Zariski's lemma

In algebra, Zariski's lemma, proved by Oscar Zariski (1947), states that, if a field $K$ is finitely generated as an associative algebra over another field $k$ , then $K$ is a finite field extension of $k$ (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:^[1] if I is a proper ideal of $k[t_{1},...,t_{n}]$ (k 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 ${\mathfrak {m}}$ , we can assume $I={\mathfrak {m}}$ is maximal. Let $A=k[t_{1},...,t_{n}]$ and $\phi :A\to A/{\mathfrak {m}}$ be the natural surjection. Since k is algebraically closed, by the lemma, $A/{\mathfrak {m}}=k$ and then for any $f\in {\mathfrak {m}}$ ,

f(\phi (t_{1}),\cdots ,\phi (t_{n}))=\phi (f(t_{1},\cdots ,t_{n}))=0

;

that is to say, $x=(\phi (t_{1}),\cdots ,\phi (t_{n}))$ is a zero of ${\mathfrak {m}}$ .)

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.^[2] Thus, the lemma follows from the fact that a field is a Jacobson ring.

Proof

Two direct proofs, one of which is due to Zariski, are given in Atiyah–MacDonald.^[3]^[4] For Zariski's original proof, see the original paper.^[5] 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 $k[x_{1},\ldots ,x_{d}]$ where $x_{1},\ldots ,x_{d}$ 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.)

Theorem — ^[2] Let A be a ring. Then the following are equivalent.

A is a Jacobson ring.
Every finitely generated A-algebra B that is a field is finite over A.

Proof: 2. $\Rightarrow$ 1.: Let ${\mathfrak {p}}$ be a prime ideal of A and set $B=A/{\mathfrak {p}}$ . We need to show the Jacobson radical of B is zero. For that end, let f be a nonzero element of B. Let ${\mathfrak {m}}$ be a maximal ideal of the localization $B[f^{-1}]$ . Then $B[f^{-1}]/{\mathfrak {m}}$ is a field that is a finitely generated A-algebra and so is finite over A by assumption; thus it is finite over $B=A/{\mathfrak {p}}$ and so is finite over the subring $B/{\mathfrak {q}}$ where ${\mathfrak {q}}={\mathfrak {m}}\cap B$ . By integrality, ${\mathfrak {q}}$ 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\supset 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

\phi :A\to K

, K an algebraically closed field, with

\phi (a)\neq 0

extends to

{\widetilde {\phi }}:B\to K

.

Indeed, choose a maximal ideal ${\mathfrak {m}}$ of A not containing a. Writing K for some algebraic closure of $A/{\mathfrak {m}}$ , the canonical map $\phi :A\to A/{\mathfrak {m}}\hookrightarrow K$ extends to ${\widetilde {\phi }}:B\to K$ . Since B is a field, ${\widetilde {\phi }}$ is injective and so B is algebraic (thus finite algebraic) over $A/{\mathfrak {m}}$ . 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 $a_{i0}\neq 0$ . Set $a=a_{10}a_{20}\dots a_{r0}$ . Then $B[a^{-1}]$ is integral over $A[a^{-1}]$ . Now given $\phi :A\to K$ , we first extend it to ${\widetilde {\phi }}:A[a^{-1}]\to K$ by setting ${\widetilde {\phi }}(a^{-1})=\phi (a)^{-1}$ . Next, let ${\mathfrak {m}}=\operatorname {ker} {\widetilde {\phi }}$ . By integrality, ${\mathfrak {m}}={\mathfrak {n}}\cap A[a^{-1}]$ for some maximal ideal ${\mathfrak {n}}$ of $B[a^{-1}]$ . Then ${\widetilde {\phi }}:A[a^{-1}]\to A[a^{-1}]/{\mathfrak {m}}\to K$ extends to $B[a^{-1}]\to B[a^{-1}]/{\mathfrak {n}}\to K$ . Restrict the last map to B to finish the proof. $\square$