Zariski's lemma

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., $d=0$. For Zariski's original proof, see the original paper.[6]

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