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.

## Proof

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.

*A*is a Jacobson ring.- 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.

## Notes

**^**Milne, Theorem 2.12- ^
^{a}^{b}Atiyah-MacDonald 1969, Ch 5. Exercise 25 **^**Atiyah–MacDonald 1969, Ch 5. Exercise 18**^**Atiyah–MacDonald 1969, Proposition 7.9**^**http://projecteuclid.org/euclid.bams/1183510605

## References

- 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