# Zariski's lemma

In algebra, Zariski's lemma, introduced by Oscar Zariski (1947), states that if K is a finitely generated algebra over a field k and if K is a field, then K is a finite field extension of k.

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 ${\displaystyle k[t_{1},...,t_{n}]}$ (k algebraically closed field), then I has a zero; i.e., there is a point x in ${\displaystyle k^{n}}$ such that ${\displaystyle f(x)=0}$ for all f in I.[2]

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.[3] 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.[4][5] For Zariski's original proof, see the original paper.[6] Another direct proof in the context 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 ${\displaystyle k[x_{1},\ldots ,x_{d}]}$ where ${\displaystyle x_{1},\ldots ,x_{d}}$ are algebraically independent over k. But since K has Krull dimension zero, the polynomial ring must have dimension zero; i.e., ${\displaystyle d=0}$.

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 — Let A be a ring. Then the following are equivalent.

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

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

1. ${\displaystyle \Rightarrow }$ 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 ${\displaystyle B\supset A}$ 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 ${\displaystyle \phi :A\to K}$, K an algebraically closed field, with ${\displaystyle \phi (a)\neq 0}$ extends to ${\displaystyle {\widetilde {\phi }}:B\to K}$.

Indeed, choose a maximal ideal ${\displaystyle {\mathfrak {m}}}$ of A not containing a. Writing K for some algebraic closure of ${\displaystyle A/{\mathfrak {m}}}$, the canonical map ${\displaystyle \phi :A\to A/{\mathfrak {m}}\hookrightarrow K}$ extends to ${\displaystyle {\widetilde {\phi }}:B\to K}$. Since B is a field, ${\displaystyle {\widetilde {\phi }}}$ is injective and so B is algebraic (thus finite algebraic) over ${\displaystyle A/{\mathfrak {m}}}$. 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 ${\displaystyle x_{1},\dots ,x_{r}}$ be the generators of B as A-algebra. Then each ${\displaystyle x_{i}}$ satisfies the relation

${\displaystyle a_{i0}x_{i}^{n}+a_{i1}x_{i}^{n-1}+\dots +a_{in}=0,\,\,a_{ij}\in A}$

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

## Notes

1. ^ Milne, Theorem 2.6
2. ^ Proof: it is enough to consider a maximal ideal ${\displaystyle {\mathfrak {m}}}$. Let ${\displaystyle A=k[t_{1},...,t_{n}]}$ and ${\displaystyle \phi :A\to A/{\mathfrak {m}}}$ be the natural surjection. By the lemma, ${\displaystyle A/{\mathfrak {m}}=k}$ and then for any ${\displaystyle f\in {\mathfrak {m}}}$,
${\displaystyle f(\phi (t_{1}),\cdots ,\phi (t_{n}))=\phi (f(t_{1},\cdots ,t_{n}))=0}$;
that is to say, ${\displaystyle x=(\phi (t_{1}),\cdots ,\phi (t_{n}))}$ is a zero of ${\displaystyle {\mathfrak {m}}}$.
3. ^ Atiyah-MacDonald 1969, Ch 5. Exercise 25
4. ^ Atiyah–MacDonald 1969, Ch 5. Exercise 18
5. ^ Atiyah–MacDonald 1969, Proposition 7.9
6. ^ http://projecteuclid.org/euclid.bams/1183510605