In algebra, Zariski's lemma, proved by, states that, if a field is finitely generated as an associative algebra over another field, then is a finite field extension of . An important application of the lemma is a proof of the weak form of Hilbert's nullstellensatz: if I is a proper ideal of , then I has a zero; i.e., there is a point x in such that for all f in I. The lemma may also be understood from the following perspective. In general, a ringR is a Jacobson ringif and only if every finitely generated R-algebra that is a field is finite over R. 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. For Zariski's original proof, see the original paper. 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 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. 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 ofB. 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: 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 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 and so we can assume B is algebraic over A. 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.
