, $g_1,\cdots,g_k,f$ have no common zeros. By the Nullstellensatz, $(g_1,\cdots,g_k,f) = 1$. This means that there exists polynomials $h_1,\cdots,h_k, h$ such that

$g_1h_1 + \cdots + g_kh_k + fh = 1$

Modulo $I(X)$ gives the desired result. $\Box$

Thus maps of the form $\frac{g}{f}$ on $X$ can also be represented by $gh$, i.e. we are still working with polynomial maps.

Definition 2 (Regular functions) Let $X \subset \mathbb{A}^n$ be an affine variety. A regular function $X$ is a polynomial map from $X$ to $k$.

It is clear that the regular functions form a ring, which is exactly $A(X)$, the affine coordinate ring.

Maps between affine varieties

Naturally, just as in the case for differentiable functions,

Definition 2 For two affine varieties $X \subset \mathbb{A}^n$, $Y \subset \mathbb{A}^m$, a map $f = (f_1,\cdots, f_m): X \rightarrow Y$ is called regular if $f_1, \cdots, f_m$ are regular functions on $X$.

This immediately gives us a notion of isomorphism: Two affine varieties $X \subset \mathbb{A}^n$ and $Y \subset \mathbb{A}^m$ are isomorphic if there exists regular maps $f: X \rightarrow Y$ and $g: Y \rightarrow X$ such that $f \circ g = 1$ and $g \circ f = 1$.

Regular maps and ring of regular functions

Notice that composition of regular maps is still regular. In particular a regular map $f: X \rightarrow Y$ induces a $k$-algebra homomorphism $f^*: A(Y) \rightarrow A(X)$.

On the other hand, a $k$-algebra homomorphism $\theta: A(Y) \rightarrow A(X)$ has to come from a regular map. For $y_1, \cdots, y_m$ are the coordinates of $Y$, and $\theta(y_1), \cdots, \theta(y_m)$ tells you what they are in terms of $x_1, \cdots, x_n$. Thus consider $f:X \rightarrow Y$ defined by

$f\left((x_1,\cdots,x_n)\right) = \left(\theta(y_1), \cdots, \theta(y_m)\right)$

Then one can see that $f^* = \theta$. In fact we have

Proposition 2 For two affine varieties $X \subset \mathbb{A}^n$ and $Y \subset \mathbb{A}^m$, we have a natural isomorphism

$\mathrm{Hom}(X,Y) \rightarrow \mathrm{Hom}_{k-algebra}(A(Y), A(X))$

Corollary 1 Two affine varieties $X \subset \mathbb{A}^n$ and $Y \subset \mathbb{A}^m$ are isomorphic if and only if $A(Y)$ is isomorphic to $A(X)$ as $k$-algebra.

Examples

1. $x^2+y^2 = 1 \subset \mathbb{A}^2$ is an affine variety.

2. $\mathcal{C}: y^2 = x^3 \subset \mathbb{A}^2$ is an affine variety.

3. (Let $k$ be algebrically closed) The map $f: \mathbb{A}^1 \rightarrow \mathcal{C} \subset \mathbb{A}^2$ defined by $f(t) = (t^2, t^3)$ is regular. It is regular, bijective, yet not an isomorphism, since a polynomial in $t^2$ and $t^3$ can’t possibly give $t$.

What algebras are coordinate rings?

This is answered by the following corollary of the Nullstellensatz.

Corollary 2 Let $k$ be algebraically closed. The association $X \rightarrow I(X)$ is a 1-1 correspondence between affine subvarieties of $\mathbb{A}^n$ and radical ideals of $k[x_1,\cdots,x_n]$. Moreover, $X$ is irreducible iff $I(X)$ is prime.

Recall that a topological space $X$ is irreducible if it cannot be written as a union of two proper closed subsets.

Proof

The 1-1 correspondence was established already.

Suppose that $X$ is reducible, i.e. $X = Y \cup Z$, $Y,Z$ being closed and proper. Then $I(X) = I(Y) \cap I(Z)$. If $I(X)$ is prime, then $I(X) = I(Y)$ or $I(Z)$, meaning $X = Y$ or $Z$, contradiction. Therefore if $X$ is reducible, $I(X)$ is not prime.

If $I(X)$ is not prime, let $fg \in I(X)$ but $f,g \notin I(X)$.  (i.e. $fg$ vanishes entirely on $X$, but neither $f$ or $g$ do). Then $X = Z\left(I(X) + (f)\right) \cup Z\left(I(X) + (g)\right)$, and both are proper, closed subsets. So if $I(X)$ is not prime, $X$ is reducible.

Corollary 3 Let $k$ be algebraically closed. Then an $k$-algebra is an affine coordinate ring if and only if it is finitely generated over $k$ and admits no nilpotents.

Quotient by group action

(Assumption: $k$ is an algebraically closed field)

Let $X \subset \mathbb{A}^n$ be an affine variety, and $G$ be a finite group of automorphisms of $X$. $G$ can also be realized as a group of $k$-automorphisms of $A(X)$. We want to give the $G$-orbits a variety structure.

