## Hilbert’s Nullstellensatz

Definition: If $I\subset R$ is an ideal, $I$ is called radical if, whenever $f^n\in I$, we have $f\in I$.
Alternatively, $I$ is radical if $R/I$ has no nilpotent elements. If $I$ is any ideal, then $\mathrm{rad}(I)$ is the set of elements $f$ such that $f^{m}\in I$ for some $m$. Finally, the radical of the zero ideal, which is the set of nilpotent elements in $R$, is called the nilradical of $R$.

Remark: We’ve seen at various times in the past that the nilpotent elements of a (commutative) ring form an ideal.

Proposition: If $I$ is a proper ideal of $R$, then the radical of $I$ is the intersection of all the prime ideals of $R$ containing $I$.

Proof: It’s enough to prove that the nilradical of $R/I$ is the intersection of all prime ideals of $R/I$. If $P\supset I$ is a prime ideal, and $f^{n}\in I$ for some $n$, choose the smallest such $n$. Then $f^{n}\in P$ so either $f^{n-1}\in P$ or $f\in P$. By minimality of $n$, this means that $f\in P$. So the nilradical is contained in every prime ideal.

For the converse, suppose that $a$ is not a nilpotent element of $R$ (and is not a unit in $R$). Then we will construct a prime ideal $P$ that does not contain $a$. Let $A$ be the set of powers of $a$: $A={a,a^2,a^3,\ldots}$ and let $S$ be the set of ideals of $R$ not meeting $A$. This is a nonempty set, since it contains the zero ideal. If $I_1\subset I_2\subset\cdots$ is a chain of ideals in $S$, then the union of the $I_{k}$ is again an ideal in $S$, so chains in $S$ have upper bounds. By Zorn’s lemma, $S$ has a maximal element $Q$. Now suppose that $x$ and $y$ are elements of $R$ and $xy\in P$. Since $P$ is maximal in $S$, we know that some power of $a^r$ is in $(x)+P$ and some power of $a^s$ is in $(y)+P$. But then $a^{r+s}$ is in $xy+P=P$ since $xy\in P$. This is a contradiction, since $P$ is in $S$. It follows that one of $x$ or $y$ must have been in $P$, so $P$ is prime.

Corollary: Prime (and maximal) ideals of $R$ are radical ideals.

### Integral Extensions

Definition: Let $S$ be a commutative $R$ algebra.

• An element $a\in S$ is integral over $R$ if it is the root of a monic polynomial in $R[x]$.
• If every element of $s$ is integral over $R$, then $S$ is called an integral extension of $R$.
• The subset of $S$ consisting of elements integral over $R$ is called the integral closure of $R$ in $S$.
• $R$ is integrally closed in $S$ if it is equal to its integral closure.
• If $R$ is an integral domain, and $R$ is integrally closed in its field of fractions, then $R$ is integrally closed (full stop) or normal. The integral closure of $R$ in its field of fractions is called its normalization.

Proposition: The following are equivalent:

• $a$ is integral over $R$.
• $R[a]$ is a finitely generated $R$ module.
• There is a subring $R\subset T\subset S$ containing $a$ wuch that $T$ is a finitely generated $R$-module

Proof: If $a$ satisfies the monic polynomial $x^n+r_{n-1}x^{n-1}+\cdots+r_0$, then any element of $R[a]$ can be written as a linear combination of $1,a,a^2,\ldots, a^{n-1}$. So $R[a]$ is finitely generated. The ring $R[a]\subset S$ is a finitely generated $R$ module inside $S$. Finally, if $a$ belongs to a finitely generated $R$ module $T$, choose generators for $T$ $t_1,\ldots, t_n$ over $R$ and consider the equations

$at_{i}=\sum r_{ij}t_{j}$

The element $a$ satisfies the (monic) characteristic polynomial made from the entries $r_{ij}$, so $a$ is integral over $R$.

Corollary: The sum and product of integral elements are integral; the integral closure of $R$ in $S$ is a subring of $S$; and if $S$ is integral over $R$ and $T$ is integral over $S$ then $T$ is integral over $R$.

