2Algebra and geometry

The exposition in this chapter is a supplement to Chapter 1 in Fulton's book.

2.1 Algebraic sets

Let $k$ denote a field, not necessarily the real numbers or complex numbers. It might be a field of positive charateristic.

Give an example of an infinite field of characteristic $p > 0$ , where $p$ is a prime number. Why is an algebraically closed field infinite? Do you know a concrete example of an algebraically closed field of characteristic $p>0$ ?

The geometry in Algebraic Geometry comes from loking at points in affine $n$ -dimensional space over $k$ . This object is defined by

$k^n = \{(a_1, \dots, a_n) \mid a_1, \dots, a_n\in k\}.$

2.1.1 From ideals to subsets

Consider finitely many polynomials $f_1, \dots, f_m\in R = k[x_1, \dots, x_n]$ . The set of solutions to

$\begin{aligned} f_1(x_1, \dots, x_n) &= 0\\ &\vdots\\ f_m(x_1, \dots, x_n) &= 0 \end{aligned}$ in $k^n$ is denoted $V(f_1, \dots, f_m)$ . If we wish to be completely formal,

$V(f_1, \dots, f_m) = \{(a_1, \dots, a_n)\in k^n \mid \forall i=1, \dots, m:\, f_i(a_1, \dots, a_n) = 0\}.$ A subset of this form is called an algebraic set. We may even generalize more and introduce

$V(I) = \{(a_1, \dots, a_n)\in k^n \mid \forall f\in I: f(a_1, \dots, a_n) =0 \}$ for an ideal $I \subset R$ .

Prove that

$V(f_1, \dots, f_m) = V(\langle f_1, \dots, f_m \rangle)$ for $f_1, \dots, f_m\in R$ .

2.1.2 The Hilbert Basissatz

A natural question arises. Suppose $I\subset R$ is an ideal. Then $V(I)$ can be viewed as the solution set to potentially infinitely many polynomial equations. Is $V(I)$ an algebraic set? Is it the solution set to finitely many polynomial equations?

The answer is affirmative and contained in the famous Hilbert Basissatz. Reading through the proof below makes me think of Gordan's famous words

das ist keine Mathematik, das ist Theologie!

In a sense, Gordan is right. The proof is divine and not very constructive.

If $I\subset R$ is an ideal, then there exists finitely many polynomials $f_1, \dots, f_m\in R$ , such that

$I = \langle f_1, f_2, \dots, f_m\rangle,$ i.e., $I$ is finitely generated.

Here is the proof from Wikipedia.

We will prove more generally, that if $R$ is a commutative ring, where every ideal is finitely generated, then the same property holds for the polynomial ring $R[x]$ .

Suppose for contradiction that there exists an ideal $I\subseteq R[x]$ , which is not finitely generated. Then we may find a sequence $f_1, f_2, \dots$ of polynomials in $I$ , such that

$0\subsetneq \langle f_1\rangle \subsetneq \langle f_1, f_2\rangle\subsetneq \cdots$
$\deg(f_1)\leq \deg(f_2)\leq \cdots$

At each step in the process, we simply pick a polynomial of minimal degree in

$I\setminus \langle f_1, \dots, f_m\rangle.$ Let $a_i\in R$ denote the leading coefficient of $f_i$ . Then the increasing chain

$\langle a_1\rangle\subseteq \langle a_1, a_2\rangle \subseteq \cdots$ of ideals in $R$ must stabilize i.e., there exists $N$ , such that

$\langle a_1, \dots, a_N\rangle = \langle a_1, \dots, a_N, a_{N+1}\rangle = \cdots$ Suppose that

$a_{N+1} = \lambda_1 a_1 + \cdots + \lambda_N a_N.$ Then

$f_{N+1} - \lambda_1 x^{\deg(f_{N+1})- \deg(f_1)} f_1 - \cdots - \lambda_N x^{\deg(f_{N+1})- \deg(f_N)} f_N \tag{2.1}$ is a polynomial of degree strictly less than $\deg(f_{N+1})$ . The polynomial (2.1) cannot belong to $\langle f_1, \dots, f_N\rangle$ . On the other hand it cannot not belong either. This is a contradiction.

