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Let $\phi: A\to B$ be a ring homomorphism and suppose $B$ is finitely generated as an $A$-module. Let $\mathfrak{p}\subseteq A$ be a prime ideal and let $K$ be the field of fractions of $A/\mathfrak{p}$. Is it true that the ring $B\otimes_A K$ has only finitely many prime ideals?

Motivation: I am trying to see why a finite morphism of schemes has finite fibres and I've been able to reduce to this problem (hopefully correctly) but I'm now unsure about how to proceed. I feel as though I'm either missing something totally obvious or else lacking in certain commutative algebra knowledge. Thanks for your help!

Alex Saad
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2 Answers2

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Yes, $C:=B\otimes _A K$ has only finitely many prime ideals.
Indeed, $C$ is finitely generated as a vector space over the field field $K$ and thus has only finitely many prime ideals because of the two following observations:

Observation 1: Every prime ideal $\mathfrak p\subset C$ is maximal.
This is equivalent to proving that the finite-dimensional domain $C/\mathfrak p$ is a field, which is shown here.

Observation 2: The ring $C$ has only finitely many maximal ideals.
Indeed if $\mathfrak m_1,\dots, \mathfrak m_r\subset C$ are maximal, the canonical map $$C\to C/\mathfrak m_1\times\dots \times C/\mathfrak m_r $$ is surjective by the Chinese remainder theorem and thus, since $\operatorname {dim}_K(C/\mathfrak m_i)\geq 1$, we have $r\leq \operatorname{dim}_K(C)$ and there are thus only finitely many maximal ideals in $C$.

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Georges Elencwajg
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  • A very very nice answer - thank you! Just to clarify the second step though: you show that given any finite collection of maximal ideals in $C$ the number of these is bounded by the dimension of $C$. But does this imply that given any collection of maximal ideals $\left{\mathfrak{m}_i\right}_i$ in $C$ there are at most $\dim_K C$ distinct maximal ideals? (i.e. Is that map still a surjection in the infinite case?) – Alex Saad Mar 13 '16 at 19:36
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Your question reduces easily to Finitely many prime ideals lying over $\mathfrak{p}$:

$B\otimes_A K\simeq B_{\mathfrak p}/\mathfrak pB_{\mathfrak p}$ where $B_{\mathfrak p}$ denotes the ring of fractions of $B$ with respect to the multiplicative set $A\setminus\mathfrak p$. Then a prime ideal in this factor ring is of the form $PB_{\mathfrak p}/\mathfrak pB_{\mathfrak p}$ with $P∩(A\setminus\mathfrak p)=∅$ (hence $P∩A⊆\mathfrak p$) and $\mathfrak pB_{\mathfrak p}⊆PB_{\mathfrak p}$ (which gives $\mathfrak p⊆P\cap A$). Then $P\cap A=\mathfrak p$.

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user26857
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  • Thank you very much for taking the time to explain the link to this other question. It's really helpful to now have 2 quite different proofs of this result! – Alex Saad Mar 13 '16 at 19:37