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\title{Lecture 5-10: Noether normalization and the proof of the Nullstellensatz}
\date{May 10, 2024}
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As previously promised, we prove the Nullstellensatz, using the Noether Normalization Lemma.
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We begin by recalling the notion of integrality from an earlier discussion of character theory.  Given commutative rings $A\subset B$ we say that $b\in B$ is \alert{integral over $A$} if we have $b^n+\sum\limits_{i=0}^{n-1}a_ib^i=0$ for some $a_0,\ldots,a_{n-1}\in A$.  We have already seen that the set of elements in $B$ integral over $A$ is a subring containing $A$; we call it the \alert{integral closure} of $A$ in $B$ (DF, p. 691) and denote it by $\overline A$.  The integral closure of $\overline A$ in $B$ is then just $\overline A$.  An integral domain is called \alert{integrally closed} or \alert{normal} if it is integrally closed in its quotient field.  For example, we previously showed that $\mathbb Z$ is integrally closed; the same argument shows that any unique factorization domain is integrally closed.
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We need a simple lemma.
\begin{alertblock}{Proposition; DF, p. 694}
If $A$ and $B$ are integral domains with $B$ integral over $A$ (that is, every element of $B$ integral over $A$), then $A$ is a field if and only if $B$ is.
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If $A$ is a field and $b\in B$ with $b\ne0$ satisfies $b^n+\sum\limits_{i=0}^{n-1}a_ib^i=0$, then by cancelling a suitable power of $b$ we may assume that $a_0\ne0$, whence $b$ has the multiplicative inverse $-a_0^{-1}(b^{n-1}+\sum\limits_{i=1}^{n-1}a_ib^{i-1})$.  Conversely, if $B$ is a field, then any $a\in A$ with $a\ne0$ has a multiplicative inverse $a^{-1}$ in $B$, which must be integral over $A$, so that $a^{-n}+\sum\limits{i=0}^{n-1}c_ia^{-i}=0$ for some $c_i\in A$.  Multiplying by $a^{n-1}$, we see that $a^{-1}\in A$, as desired.
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Now we are ready to prove
\begin{alertblock}{Theorem; Noether Normalization Lemma; DF, p. 699}
Let $k$ be a field and suppose that $A=k[r_1,\ldots,r_n]$ is a finitely generated $k$-algebra.  Then for some $m\le n$ there are elements $y_1\ldots,y_m$ algebraically independent over $k$ such that $A$ is integral and finitely generated as a module over the $k$-subalgebra $k[y_1,\ldots,y_m]$.
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\begin{proof}
By induction on $n$.  If the $r_i$ are already algebraically independent then there is nothing to prove.  Otherwise there is a nonzero $f$ in the polynomial ring $P_n=k[x_1,\ldots,x_n]$, say of degree $d$, with $f(r_1,\ldots,r_n)=0$.  Renumbering the variables if necessary, we may assume that $f$ is a nonconstant polynomial in $x_n$ with coefficients in $k[x_1\ldots,x_{n-1}]$.  We now change variables, transforming $f$ into a monic polynomial in $x_n$ whose coefficients lie in a subring of $A$ generated by $n-1$ elements.
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\begin{proof}
Define integers $\alpha_i=(1+d)^i$ and new variables $X_i=x_i-x_n^{\alpha_i}$ for $1\le i\le n-1$.  Let $g(X_1,\ldots,X_{n-1},x_n) = f(X_1+x_n^{\alpha_1},\ldots,X_{n-1}+x_n^{\alpha_{n-1}},x_n)$, so that $g$ is a polynomial in the $X_i$ and $x_n$.  Each monomial term of $f$ contributes a single term of the form a constant times $x_n^e$ to $g$ for some $e$.  The choice of $\alpha_i$ guarantees that different terms of $f$ give different values of $e$.  Hence if $N$ is the highest power of $x_n$ that occurs in $g$ then we have $g=cx_n^N+\sum\limits_{i=0}^{N-1} h_i(X_1,\ldots,X_{n-1})x_n^i$ for some nonzero $c$.  Setting $s_i=r_i-r_n^{\alpha_i}$, we get $\frac 1 cg(s_1\ldots,s_{n-1},r_n) = \frac 1 cf(r_1,\ldots,r_n)=0$, so that $r_n$ is integral over $B=k[s_1,\ldots,s_{n-1}]$.  Each $r_i$ for $1\le i\le n-1$ is also integral over $B[r_n]$, so $A$ is integral over $B[r_n]$ and thus also over $B$.  An appeal to the inductive hypothesis completes the proof.
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It follows at once form this lemma and the preceding one that \alert{a finitely generated $k$-algebra $A$ that is a field is a finite extension of $k$}, since Noether normalization implies that $A$ is integral over a polynomial ring $k[y_1,\ldots,y_m]$, which is a field if and only if $m=0$.  Then we get
\begin{alertblock}{Theorem:  Weak Nullstellensatz, DF, p. 700}
If $k$ is algebraically closed and $I$ is a proper ideal in a polynomial ring $P_n=k[x_1,\ldots,x_n]$ then $\mathcal V(I)\ne\emptyset$.
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\begin{proof}
Enlarge $I$ to a maximal ideal $M$ of $P_n$.  Then the coordinate ring $P_n/M$ is a finitely generated $k$-algebra that is a field, whence by algebraic closure it must be isomorphic to $k$.  If the surjection from $P_n$ to $k$ sends the variable $x_i$ to $a_i\in k$, then $\mathcal Z(M)$ is the point $(a_1,\ldots,a_n)\in k^n$, whence $\mathcal Z(M)$ and $\mathcal Z(I)$ are nonempty.
