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

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

  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

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