2.1.3 From subsets to ideals

Let $X \subset k^n$ . Then $I(X)$ denotes the ideals of polynomials vanishing on $X$ i.e.,

$I(X) = \{f\in R \mid \forall (a_1, \dots, a_n)\in X: f(a_1, \dots, a_n) = 0\}.$

Let $J$ be an ideal in a commutative ring $R$ . Then

$\operatorname{rad}(J) = \{r\in R \mid \exists N\in \mathbb{N}: n\geq N \implies r^n \in J\}.$

Prove that $\operatorname{rad}(J)$ is an ideal. Show that $J \subset \operatorname{rad}(J)$ and given an example of an ideal, where $J\subsetneq \operatorname{rad}(J)$ .

Let $J$ be an ideal in a commutative ring $R$ . Prove that the radical $\operatorname{rad}(J)$ is the intersection of the prime ideals $P\supset J$ containing $J$ .

Suppose that $f^n\not\in J$ for $n\in \mathbb{N}$ and consider

$S = \{f^n \mid n\in \mathbb{N}\}$ and the set of ideals

$\mathcal{I} = \{\mathfrak{a}\supseteq J \mid \mathfrak{a} \cap S = \emptyset\}.$ Show that $\mathcal{I}$ contains a maximal element $\mathfrak{p}$ with respect to inclusion and that $\mathfrak{p}$ is a prime ideal containing $J$ .

Apply Zorn's lemma to $\mathcal{I}$ to conclude the existence of a maximal element $\mathfrak{p}\in \mathcal{I}$ . If $\mathfrak{p}$ is not a prime ideal, there exists $x, y\not\in \mathfrak{p}$ , such that $x y \in \mathfrak{p}$ . Show that this leads to a contradiction.

2.1.4 Eisenbud-Evans

So given an algebraic subset in $k^n$ it is always a finite intersection of algebraic hypersurfaces $V(f)$ for $f\in k[x_1, \dots, x_n]$ . This is a consequence of Theorem 2.4. It turns out that $n$ hypersurfaces is enough! This is a famous result by Eisenbud and Evans. Graham Evans was a professor at the University of Illinois, where I got my phd. He passed away recently and Eisenbud (along with Phil Griffith) wrote obituaries.

Here is an excerpt from the Eisenbud-Evans Inventiones paper.

The paper itself is very readable and starts with nice summary of the history:

Then they move on to the proof itself. Here one needs to know some basic commutative algebra.

2.1.5 The Zariski topology on $k^n$

Suppose that $I$ and $J$ are ideals in a commutative ring. What is the definition of the ideal $I J$ ?

$\{x y \mid x\in I, y\in J\}$

No! It is a bit more subtle.

Give an example of two ideals $I, J$ in a ring $R$ , where

$I J \neq \{x y \mid x\in I, y\in J\}.$

The following result is not too hard to prove (do it!). Especially the last item is fun. It combines the other items.

Let $X_1\subseteq X_2$ be subsets of $k^n$ , $S_1\subseteq S_2$ subsets of $R$ with $J_1=\langle S_1\rangle$ and $J_2 = \langle S_2\rangle$ . Then

$V(S_1)\supseteq V(S_2)$
$I(X_1)\supseteq I(X_2)$
$V(J_1)\cup V(J_2) = V(J_1 J_2)$
$\bigcap_i V(J_i) = V\left(\sum_i J_i\right)$ for a family $(J_i)$ of ideals in $R$ .
The algebraic subsets form a topology (of closed subsets) on $k^n$ .
$V(I(X))\supseteq X$ , where $X$ is a subset of $k^n$ .
$I(V(S))\supseteq S$ , where $S$ is a subset of $R$ .
$V(I(V(S)) = V(S)$ , where $S$ is a subset of $R$ .
$I(V(J)) \supseteq \operatorname{rad}(J)$ , where $J$ is an ideal of $R$ .
$V(I(X)) = \overline{X}$ i.e., the closure of $X$ .

The topology alluded to in the above result is called the Zariski topology on $k^n$ . It is named after the legendary algebraic geometer Oscar Zariski.