Corollary: Let $\tilde{R}$ be the integral closure of $R$ in $S$. Then $\tilde{R}$ is integrally closed.

Proof: If $x\in S$ is integral over $\tilde{R}$, then since $\tilde{R}$ is integral over $R$, we have $x$ is integral over $R$ so belongs to $\tilde{R}$.

Proposition: Suppose that $S$ is an $R$-algebra that is integral over $R$. Then $R$ is a field if and only if $S$ is a field.

Proof: Suppose first that $R$ is a field. Choose $s\in S$. Then

$s^{n}+r_{n-1}s^{n-1}+\cdots+r_{0}=0$

where we can assume $r_{0}\not=0$. Then

$s(s^{n-1}+r_{n-1}s^{n-2}+\cdots+r_{1})=-r_{0}.$

Since $-r_0\not=0$, we can divide the polynomial on the right by $-r_{0}$ to obtain a multiplicative inverse for $s$.

Now suppose that $S$ is a field. If $r\in R$, then $r\in S$, so $r^{-1}\in S$. We have

$r^{-m}+r_{m-1}r^{-m-1}+\cdots+r_{0}=0$

so by clearing demoninators we can write $r^{-1}$ as an element of $R$.

### Noether Normalization

Definition: Elements $x_1,\ldots, x_n$ in a $k$-algebra $S$ are called algebraically independent if there are no nonzero polynomial relations among them: there are no polynomials $p$ so that $p(x_1,\ldots, x_n)=0$. In other words, they generate a copy of $k[x_1,\ldots, x_n]\subset S$.

Theorem: (Noether Normalization) Let $k$ be a field and let $A$ be a finitely generated $k$-algebra. Then there are algebraically independent elements $y_1,\ldots, y_q$ in $A$ such that $A$ is integral over $k[y_1,\ldots, y_q]$.

Proof: The proof is by induction and is (more or less) algorithmic. Start with generators $x_1,\ldots, x_n$ for $A$. If they are algebraically independent, you’re done. Otherwise you have a polynomial relation

$p(x_1,\ldots, x_n)=0.$

This is a sum of monomials $x_1^{a_1}\cdots x_n^{a_n}$. The degree of $p$ is the largest of the sums of these exponents; call that $d$. Then let $\alpha$ be any integer bigger than $d$ ($d+1$ works fine).

Introduce new coordinates $X_i$ (for $i=1,\ldots, n-1$) by:

$\begin{array}{rcl} x_1 &=& X_1 + x_n^{\alpha}\\ x_2 &=& X_2 + x_n^{\alpha^2}\\ \vdots &=& \vdots\\ x_{n-1} &=& X_{n-1} + x_n^{\alpha^{n-1}}\\ \end{array}$

If we substitute the new coordinates, we get $p(X_1+x_m^{\alpha},\cdots, X_{n-1}+x_n^{\alpha^{n-1}},x_n)=0$. But a monomial $x_1^{a_1}\cdots x_n^{a_n}$ will contribute a term

$x_n^{a_n+a_{n-1}\alpha^{n-1}+a_{n-2}\alpha^{n-2}+\cdots+a_1\alpha}$

and since we choose $\alpha$ bigger than $d$ we have all $a_i<\alpha$. In other words, all of these exponents of $x_n$ are distinct (they are different in base $\alpha$).

It follows that the polynomial $p(X_1+x_m^{\alpha},\cdots, X_{n-1}+x_n^{\alpha^{n-1}},x_n)$ has the form

$p(X_1+x_m^{\alpha},\cdots, X_{n-1}+x_n^{\alpha^{n-1}},x_n)=cx_m^{N}+\sum H_{i}(X_1,\ldots, X_{n-1})x_m^{i}$

and so $x_m$ is integral over the subring $B=k[X_1,\ldots, X_{m-1}]$. But then $x_i$ for $i=1,\ldots,n-1$ are integral over $B[x_{m}]$ because they satisfy the equations $x_i-X_{i}-x_{n}^{\alpha^{i}}$. Therefore $A$ is integral over $B$ (which has fewer generators). Continue by induction.

