Zariski's lemma

Proofs

Two direct proofs are given in Atiyah–MacDonald; the one is due to Zariski and the other uses the Artin–Tate lemma. 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 k[x_1, \ldots , x_d] where x_1, \ldots , x_d 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., 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.)

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

Proof: 2. \Rightarrow 1.: Let \mathfrak{p} be a prime ideal of A and set 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 \mathfrak{m} be a maximal ideal of the localization B[f^{-1}]. Then 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 B = A/\mathfrak{p} and so is finite over the subring B/\mathfrak{q} where \mathfrak{q} = \mathfrak{m} \cap B. By integrality, \mathfrak{q} is a maximal ideal not containing f.

1. \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 B \supseteq 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 \phi: A \to K, K an algebraically closed field, with \phi(a) \ne 0 extends to \widetilde{\phi}: B \to K.

Indeed, choose a maximal ideal \mathfrak{m} of A not containing a. Writing K for some algebraic closure of A/\mathfrak{m}, the canonical map \phi: A \to A/\mathfrak{m} \hookrightarrow K extends to \widetilde{\phi}: B \to K. Since B is a field, \widetilde{\phi} is injective and so B is algebraic (thus finite algebraic) over 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 x_1, \dots, x_r be the generators of B as A-algebra. Then each x_i satisfies the relation :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 a_{i0} \ne 0. Set a = a_{10}a_{20} \dots a_{r0}. Then B[a^{-1}] is integral over A[a^{-1}]. Now given \phi: A \to K, we first extend it to \widetilde{\phi}: A[a^{-1}] \to K by setting \widetilde{\phi}(a^{-1}) = \phi(a)^{-1}. Next, let \mathfrak{m} = \operatorname{ker}\widetilde{\phi}. By integrality, \mathfrak{m} = \mathfrak{n} \cap A[a^{-1}] for some maximal ideal \mathfrak{n} of B[a^{-1}]. Then \widetilde{\phi}: A[a^{-1}] \to A[a^{-1}]/\mathfrak{m} \to K extends to B[a^{-1}] \to B[a^{-1}]/\mathfrak{n} \to K. Restrict the last map to B to finish the proof. \square

Notes

Sources

• {{Cite book| title = Introduction to Commutative Algebra | author1-link = Michael Atiyah | author2-link = Ian G. Macdonald
• {{Cite web| title = Algebraic Geometry | author-link = James Milne (mathematician)
• {{Cite journal | title = A new proof of Hilbert's Nullstellensatz | doi-access = free