Compared to the usual topology on for example $\mathbb{C}$ it is incredibly coarse. The closed subsets of $\mathbb{C}$ are precisely the finite ones (why?). It is not Hausdorff, since two non-empty open subsets always have non-trivial intersection. Since the open subsets are the cofinite ones, the Zariski topology makes $\mathbb{C}$ into a compact topological space.

2.2 Irreducible components

In the previous chapter we defined what it means for a topological space to be irreduible.

An algebraic subset $V$ is irreducible if and only if $I(V)$ is a prime ideal.

To show that every algebraic subset is a finite union of irreducible subsets, we need the following result on noetherian rings.

Every non-empty collection of ideals in a noetherian ring $R$ has a maximal element.

This translates into the fact that every collection of algebraic subsets has a minimal member! Now the proof of the following theorem becomes accessible. In fact the idea of the proof is very similar to the proof of the fundamental theorem of arithmetic! Just think about it: let $m$ be the smallest natural number not having a prime factorization etc.

Every algebraic subset $V$ has a unique decomposition into irreducible subsets $V_1, \dots, V_m$ , such that

$V = V_1 \cup \cdots \cup V_m$ and $V_i\nsubseteq V_j$ for $i\neq j$ .

Try to prove this using the above hints. Now, just because there is a slick proof does NOT mean that the irreducible components are easy to find.

2.3 Primary decomposition

According to Theorem 2.15 an algebraic subset $V\subset k^n$ is the union of irreducible algebraic subsets in an essentially unique way. If we translate this to algebra, it means

$I(V) = I(V_1) \cap \cdots \cap I(V_m)$ or that a radical ideal is the intersection of the minimal prime ideals above it. We are interested in a generalization of this result. We want it to hold for all ideals, not just radical ideals.

A first guess looking at the non-radical ideal $(x^2)\subset \mathbb{C}[x]$ could be that every ideal is the intersection of powers of prime ideals. The exercise below shows that this is wrong.

let $I = (x^2, y) \subset \mathbb{C}[x, y]$ . Show that $I$ is not an intersection of powers of prime ideals.

The right ideals in this context are primary ideals.

An ideal $I\subset R$ in a commutative ring $R$ is called irreducible if $I = J_1 \cap J_2 \implies I = J_1 \lor I = J_2$ , where $J_1, J_2\subset R$ are ideals of $R$ .

It is called primary if $\forall a, b\in R$ ,

$a b \in I\quad\land\quad a\not\in I \quad \implies \quad \exists n\in \mathbb{N}:\quad b^n\in I.$

The third part of the exercise below was inspired by a great in class interaction.

Show that an ideal $I\subset R$ is primary if and only if the zero divisors in $R/I$ are nilpotent.
Show that the radical of a primary ideal is a prime ideal.
Suppose that $Q$ is a primary ideal with $\operatorname{rad}(Q) = P$ . Prove that if $a b\in Q$ and $a\not\in P$ , then $b\in Q$ .
Suppose that $Q_1$ and $Q_2$ are primary ideals with $\operatorname{rad}(Q_1) = \operatorname{rad}(Q_2) = P$ . Prove that $Q_1 \cap Q_2$ is a primary ideal with $\operatorname{rad}(Q_1 \cap Q_2) = P$ .

Is a power of a prime ideal a primary ideal? Is an ideal $I$ a primary ideal if $\operatorname{rad}(I)$ is a prime ideal?

Consider $R = k[x, y, z]/(x y - z^2)$ along with the prime ideal $\mathfrak{p} = ([x], [z])$ and $\mathfrak{q} = \mathfrak{p}^2$ . Focus on the identity $[x] [y] = [z]^2$ in $R$ .

Is an ideal $I$ a primary ideal if $\operatorname{rad}(I)$ is a maximal ideal?

Show that an irreducible ideal $I \subset R$ in a noetherian ring $R$ is primary.

