Zariski's lemma

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In algebra, Zariski's lemma, introduced by Oscar Zariski, states that if K is a finitely generated algebra over a field k and if K is a field, then K is a finite field extension of k.

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[t_1, ..., t_n] (k 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.[2]

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.[3] Thus, the lemma follows from the fact that a field is a Jacobson ring.

Proof[edit]

Two direct proofs, one of which is due to Zariski, are given in Atiyah–MacDonald.[4][5] 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 algebraically independent over k. But since K has Krull dimension zero, the polynomial ring must have dimension zero; i.e., d=0.

In fact, the lemma is a special case of the general formula \dim A = \operatorname{tr.deg}_k A for a finitely generated k-algebra A that is an integral domain, which is also a consequence of the normalization lemma.

Notes[edit]

  1. ^ Milne, Theorem 2.6
  2. ^ Proof: it is enough to consider a maximal ideal \mathfrak{m}. Let A = k[t_1, ..., t_n] and \phi: A \to A / \mathfrak{m} be the natural surjection. By the lemma, A / \mathfrak{m} = k and 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}.
  3. ^ Atiyah-MacDonald 1969, Ch 5. Exercise 25
  4. ^ Atiyah–MacDonald 1969, Ch 5. Exercise 18
  5. ^ Atiyah–MacDonald 1969, Proposition 7.9

References[edit]