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I assume each algebra here is commutative with a unit.

Let $\nu : A \hookrightarrow B$ be a finite embedding of algebras (that is $B$ is finitely-generated as an $A$-module). I want to show the following:

  1. That the map $\nu^{-1} : \mathrm{Spec}(B) \to \mathrm{Spec}(A)$ is onto and has finite fibers. That is, every prime ideal in $A$ has a preimage which is a prime ideal in $B$, and there are only finitely many such ideals in $B$.

  2. An ideal $P \lhd B$ is maximal if and only if $\nu^{-1}(P) \lhd A$ is maximal.

My attempts:

My thoughts on 1) where to extend the ideal $P \lhd A$ to $B$ via $P \mapsto B\nu(P)$ and then take $\nu^{-1}(B \nu(P))$ and show it is equal to $P$, however, there is no $\nu^{-1}(B)$ since $\nu$ is not surjective.

We note that $B\nu(P)$ is not necessarily a prime ideal. Therefore we must take the maximal ideal $I$ in $B$ containing $B\nu(P)$ and such that $I \cap \nu(A) = \nu(P)$. We need to prove that $I$ is prime ideal and that $\nu^{-1}(I) = P$.

My thoughts on 2) where to use the characteriziation that $P \lhd B$ is maximal if and only if $B/P$ is a field. However, lack of surjectivity still causing me troubles. By using a Proposition that say that if $\beta$ is algebraic over a field $F$ then $F[\beta]$ is a field, using the fact that $B$ is finite over $A$ I can prove that $$ A/\nu^{-1}(P) \mbox{ is a field } \Rightarrow B/P \mbox{ is a field } $$ since the quotient $B/P$ is finite over $A/\nu^{-1}(P)$ and has a finite number of generators, all are algebraic. Now I need to prove the other direction. Any thoughts?

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Does it obvious that $I \cap \nu(A) = \nu(P)$ and in particular $B\nu(P) \cap \nu(A) = \nu(P)$ or it needs a proof? If so, how do I prove it? – LinAlgMan Nov 18 '12 at 19:03
up vote 1 down vote accepted

Finite algebras are integral and the results you want (except for finiteness of fibers) are valid more generally if you replace "finite" by "integral": the proofs are in Atiyah-Macdonald's Introduction to Commutative Algebra, Corollary 5.8 and Theorem 5.10.

For the finiteness of the fiber over the prime ideal $\mathfrak p\subset A$, use that this fiber is in bijection with the primes of the ring $B\otimes_A Frac (A/\mathfrak p)$, which is also finite over $Frac (A/\mathfrak p)$, to reduce to the case where $A$ is a field.
Then use that $B$ is trivially artinian to conclude that $B$ has only finitely many primes (which are maximal) thanks to Proposition 8.3 of our reference.

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Thank you. I guess in frac you mean the field of fractions. Also, could you elaborate on the bijection between the fibers and the tensor product? I know (in the definition level) what is an Artinian ring, but we havn't cover it in class. I do have Atiyah-McDonalds and read the above mentioned propositions. – LinAlgMan Nov 19 '12 at 18:36
Dear LinAlgMan, I recommend you try to solve Exercise 21 of Chapter 3 in Atiyah-Macdonald: part iv) gives the required bijection (which is even a homeomorphism). – Georges Elencwajg Nov 22 '12 at 19:18

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