We may assume that $I = (0)$ is an irreducible ideal by moving to $R/I$ and in there prove that $(0)$ is a primary ideal. Assume that $a\neq 0$ and $a b = 0$ . The annihilator $\operatorname{ann}(x) = \{r\in R \mid r x = 0\}$ is an ideal for $x\in R$ . Apply the ascending chain

$\operatorname{ann}(b) \subset \operatorname{ann}(b^2) \subset \cdots$ of ideals to prove that $b^n = 0$ for some $n$ by showing that

$(a) \cap (b^n) = (0).$

The result below is often referred to as the Lasker-Noether theorem. It may be viewed as a purely algebraic analogue of the fact that every integer is a product of prime numbers.

Let $I\subset R$ be an ideal in a noetherian commutative ring $R$ . Then there exists irredundant primary ideals $Q_1, \dots, Q_r$ in $R$ , such that

$I = Q_1 \cap \cdots \cap Q_r$ and the prime ideals $\operatorname{rad}(Q_i)$ are distinct.

First prove that $I$ can be written as the intersection of finitely many irreducible ideals: suppose there exists ideals without a finite irreducible decomposition. Pick a maximal one among these (every non-empty set of ideals in a noetherian ring has a maximal element with respect to inclusion). Proceed like in the proof of the existence of prime factorizations of natural numbers.

Use Exercise 2.20.

Then move on to prove the last statement in the theorem by applying Exercise 2.18 (ⅳ.).

2.3.1 Monomial ideals

For $v\in \mathbb{N}^n$ . We use the notation $x^v$ for the corresponding monomial in $k[x_1, \dots, x_n]$ i.e., if for example $v = (1, 2, 3)$ , then $x^v = x_1\, x_2^2\, x_3^3$ .

An ideal $I$ in $k[x_1, \dots, x_n]$ is called a monomial ideal if it is generated by monomials:

$I = \langle x^{v_1}, \dots, x^{v_m}\rangle. \tag{2.2}$ A set of monomial generators as in (2.2) is called minimal if

$x^{v_i} \nmid x^{v_j}$ for $i\neq j$ .

Let $I = \langle x^{v_1}, \dots, x^{v_m}\rangle$ be a monomial ideal.

Prove that a monomial $x^v\in I$ if and only if $x^{v_i} \mid x^v$ for some $i = 1, \dots, m$ .
Prove that a polynomial $f\in k[x_1, \dots, x_n]$ belongs to $I$ if and only if every term (monomial with non-zero coefficient) in $f$ belongs to $I$ .
Prove that a monomial ideal is uniquely given by a set of minimal generators.
Prove that the intersection of two monomial ideals is a monomial ideal.
Prove that the product of two monomial ideals is a monomial ideal.
Suppose that $I$ and $J$ are monomial ideals. Is $I J = I \cap J$ ?

Irreducible decompositions (and thereby primary decompositions) of monomial ideals can be computed using the result below.

Let $I$ be a monomial ideal and $m\in I$ a minimal generator with $m = m_1 m_2$ , where $m_1$ and $m_2$ are relatively prime monomials. Then

$I = (I + \langle m_1\rangle)\cap (I+\langle m_2 \rangle). \tag{2.3}$ A monomial ideal is irreducible if and only if it is generated by powers of a subset of the variables.

If $m_1$ and $m_2$ are relatively prime monomials, then $m_1\mid x^v$ and $m_2\mid x^v$ if and only if $m_1 m_2 \mid x^v$ . This proves (2.3).

We have not proved above that a monomial ideal is irreducible if and only if it is generated by powers of a subset of the variables like for example

$(x^2, y^3, z^5) \subset k[x, y, z, w]. \tag{2.4}$ A light version of this result says the following. Suppose that $M$ is a monomial ideal in $R = k[x_1, \dots, x_n]$ generated by powers of some of the variables (as in (2.4)). Suppose that

$M = M_1 \cap M_2,$ where $M_1$ and $M_2$ are monomial ideals. Then $M = M_1$ or $M = M_2$ . Prove this.

Extra credit

Show in the above setting, that if $M = J_1 \cap J_2$ , where $J_1$ and $J_2$ are arbitrary ideals, that $M = J_1$ or $M = J_2$ .

