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 algebraically closed field), then I has a zero; i.e., there is a point x in such that for all f in I. (Proof: replacing I by a maximal ideal , we can assume is maximal. Let and be the natural surjection. Since k is algebraically closed, by the lemma, and then for any ,


that is to say, is a zero of .)

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, 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 where 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. 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. 1.: Let 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. Let be a maximal ideal of the localization . Then is a field that is a finitely generated A-algebra and so is finite over A by assumption; thus it is finite over and so is finite over the subring where . By integrality, is a maximal ideal not containing f.

1. 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 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 , K an algebraically closed field, with extends to .

Indeed, choose a maximal ideal of A not containing a. Writing K for some algebraic closure of , the canonical map extends to . Since B is a field, is injective and so B is algebraic (thus finite algebraic) over . 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 be the generators of B as A-algebra. Then each satisfies the relation

where n depends on i and . Set . Then is integral over . Now given , we first extend it to by setting . Next, let . By integrality, for some maximal ideal of . Then extends to . Restrict the last map to B to finish the proof.



  • M. Atiyah, I.G. Macdonald, Introduction to Commutative Algebra, Addison–Wesley, 1994. ISBN 0-201-40751-5
  • James Milne, Algebraic Geometry
  • Zariski, Oscar (1947), "A new proof of Hilbert's Nullstellensatz", Bull. Amer. Math. Soc., 53: 362–368, doi:10.1090/s0002-9904-1947-08801-7, MR 0020075
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