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 an algebraically closed field), then I has a zero; i.e., there is a point x in
(Proof: replacing I by a maximal ideal
be the natural surjection.
is a finite extension.
Since k is algebraically closed that extension must be k. Then for any
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.
Two direct proofs are given in Atiyah–MacDonald;[3][4] the one is due to Zariski and the other uses the Artin–Tate lemma.
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
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.,
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.
be a prime ideal of A and set
We need to show the Jacobson radical of B is zero.
For that end, let f be a nonzero element of B.
be a maximal ideal of the localization
and so is finite over the subring
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: Indeed, choose a maximal ideal
Writing K for some algebraic closure of
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).
satisfies the relation where n depends on i and
Restrict the last map to B to finish the proof.