I have only seen a proof of this once in the literature (see [1, Theorem 3.2.4][1]: W. Frank Moore, Marks Rogers and Sean Sather-Wagstaff. Monomial Ideals and Their Decompositions) and it seemed rather complicated. Would be nice with a short (and natural?) proof.

Give an example of a primary ideal that is not irreducible.

Show that

$(x^2, x y) = (x) \cap (x^2, y) = (x) \cap (x, y)^2.$ Why is this example worth noticing? What does Macaulay2 prefer?

Primary decomposition is built into Macaulay2:

2.4 Computation

Consider

$X = \{(t, t^2, t^3) \mid t\in k\} \subset k^3.$

Below you see a bit of Macaulay2 magic used to compute $I(X)$ .

The magic technology in the above Macaulay2 window involves the command eliminate(t, I). What does this command do? In the situation above, it simply computes the ideal

$(x - t, y - t^2, z - t^3) \cap k[x, y, z].$ This operation is called eliminating $t$ .

To find the polynomials vanishing on $(t, t^2, t^3)$ i.e.,

$I(\{(t, t^2, t^3) \mid t\in k\})$ is the same as finding the kernel of the $k$ -algebra homomorphism $\varphi: k[x, y, z] \rightarrow k[t]$ given by

$\begin{aligned} \varphi(x) &= t\\ \varphi(y) &= t^2\\ \varphi(z) &= t^3. \end{aligned}$

Surprisingly,

$\operatorname{Ker}(\varphi) = k[x, y, z] \cap (x - t, y - t^2, z - t^3).$

Here is how to prove this. For a polynomial $f(x, y, z) \in k[x, y, z, t]$ we may use the division algorithm with respect to the lexicographic term order $x > y > z > t$ to write

$f(x, y, z) = \lambda_1(x, y, z, t) (x - t) + \lambda_2(x, y, z, t) (x - t^2) + \lambda_3(x, y, z, t) (x - t^3) + R(t),$ where $R(t)\in k[t]$ . This shows that $f(t, t^2, t^3) = 0$ if and only if $f(x, y, z)$ is in the ideal generated by $x - t, y - t^2$ and $z - t^3$ .

Let

$X = \{(t, t^2, t^3) \mid t\in k\}\subseteq k^3$ where $k$ is a field. Show that

$I(X) = (y^2 - x z, x y - z, x^2 - y)$ and that $I(X)$ can be generated by two polynomials.

Let

$X = \{(t^3, t^4, t^5) \mid t\in k\},$ where $k$ is a field. Show that

$I(X) = (x^3 - y z, y^2 - x z, z^2 - x^2 y)$ and that $I(X)$ cannot be generated by less than three polynomials.

Consider the ideal $I = I(X)$ from Exercise 2.31. Show that $I$ is a prime ideal, but $I^2$ is not a primary ideal. Use Macaulay2 to compute a primary decomposition of $I^2$ .

Consider

$f_2^2 - f_1 f_3 = x (x^5 - 3 x^2 y z + x y^3 + z^3)\in I^2,$ where

$\begin{aligned} f_1 &= y^2 - x z\\ f_2 &= y z - x^3\\ f_3 &= z^2 - x^2 y. \end{aligned}$ Then use Exercise 2.18 (ⅲ.).

2.5 The Hilbert Nullstellensatz

2.5.1 Ring extensions

Let $K\subseteq L$ be a field extension. An element $\alpha\in L$ is called algebraic over $K$ if $f(\alpha) = 0$ for some $f\in K[x]$ . If every element in $L$ is algebraic over $K$ , then $L$ is called an algebraic extension of $K$ .

$L$ is naturally a vector space over $K$ and we let $[L:K] = \dim_K L$ . If $[L:K]<\infty$ then $L$ is an algebraic field extension (why?), but not the other way round (example?).

We wish to generalize this to integral domains $R\subseteq S$ , but here we have to be a bit more careful. We call an element $s\in S$ integral over $R$ if there exists a monic polynomial $f\in R[x]$ , such that $f(s) = 0$ i.e.,

$s^n + a_{n-1} s^{n-1} + \cdots + a_1 s + a_0 = 0$ for suitable $n\in\mathbb{N}$ and $a_0, \dots, a_{n-1}\in R$ . If every $s\in S$ is integral over $R$ , then $S$ is called an integral extension of $R$ .

