Zariski's lemma

In algebra, Zariski's lemma, proved by Template:Harvs, states that, if a field Template:Math is finitely generated as an associative algebra over another field Template:Math, then Template:Math is a finite field extension of Template:Math (that is, it is also finitely generated as a vector space).

An important application of the lemma is a proof of the weak form of Hilbert's Nullstellensatz:Template:Sfn if I is a proper ideal of $k[t_1,...,t_n]$ (k an algebraically closed field), then I has a zero; i.e., there is a point x in $k^n$ such that $f(x)=0$ for all f in I. (Proof: replacing I by a maximal ideal $𝔪$, we can assume $I=𝔪$ is maximal. Let $A=k[t_1,...,t_n]$ and $\varphi :A\to A/𝔪$ be the natural surjection. By the lemma $A/𝔪$ is a finite extension. Since k is algebraically closed that extension must be k. Then for any $f\in 𝔪$,

$f(\varphi (t_1),\cdots ,\varphi (t_n))=\varphi (f(t_1,\cdots ,t_n))=0$;

that is to say, $x=(\varphi (t_1),\cdots ,\varphi (t_n))$ is a zero of $𝔪$.)

The lemma may also be understood from the following perspective. In general, a ring R is a Jacobson ring if and only if every finitely generated R-algebra that is a field is finite over R.Template:Sfn Thus, the lemma follows from the fact that a field is a Jacobson ring.

Proofs

Two direct proofs are given in Atiyah–MacDonald;Template:SfnTemplate:Sfn the one is due to Zariski and the other uses the Artin–Tate lemma. For Zariski's original proof, see the original paper.Template:Sfn Another direct proof in the language of Jacobson rings is given below. The lemma is also a consequence of the Noether normalization lemma. Indeed, by the normalization lemma, K is a finite module over the polynomial ring $k[x_1,\dots ,x_d]$ where $x_1,\dots ,x_d$ are elements of K that are algebraically independent over k. But since K has Krull dimension zero and since an integral ring extension (e.g., a finite ring extension) preserves Krull dimensions, the polynomial ring must have dimension zero; i.e., $d=0$.

The following characterization of a Jacobson ring contains Zariski's lemma as a special case. Recall that a ring is a Jacobson ring if every prime ideal is an intersection of maximal ideals. (When A is a field, A is a Jacobson ring and the theorem below is precisely Zariski's lemma.)

Proof: 2. $\Rightarrow$ 1.: Let $𝔭$ be a prime ideal of A and set $B=A/𝔭$. We need to show the Jacobson radical of B is zero. For that end, let f be a nonzero element of B. Let $𝔪$ be a maximal ideal of the localization $B[f^{-1}]$. Then $B[f^{-1}]/𝔪$ is a field that is a finitely generated A-algebra and so is finite over A by assumption; thus it is finite over $B=A/𝔭$ and so is finite over the subring $B/𝔮$ where $𝔮=𝔪\cap B$. By integrality, $𝔮$ is a maximal ideal not containing f.

1. $\Rightarrow$ 2.: Since a factor ring of a Jacobson ring is Jacobson, we can assume B contains A as a subring. Then the assertion is a consequence of the next algebraic fact:

(*) Let $B\supseteq A$ be integral domains such that B is finitely generated as A-algebra. Then there exists a nonzero a in A such that every ring homomorphism $\varphi :A\to K$, K an algebraically closed field, with $\varphi (a)\ne 0$ extends to $\tilde{\varphi }:B\to K$.

Indeed, choose a maximal ideal $𝔪$ of A not containing a. Writing K for some algebraic closure of $A/𝔪$, the canonical map $\varphi :A\to A/𝔪\hookrightarrow K$ extends to $\tilde{\varphi }:B\to K$. Since B is a field, $\tilde{\varphi }$ is injective and so B is algebraic (thus finite algebraic) over $A/𝔪$. We now prove (*). If B contains an element that is transcendental over A, then it contains a polynomial ring over A to which φ extends (without a requirement on a) and so we can assume B is algebraic over A (by Zorn's lemma, say). Let $x_1,\dots ,x_r$ be the generators of B as A-algebra. Then each $x_i$ satisfies the relation

$a_{i0}{x}_i^n+a_{i1}{x}_i^{n-1}+\dots +a_{in}=0,a_{ij}\in A$

where n depends on i and $a_{i0}\ne 0$. Set $a=a_{10}{a}_{20}\dots a_{r0}$. Then $B[a^{-1}]$ is integral over $A[a^{-1}]$. Now given $\varphi :A\to K$, we first extend it to $\tilde{\varphi }:A[a^{-1}]\to K$ by setting $\tilde{\varphi }(a^{-1})=\varphi (a{)}^{-1}$. Next, let $𝔪=\ker \tilde{\varphi }$. By integrality, $𝔪=𝔫\cap A[a^{-1}]$ for some maximal ideal $𝔫$ of $B[a^{-1}]$. Then $\tilde{\varphi }:A[a^{-1}]\to A[a^{-1}]/𝔪\to K$ extends to $B[a^{-1}]\to B[a^{-1}]/𝔫\to K$. Restrict the last map to B to finish the proof. $◻$

Sources

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