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Finally we are ready to prove the full Nullstellensatz.  
\begin{alertblock}{Theorem:  Nullstellensatz}
If $k$ is algebraically closed and $I\subset P_n$ is a proper ideal, then $\mathcal I(\mathcal V(I)) = \sqrt I$.  In particular the maps $\mathcal Z,\mathcal I$ define inverse inclusion-reversing bijections between Zariski closed subsets of $k^n$ and radical ideals in $P_n$.
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\begin{proof}
Clearly $\sqrt I\subseteq\mathcal I(\mathcal Z(I))$, so it remains to prove the reverse inclusion.  Let $f_1,\ldots,f_m$ be a finite set of generators of $I$ and let $g\in\mathcal I(\mathcal Z(I))$.  Introduce a new variable $x_{n+1}$ and consider the ideal $I'$ generated by $f_1,\ldots,f_m$ and $x_{n+1}g-1$ in $P_{n+1}$.  At any point of $\mathbf A^{n+1}$ where the $f_i$ vanish so too does $g$, whence $x_{n+1}g-1$ does not vanish.  Hence $\mathcal Z(I')=\emptyset$, whence $I'$ must be all of $P_{n+1}$.  Now we have an equation $1=a_1f_1+\cdots+a_mf_m+a_{m+1}(x_{n+1}g-1)$ for some $a_i\in P_{n+1}$.  Setting $y=\frac 1 {x_{n+1}}$ and multiplying by a high power of $y$ we get $y^N = c_1f_1+\cdots+c_mf_m+c_{m+1}(g-y)$ for some $c_i\in k[x_1,\ldots,x_n,y]$.  Substituting $g$ for $y$ in this last equation shows that $g^N\in I=(f_1,\ldots,f_m)$, so that $g\in\sqrt I$, as desired.
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A side benefit of Noether normalization is that it gives us another way to compute the dimension of an algebraic set $V$:  writing the coordinate ring $k[V]$ as a finitely generated integral extension of a polynomial ring $k[y_1,\ldots,y_d]$, we take the dimension of $V$ to be $d$.  This is justified since if $V$ is irreducible then the quotient field of $k[V]$ is a finite extension of the rational function field in $d$ variables over $k$, so has transcendence degree $d$.  But now it turns out that there is more that we can say about the morphism $V\rightarrow\mathbf A^d$ giving rise to the inclusion $k[y_1,\ldots,y_d]\subset k[V]$.  
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As an example of this, consider again the subvariety $V$ of say $\mathbb C^2$ defined by the equation $x^3-y^2=0$.  We have seen that the coordinate ring of $V$ may be identified with the subring $R=\mathbb C[t^2,t^3]$ of $S=\mathbb C[t]$; the inclusion of $R$ into $S$ corresponds to the morphism $t\mapsto(t^2,t^3)$ of $\mathbf A^1(=\mathbb C)$ to $V$, which is bijective.  But we also have the maps $S\rightarrow R$ sending $t$ to $t^2$, or $t$ to $t^3$; these correspond to the projections $\pi_1,\pi_2$ from $V$ onto its first and second coordinates.  These maps are generically two-to-one and three-to one, respectively, though in both cases there is only one preimage of 0, namely the origin $(0,0)$.  Thus these maps are not covering maps of topological spaces; we call them \alert{ramified finite covers}, since not all fibers have the same size.
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More generally, we want to study the relationship between ideals of a ring $R$ and those of a ring $S$ containing $R$.  We call the ring $S$ an \alert{extension} of $R$.If $I$ is an ideal of $R$ then it generates an ideal $I^e=IS$ of $S$, called the \alert{extension} of $I$; similarly, given an ideal $J$ of $S$, its \alert{contraction} $J^c=J\cap R$ is an ideal of $R$.  Clearly any ideal $I$ of $R$ lies in the contraction $I^{ec}$ of its extension to $S$ and any ideal $J$ of $S$ contains the extension $J^{ce}$ of its contraction to $R$, but in general we do not get equality in either case.  The contraction $P=Q^c$ of a prime ideal $Q$ in $S$ is prime in $R$, since the quotient $R/P$, as a subring of $S/Q$, cannot have zero divisors if $S/Q$ does not.  On the other hand, the contraction of a maximal ideal in $S$ need not be maximal in $R$; nor is it true that every prime ideal of $R$, or even every maximal ideal, is the contraction of some ideal in $S$.  We focus on maximal ideals here because the Nullstellensatz implies that maximal ideal in the coordinate ring $k[V]$ of an algebraic set $V$ correspond bijectively to points of $V$.
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 If $S$ is integral over $R$, however, then we have more control over the situation.  Given a prime ideal $P$ of $R$ that is the contraction $Q^c$ of a prime ideal $Q$ of $S$, we know that $P$ is maximal if and only if $Q$ is maximal by a previous result, since $S/Q$ is integral over $R/P$ and both are integral domains.  If in addition $S$ is finitely generated as a ring over $R$, say by $s_1,\ldots,s_m$, then given any homomorphism $\pi$ from $R$ with kernel $P$ to a field $K$ there are only finitely many ways to extend  $\pi$ to $S$, since each generator must go to a root of a monic polynomial with specified coefficients.  It follows that \alert{there are only finitely many ideals $Q$ whose contraction is a fixed maximal ideal of $P$ of $R$, all of them maximal} (DF, Corollary 27, p. 695).  We will see next time that there is always at least one such ideal $Q$.
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