If $S$ is finitely generated as an $R$ -module, then $R\subseteq S$ is an integral extension.

Suppose that $S = R s_1 + \cdots + R s_n$ and let $s\in S$ . Then we may write

$\begin{aligned} s s_1 &= \lambda_{11} s_1 + \cdots + \lambda_{1n} s_n\\ &\vdots\\ s s_n &= \lambda_{n1} s_1 + \cdots + \lambda_{nn} s_n \end{aligned}$ This translates into

$\begin{pmatrix} \lambda_{11} - s & \cdots & \lambda_{1n}\\ \vdots & \ddots &\vdots\\ \lambda_{n1} & \cdots & \lambda_{nn} - s \end{pmatrix} \begin{pmatrix} s_1\\ \vdots\\ s_n \end{pmatrix} = \begin{pmatrix} 0\\ \vdots\\ 0 \end{pmatrix}.$ By linear algebra the determinant of the above matrix is zero, but this determinant is a monic polynomial (up to a sign) of degree $n$ in $s$ .

Let us record the following result.

Let $R\subseteq L$ be an integral ring extension, where $L$ is a field. Then $R$ is a field.

If $r\in R\setminus\{0\}$ , then $r^{-1}$ satisfies

$(r^{-1})^n + a_{n-1} (r^{-1})^{n-1} + \cdots + a_1 r^{-1} + a_0 = 0.$ Multiplying the above by $r^n$ gives that $r$ is invertible in $R$ .

Our main result is the following.

Let $K\subseteq L$ be a field extension and suppose that

$L = K[\alpha_1, \dots, \alpha_n]$ for suitable $\alpha_1, \dots, \alpha_n\in L$ . Then $\alpha_1, \dots, \alpha_n$ are all algebraic over $K$ .

For $n=1$ the result is true (why?). Since $L = K(\alpha_1)[\alpha_2, \dots, \alpha_n]$ , we may assume by induction that $\alpha_2, \dots, \alpha_n$ are algebraic over $K(\alpha_1)$ . If $\alpha_1$ is algebraic over $K$ we are done, because this implies that $\alpha_2, \dots, \alpha_n$ are algebraic over $K$ .

Assume this is not the case. Then there exists $f\in K[\alpha_1]$ , such that the elements $\alpha_2, \dots, \alpha_n$ are integral over the subring $R = K[\alpha_1, 1/f]$ . This implies that the ring extension $R\subseteq L$ is finite and hence integral by Lemma 2.33. Therefore $R$ is a field by Lemma 2.34.

But for an element $f\in K[x]$ in a polynomial ring, $K[x, 1/f]$ is never a field.

2.5.2 The weak Hilbert Nullstellensatz

The major surprise is the following result:

Let $K$ be an algebraically closed field and $J\subseteq K[x_1, \dots, x_n]$ an ideal. Then

$V(J) = \emptyset \iff 1\in J.$

Theorem 2.36 is sometimes referred to as the weak Nullstellensatz, which I think is a bit of a misnomer. There is also a much more difficult version called the effective Nullstellensatz: if $1\in J = (f_1, \dots, f_m)$ , then

$1 = a_1 f_1 + \cdots + a_m f_m,$ for suitable $a_1, \dots, a_m\in K[x_1, \dots, x_n]$ . If $\deg(f_i) \leq d$ , what can be said about $\deg(a_i)$ above for $i = 1, \dots, n$ ? This question can be traced back to the classical paper from 1926 by Grethe Hermann.

If $1\in J$ , then clearly $V(J)=\emptyset$ . If on the other hand $1\notin J$ , then $J$ is a proper ideal contained in a maximal ideal $M$ . Consider the field

$L = K[x_1, \dots, x_n]/M = K[\alpha_1, \dots, \alpha_n],$ where $\alpha_1 = [x_1], \dots, \alpha_n=[x_n]$ By Theorem 2.35, $\alpha_i$ is algebraic over $K$ . Since $K$ is algebraically closed, it follows that $\alpha_i = [x_i] = \lambda_i\in K$ . Therefore we get

$(x_1 - \lambda_1, \dots, x_n - \lambda_n) \subseteq M,$ but the ideal to the left is already a maximal ideal, so equality must hold. Therefore

$J \subseteq (x_1 - \lambda_1, \dots, x_n-\lambda_n)$ and hence

$(\lambda_1, \dots, \lambda_n)\in V(J).$ Therefore $V(J) \neq \emptyset$ .

