Random rambling about k algebras
11 Mar 2013In trying to verify II.3.7 in [H], we came upon a discussion which I would like to summarize here:
Let k be a field, then any k-algebra R with finitely many prime ideals has dimension 0. In general, a ring R with finitely many prime ideals is not necessarily Artin. The case here follows from the following facts.
1) Noether normalization: any k-algebra R is finite (as a module) over k[x_1,...,x_n] for some n. In the case where k is infinite see [AM, Ex 5.16], and the general case, see [E, Sect. 8.2.1].
2) k[x] (a PID and thus a UFD) necessarily has infinitely many prime ideals, since it has infinitely many irreducible polynomials, each of which generates a prime. (The Euclid’s argument used to prove this is due to EK: if there were only finitely many irreducible polynomials f_1,...,f_k, consider 1 + \prod_k f_k, which is either irreducible or contains an irreducible factor not one of f_1,...,f_k.)
It follows from (1) and (2) that if R is finite over k[x_1,...,x_n], and if R has only finitely many primes, then n = 0 and R is finite over k. That is, R is a finite dimensional vector space.
In this way, for any domain B, and A finitely generated B-domain. If A \otimes_B K(B) is a K(B)-algebra with finitely many primes, by Noether normalization, A \otimes_B K(B) is finite over K(B)[x_1,\cdot\cdot\cdot,x_n]; by the above, it must be integral over K(B). That is, A \otimes_B K(B) is a field equal to K(A).
In preparation for what is to come, we prove that for A and B as above, there exists some multiplicative set S \subset B such that A \otimes_B S^{-1}B is a finite S^{-1}B module.
To see this, notice that A is a finitely generated B-algebra. Write A = B[x_1,...,x_n]/I for some ideal I in the polynomial ring with coefficients in B. Certainly, each x_i is an element of K(A) and satisfies some monic polynomial with coefficient in K(B):