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, K is finitely generated as a module (in other words, K is a finite dimensional vector space) over k. An important application of the lemma is a proof of the weak form of Hilbert's Nullstellensatz: if I is a proper ideal of k [ t 1 , . . . , t n ] {\displaystyle 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 {\displaystyle k^{n}} such that f ( x ) = 0 {\displaystyle f(x)=0} for all f in I. (Proof: replacing I by a maximal ideal m {\displaystyle {\mathfrak {m}}} , we can assume I = m {\displaystyle I={\mathfrak {m}}} is maximal. Let A = k [ t 1 , . . . , t n ] {\displaystyle A=k[t_{1},...,t_{n}]} and ϕ : A → A / m {\displaystyle \phi :A\to A/{\mathfrak {m}}} be the natural surjection. By the lemma A / m {\displaystyle A/{\mathfrak {m}}} is a finite extension. Since k is algebraically closed that extension must be k. Then for any f ∈ m {\displaystyle f\in {\mathfrak {m}}} ,
f ( ϕ ( t 1 ) , ⋯ , ϕ ( t n ) ) = ϕ ( f ( t 1 , ⋯ , t n ) ) = 0 {\displaystyle f(\phi (t_{1}),\cdots ,\phi (t_{n}))=\phi (f(t_{1},\cdots ,t_{n}))=0} ; that is to say, x = ( ϕ ( t 1 ) , ⋯ , ϕ ( t n ) ) {\displaystyle x=(\phi (t_{1}),\cdots ,\phi (t_{n}))} is a zero of m {\displaystyle {\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. Thus, the lemma follows from the fact that a field is a Jacobson ring.
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 , … , x d ] {\displaystyle k[x_{1},\ldots ,x_{d}]} where x 1 , … , x d {\displaystyle 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 {\displaystyle 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.)
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