2.5.3 The Rabinowitz trick

The following result is called the strong Nullstellensatz (the result above is called the weak Nullstellensatz). The slick proof uses the socalled Rabinowitz trick published in a one page paper on Mathematische Annalen in 1929.

Let $K$ be an algebraically closed field and $J\subseteq K[x_1, \dots, x_n]$ an ideal. Then

$I(V(J)) = \operatorname{rad}(J).$

Suppose that $g\in I(V(J))$ . Then we consider the ideal

$I = (J, x_{n+1} g - 1) \subset K[x_1, \dots, x_n, x_{n+1}].$

Notice that

$V(I) = \emptyset.$

Therefore

$1 = \lambda_1 f_1 + \cdots + \lambda_m f_m + \mu (x_{n+1} g - 1).$ Now put $x_{n+1} = 1/g$ and multiply with $g^N$ for $N\gg 0$ .

Here is an interesting application of Theorem 2.38, where linear algebra is used to bound the number of solutions to certain systems of polynomial equations. We will have more to say about this later.

Let $I$ be an ideal in $R = K[x_1, \dots, x_n]$ , where $K$ is any field. Then $V(I)$ is finite if $R/I$ is a finite dimensional vector space over $K$ . In this case,

$|V(I)| \leq \dim_K R/I. \tag{2.5}$ If $K$ is algebraically closed then $R/I$ is finite dimensional if $V(I)$ is finite. If $I = \operatorname{rad}(I)$ , then $=$ holds in (2.5).

Given $P_1, \dots, P_r\in V(I)$ , we may find polynomials, such that $F_i(P_j) = \delta_{ij}$ . In one variable you know this from Lagrange interpolation: for points $x_1, \dots, x_r\in \mathbb{R}$

$F(x) = \frac{(x-x_2)(x-x_3)\cdots (x-x_r)}{(x_1 - x_2)(x_1 - x_3) \cdots (x_1 - x_r)}$ is a polynomial in $\mathbb{R}[x]$ satisfying $F(x_1) = 1$ and $F(x_i) = 0$ for $i > 1$ . What is the generalization to several variables?

The residue classes of these polynomials in $R/I$ are linearly independent over $K$ : suppose that

$\lambda_1 [F_1] + \cdots + \lambda_r [F_r] = [\lambda_1 F_1 + \cdots \lambda_r F_r] = [0]$ in $R/I$ . Then $\lambda_1 F_1 + \cdots + \lambda_r F_r \in I$ with

$(\lambda_1 F_1 + \cdots + \lambda_r F_r)(P_i) = 0$ for $i = 1, \dots, r$ . You do the rest of the math. This proves that

$|V(I)| \leq \dim_K R/I.$

Suppose on the other hand, that $V(I) = \{P_1, \dots, P_r\}$ with $P_i = (a_{i1}, \dots, a_{in})$ . Let

$F_j = (x_j - a_{1j})(x_j - a_{2j}) \cdots (x_j - a_{nj})$ for $j = 1, \dots, n$ . Then $F_j\in I(V(I))$ . Therefore $F_j^{N_j}\in I$ for $N_j\gg 0$ by HNS. As the leading term of $F_j$ is a power of $x_j$ , this gives that $R/I$ is finite dimensional, since we have a surjective vector space homomorphism

$R/(F_1^{N_1}, ,\dots, F_n^{N_n}) \rightarrow R/I.$

Suppose that $I = \operatorname{rad}(I)$ and consider the map $R/I \rightarrow K^r$ given by

$f \mapsto (f(P_1), \dots, f(P_r)),$ where $V(I) = \{P_1, \dots, P_r\}$ . We have already proved that this map is surjective. If $f$ maps to zero, then $f\in I(V(I)) = \operatorname{rad}(I) = I$ .

