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Let $\phi: B \rightarrow A$ be a ring homomorphism and so $A$ becomes a $B$-module via this map. Let $\mathfrak p$ be a prime ideal of $B$. Is there a simple description of $A \otimes_B \textrm{Frac} (B/\mathfrak p)$? For instance, when $\mathfrak p$ is maximal then this equals $A/ (\phi ( \mathfrak p) \cdot A)$ but I am not sure how to compute this in general. Could somebody please help me out?

dekimashita
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    In general, this is the localization $(A/\mathfrak{p}A)_{(0)}$ where $(0)$ is the zero ideal of $B/\mathfrak{p}$. Can you prove it? Also, if you know any algebraic geometry: can you give a geometric interpretation of this result? – diracdeltafunk Sep 09 '20 at 06:30
  • Since you are localizing $A/\mathfrak p A$, could you clarify the statement that (0) is a prime ideal of $B/ \mathfrak p$? – dekimashita Sep 09 '20 at 06:52
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    Not a typo – $A/\mathfrak{p}A$ has a natural $B/\mathfrak{p}$-module structure, so it makes sense to localize $A/\mathfrak{p}A$ with respect to any multiplicative subset of $B/\mathfrak{p}$, including $(B/\mathfrak{p}) \setminus {0}$. It might be a good idea to begin by defining this $B/\mathfrak{p}$-module structure. – diracdeltafunk Sep 09 '20 at 06:58
  • Using $S^{-1}M=S^{-1}R \otimes_R M$, we get $(A/\mathfrak p A){(0)}=\textrm{Frac} (B/\mathfrak p) \otimes{B/\mathfrak p} A/ \mathfrak p A$. How do I proceed further? – dekimashita Sep 10 '20 at 04:56

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$\require{AMScd}$ First, observe that $A/\mathfrak{p}A\cong A\otimes_B B/\mathfrak{p}$ (see here if you're not familiar). Now, abstract nonsense tells us that $A\otimes_B\operatorname{Frac}(B/\mathfrak{p})\cong A/\mathfrak{p}A\otimes_{B/\mathfrak{p}}\operatorname{Frac}(B/\mathfrak{p}),$ because both inner squares in the commutative diagram \begin{CD} B @>>> B/\mathfrak{p} @>>> \operatorname{Frac}(B/\mathfrak{p}) \\ @VVV \lrcorner @VVV \lrcorner @VVV\\ A @>>> A/\mathfrak{p}A @>>> A/\mathfrak{p}A\otimes_{B/\mathfrak{p}}\operatorname{Frac}(B/\mathfrak{p}) \end{CD} are pushouts (this implies the outer square is a pushout as well). You've already seen why $A/\mathfrak{p}A\otimes_{B/\mathfrak{p}}\operatorname{Frac}(B/\mathfrak{p})\cong S^{-1}(A/\mathfrak{p}A),$ where $S= B/\mathfrak{p}\setminus\{0\},$ so this proves the desired isomorphism.

Stahl
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