What is the natural definition of this? Well, mimicking quotient topology, we would want a natural regular map $X \rightarrow Y$ if $Y$ is the variety representing the $G$-orbits. As we already know that the affine coordinate ring is an invariant, we can first study the functions on $Y$, and try to see if there is any nice $k$-algebra homomorphism $A(Y) \rightarrow A(X)$.

The natural class of functions on $Y$ should be those functions on $X$, who have consistent value on each $G$-orbit. This is exactly $A(X)^G$, the $G$-invariants of $A(X)$.

The first question is, can $A(X)^G$ possibly be an affine coordinate ring? It is clear that $A(X)^G$ has no nilpotents, being a subalgebra of $A(X)$.

Proposition 3 Let $m$ be the order of $G$. Suppose that $char \, k \nmid m$, then $A(X)^G$ is a finitely generated $k$-algebra.

We have a natural inclusion map $A(X)^G \rightarrow A(X)$, which tells us that if $Y$ is an affine variety such that $A(Y) = A(X)^G$, then we have a map $\theta: X \rightarrow Y$ induced by the above inclusion. We then want to see if $Y$ really represents the orbits.

Proposition 4 $x,y$ are in the same $G$-orbit if and only if $\theta(x) = \theta(y)$. Furthermore, $\theta$ is onto.

Proof of proposition 3

Define the averaging operator

$\displaystyle H(x) = \frac{1}{m} \sum_{f \in G} f(x)$

for $x \in A(X)$.This is a map from $A(X)$ to $A(X)^G$.

Let $x_1, \dots, x_n$ be the generators of $A(X)$, this means that for arbitrary $x \in A(X)$, we have an expression

$x = \sum a_{i_1\cdots i_n} x_1^{i_1}\cdots x_n^{i_n}$

where $a_{i_1 \cdots i_n} \in k$. Since $x_1^{i_1}\cdots x_n^{i_n}$ are not fixed by $G$, we hope that we can get something after averaging. Yet $H$ does NOT behave well with respect to multiplication.

So first let us deal with the simple case, WLOG, $x = x_1^{q}$. By definition,

$\displaystyle H(x) = \frac{1}{m} \sum_{f \in G} f(x_1)^q$

So this is like a power sum. Then by Newton’s identity, we would want to consider the polynomial $p(t) = \prod_{f \in G} (t - x_i)$. The coeffcients of the polynomial lie in $A(X)^G$, and $x_1^q$ can be expressed as a sum of (coefficients of $p(t)$ * $x_1^i$), where $1 \leq i \leq m$, the order of $G$.

Learning from the monomial case, we tackle the case of a general multinomial in the same fashion. WLOG consider $x_1^{i_1}\cdots x_n^{i_n}$, with $i_1 > m$. Using the same idea, we see that it can be expressed as the sum of (coefficients of $p(t)$ * $x_1^i\cdots x_n^{i_n}$), where $1 \leq i \leq m$, the order of $G$.

This shows that in general, an arbitrary monomial can be written as the sum of (coefficients of $p(t)$ * $x_1^{e_1}\cdots x_n^{e_n}$), where $1 \leq e_1,\cdots,e_n \leq m$, the order of $G$. Notice that $H$ is additive, and that there are only finitely many elements in $A(X)^G$ to be of the form (coefficients of $p(t)$ * $x_1^{e_1}\cdots x_n^{e_n}$), where $1 \leq e_1,\cdots,e_n \leq m$, we see that after taking $H$, these are the desired generators of $A(X)^G$. $\Box$

Proof of proposition 4

(I can’t find a good proof for surjectivity)

If $x,y$ are in the same $G$-orbit,

recall that how $\theta$ is constructed. We take a set of generators of $A(X)^G$, embed it into $A(X)$, and use these regular functions to define $\theta$. This means that the coordinate functions are all $G$-invariant. Therefore $\theta(x) = \theta(y)$ for $x,y$ are in the same orbit.

If $x,y$ are not in the same $G$-orbit,

we want to show that for some set of generators of $A(X)^G$, one of them would have different values at $x$ and $y$. This is equivalent to finding a $G$-invariant polynomial that has different values at $x$ and $y$.

Construct a polynomial $p$ such that $p(f(x)) = 1$ and $p(f(y)) = 0$ for all $f \in G$. (This can be done using Lagrange interpolation, analogous to the one-variable case) Then we symmetrize it by considering

$q(x) = \displaystyle \frac{1}{m} \sum_{f \in G} p(f(x))$

giving us one of the desired polynomials that separate $x$ and $y$. $\Box$

For surjectivity, one can proceed using commutative algebra, but I still can’t see conceptually why it is onto.

The inclusion $A(X)^G \rightarrow A(X)$ is integral, because for any $x \in A(X)$, the polynomial $\displaystyle \prod_{f \in G} (t - f(x))$ is one integral polynomial for $x$ with coefficients in $A(X)^G$. It is not difficult to show that $Spec(A(X)) \rightarrow Spec(A(X)^G)$ is a closed map. Since $A(X)^G \rightarrow A(X)$ is an inclusion, the Spec map is dense, thus onto.