Give an example of an ideal $I\subset R = K[x_1, \dots, x_n]$ , where $V(I)$ is finite, but $R/I$ is infinite dimensional.

Consider $I = (y^2 - x^3 + x, y^3 - x^2) \subset R = K[x, y]$ . Use the window below

to find a polynomial $f\not\in I$ , such that $f^m\in I$ for some $m>1$ (verify that $f\not\in I$ and $f^m\in I$ ).

Find a vector space basis for $R/I$ and use this basis to compute

$K[y] \cap I.$

Show that $x^3 - y^2 - x$ and $y^3 - x^2$ form a Groebner basis for $I$ with respect to the degree lexicographic term order. Use this to prove that

$1, x, y, x^2, x y, y^2, x^2 y, x y^2, x^2 y^2$ is a vector space basis for $R/I$ . Once you have this basis, use it to find a linear relation between

$1, y, y^2, y^3, \dots$

Use this to solve the equations

$\begin{aligned} y^2 - x^3 + x &= 0\\ y^3 - x^2 &= 0. \end{aligned}$

It seems that there are $7$ solutions according to the homotopy methods of numerical algebraic geometry.

Notice the rounding problems for the solution $(0, 0)$ when evaluating the window below. It seems that three homotopy paths end up in $(0, 0)$ .

2.6 The combinatorial Nullstellensatz

The following result is due to Noga Alon.

Let $F$ be any field (not necessarily algebraically closed). Suppose that $S_1, \dots, S_n\subset F$ are finite subsets of $F$ and

$g_i(x_i) = \prod_{s\in S_i} (x_i - s)$ for $i = 1, \dots, n$ . If $f\in F[x_1, \dots, x_n]$ and

$f(s_1, \dots, s_n) = 0$ for $s_1\in S_1, \dots, s_n\in S_n$ , then

$f = h_1 g_1 + \cdots + h_n g_n,$ where $\deg(h_i) \leq \deg(f) - \deg(g_i)$ .

It seems that one may prove this dividing $f$ by $(g_1, \dots, g_n)$ using the division algorithm and then applying the following result to the remainder.

Let $P\in F[x_1, \dots, x_n]$ , where $F$ is a field. If $S_i\subset F$ with $|S_i| > \deg_{x_i}(P)$ and

$P(x_1, \dots, x_n) = 0\qquad\text{for every }\qquad(x_1, \dots, x_n)\in S_1\times \cdots \times S_n,$ then $P = 0$ .

For $n=1$ this is the statement that a non-zero polynomial can have at most $\deg(P)$ roots. Write

$P = \sum_{i=0}^{N} P_i(x_1, \dots, x_{n-1}) x_n^i,$ where $\deg_{x_n}(P)$ . Fix $(s_1, \dots, s_{n-1})\in S_1\times \cdots \times S_{n-1}$ and put

$f = \sum_{i=0}^{N} P_i(s_1, \dots, s_{n-1}) x_n^i\in F[x_n].$ If non-zero this polynomial must have $>N$ roots, which is a contradiction. Therefore $P_i(s_1, \dots, s_{n-1}) = 0$ for every $(s_1, \dots, s_{n-1})\in S_1 \times \cdots \times S_{n-1}$ and the result follows by induction.

Let $K$ be a finite field and $F\in K[x_1, \dots, x_n]$ a homogeneous polynomial with $\deg(F) < n$ . Prove that $F$ has a non-trivial zero. Is this result true if $K$ is an infinite field?

Hint

Suppose that $K$ has $p$ elements, where $p$ is a prime. Assume that $V(F) = \{(0, \dots, 0)\}$ and apply the combinatorial Nullstellensatz to the polynomial

$f = 1 - F(x_1, \dots, x_n)^{p-1} + (-1)^{n+1} (x_1^{p-1} -1)\cdots (x_n^{p-1}- 1)$ to reach a contradiction through

$f = h_1 (x_1^p - x_1) + \cdots + h_n(x_n^p - x_n)$ and $\deg(h_i) \leq n(p-1) - p$ .