Zariski's lemma
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 (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: if "I" is a proper ideal of formula_1 ("k" algebraically closed field), then "I" has a zero; i.e., there is a point "x" in formula_2 such that formula_3 for all "f" in "I". (Proof: replacing "I" by a maximal ideal  formula_4, we can assume formula_5 is maximal. Let formula_6 and formula_7 be the natural surjection. By the lemma formula_8 is a finite extension. Since "k" is algebraically closed that extension must be "k". Then for any formula_9, that is to say, formula_11 is a zero of formula_4.) 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". 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 formula_13 where formula_14 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., formula_15 . 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.) Proof: 2. formula_16 1.: Let formula_17 be a prime ideal of "A" and set formula_18. We need to show the Jacobson radical of "B" is zero. For that end, let "f" be a nonzero element of "B". Let formula_4 be a maximal ideal of the localization formula_20. Then  formula_21 is a field that is a finitely generated "A"-algebra and so is finite over "A" by assumption; thus it is finite over formula_18 and so is finite over the subring formula_23 where formula_24. By integrality, formula_25 is a maximal ideal not containing "f". 1. formula_16 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 formula_4 of "A" not containing "a". Writing "K" for some algebraic closure of formula_32, the canonical map formula_33  extends to formula_30. Since "B" is a field, formula_35 is injective and so "B" is algebraic (thus finite algebraic) over formula_32. 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 formula_37 be the generators of "B" as "A"-algebra. Then each formula_38 satisfies the relation where "n" depends on "i" and formula_40. Set formula_41. Then formula_42 is integral over formula_43. Now given formula_28, we  first extend it to formula_45 by setting formula_46. Next, let formula_47. By integrality, formula_48 for some maximal ideal formula_49 of formula_42. Then formula_51 extends to formula_52. Restrict the last map to "B" to finish the proof. formula_53