Zariski's lemma

In algebra, Zariski's lemma, proved by {{harvs|txt|first=Oscar|last= Zariski|authorlink=Oscar Zariski|year=1947}}, states that, if a field {{math|K}} is finitely generated as an associative algebra over another field {{math|k}}, then {{math|K}} is a finite field extension of {{math|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:{{sfn|Milne|2017|loc=Theorem 2.12}} if I is a proper ideal of $k[t_1, ..., t_n]$ (k an algebraically closed field), then I has a zero; i.e., there is a point x in $k^n$ such that $f(x) = 0$ for all f in I. (Proof: replacing I by a maximal ideal $\mathfrak{m}$ , we can assume $I = \mathfrak{m}$ is maximal. Let $A = k[t_1, ..., t_n]$ and $\phi: A \to A / \mathfrak{m}$ be the natural surjection. By the lemma $A / \mathfrak{m}$ is a finite extension. Since k is algebraically closed that extension must be k. Then for any $f \in \mathfrak{m}$ ,

: $f(\phi(t_1), \cdots, \phi(t_n)) = \phi(f(t_1, \cdots, t_n)) = 0$ ;

that is to say, $x = (\phi(t_1), \cdots, \phi(t_n))$ is a zero of $\mathfrak{m}$ .)

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.{{sfn|Atiyah|MacDonald|1969|loc=Ch 5. Exercise 25}} Thus, the lemma follows from the fact that a field is a Jacobson ring.

Proofs

Two direct proofs are given in Atiyah–MacDonald;{{sfn|Atiyah|MacDonald|1969|loc=Ch 5. Exercise 18}}{{sfn|Atiyah|MacDonald|1969|loc=Proposition 7.9}} the one is due to Zariski and the other uses the Artin–Tate lemma. For Zariski's original proof, see the original paper.{{sfn|Zariski|1947|pp=362–368}} 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.)

{{math theorem|math_statement={{sfn|Atiyah|MacDonald|1969|loc=Ch 5. Exercise 25}} Let A be a ring. Then the following are equivalent.

A is a Jacobson ring.
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

| last1 = Atiyah | first1 = Michael

| last2 = MacDonald | first2 = Ian G.

| author1-link = Michael Atiyah

| author2-link = Ian G. Macdonald

| year = 1969

| publisher = Addison–Wesley

| series = Addison-Wesley Series in Mathematics

| isbn = 0-201-40751-5

}}

{{Cite web| title = Algebraic Geometry

| last = Milne | first = James

| author-link = James Milne (mathematician)

| url = http://www.jmilne.org/math/CourseNotes/ag.html

| date = 19 March 2017 | access-date = 1 February 2022

}}

{{Cite journal | title = A new proof of Hilbert's Nullstellensatz

| last = Zariski | first = Oscar

| journal = Bulletin of the American Mathematical Society

| date = April 1947 | volume = 53 | issue = 4 | pages = 362–368

| url = http://projecteuclid.org/euclid.bams/1183510605

| doi = 10.1090/s0002-9904-1947-08801-7 | mr = 0020075

| doi-access = free

}}

Category:Lemmas in algebra

Category:Theorems about algebras