Theorem: (the “weak” nullstellensatz) Let $k$ be an algebraically closed field and let $A=k[x_1,\ldots, x_n]$. Then the maximal ideals $M$ of $A$ are all of the form $$M=(x-a_1,\ldots, x-a_n)$$ where the $a_i\in k$.

Corollary: The correspondence between ideals and algebraic sets gives a bijection between points and maximal ideals of $\mathbb{A}^{n}_{k}$.

Corollary: Let $f_1,\ldots, f_k\in A$. Then either the $f_{i}$ have a common zero, or there are polynomials $g_1,\ldots, g_k$ in $A$ such that

$1=\sum g_{i}f_{i}.$

Proof: (of the theorem) Clearly an ideal of the form $(x_1-a_1,\ldots, x_n-a_n)$ is maximal, so suppose $M$ is a maximal ideal of $A$. Let $E=A/M$. Then $E$ is a finitely generated $k$-algebra, so there are algebraically independent elements $y_1,\ldots, y_k$ such that $E$ is integral over $k[y_1,\ldots, y_k]$. But $E$ is a field, so $k[y_1,\ldots, y_k]$ is a field. But this can only happen if $k=0$. Then $E/k$ is a finite integral (i.e. algebraic) extension of $k$, and $k$ is algebraically closed, so $E=k$. This means that each of the generators $x_i$ is congruent mod $M$ to an element of $k$, or in other words $M$ is of the desired form.

For the corollaries, any proper ideal $I$ of $A$ is contained in a maximal ideal $M$, so if $X(I)$ contains the point corresponding to $M$. So the points of $X(I)$ correspond to the maximal ideals containing $I$.

Finally, if the $f_i$ have no common zero, then they must not generate a proper ideal, so the ideal they generate contains $1$.

Theorem: (Nullstellensatz, “strong” form) Let $k$ be an algebraically closed field. Then if $J\subset A$ is any ideal, $I(X(J))=\mathrm{rad}(J)$. Thus (assuming $k$ is algebraically closed) there are mutually inverse bijections between algebraic sets in $\mathbb{A}^{n}_{k}$ and radical ideals in $A$.

Proof: We know that $\mathrm{rad}(J)\subset I(X(J))$ so we need to prove the opposite. We know that $J$ is finitely generated, say by $f_1,\ldots, f_k$. Let $g$ be any polynomial vanishing on $X(J)$. Make a new ring $A’$ by introducing a new variable $x_{n+1}$ and a new ideal $J’\in A’$ by adding the relation $gx_{n+1}-1$. (Notice that this means that $x_{n+1}$ is $1/g$.) If the elements of $J’$ had a common zero, all of the $f_{i}$ would vanish at that point and since $g\in I(X(J))$ so would $g$. But that doesn’t happen, so $J’$ can’t be a proper ideal and so we have an equation

$1=h_1 f_1 +\cdots + h_k f_k + h_{k+1}(x_{n+1}g-1)$

We can divide this equation by a high power of $x_{n+1}$ so that the powers of $h_{n+1}$ in the coefficients $h_{i}$ are all negative. In other words, writing $y=1/x_{n+1}$, we get an equation

$y^{N} = b_1 f_1 +\cdots +b_k f_k + b_{k+1}(g-y)$

where the $b_{i}$ are polynomials in $x_1,\ldots, x_n$ and $y$. This is an equation in $A’$, where $g=1/x_{n+1}$, so this means (substituting $g$ for $y$) that we have an equation showing that $g^{N}$ is in the ideal generated by the $f_{i}$, so $g\in\mathrm{rad}(J)$.

Corollary: If $k$ is not algebraically closed, then we can still conclude that a set of polynomials that generates a proper ideal of $k[x_1,\ldots, x_n]$ must have common zeros in the algebraic closure of $k$.