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Let $f:A\rightarrow B$ be a homomorphism of commutative rings, and $M$ a finite $B$-module. If $a\in A$ and $M_a$ is a free $A_a$-module, then for a prime ideal $\mathfrak q$ of $B$ with $f(a)\notin\mathfrak q$, how to prove that $M_{\mathfrak q}$ is a flat module over $A$ ?

nick
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    The $A$-algebra $B$ and the finiteness condition on $B$ have nothing to do here, forget about them. The deleted answer of @Rankeya is correct but can also be simplified by forgetting $B$. –  Apr 24 '13 at 05:55
  • Dear QiL'8, the assumption that $M$ is a finite $B$-module is not necessary, this question is from the proof of Theorem 24.3 in Matsumura's book "commutative ring theory". – nick Apr 24 '13 at 08:33

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Note that $M_a$ is free over $A_a$ implies that $M_a$ is flat over $A_a$ since free modules are flat. Since $A_a$ is a flat $A$ module (exactness of localization), we get that $M_a$ is flat over $A$. Now, $M_a = A_a \otimes_A M = A_a \otimes_A (B \otimes_B M) = (A_a \otimes_A B) \otimes_B M = B_{f(a)} \otimes_B M = M_{f(a)}$. Thus, $M_a$ is a $B_{f(a)}$-module. Then consider the ring map $A \rightarrow B_{f(a)}$ (this is the composition $A \rightarrow B \rightarrow B_{f(a)}$). Since $f(a) \notin \mathfrak{q}$, we get that $\mathfrak{q}B_{f(a)}$ is prime in $B_{f(a)}$. Let $\mathfrak{p}$ be a prime in $A$ such that $\mathfrak{q}B_{f(a)}$ lies over $\mathfrak{p}$. Then it suffices to show that $M_{\mathfrak{q}} = (M_a)_{\mathfrak{q}B_{f(a)}}$ is flat over $A_\mathfrak{p}$ (because $A_{\mathfrak{p}}$ is flat over $A$).

This latter assertion is shown in http://stacks.math.columbia.edu/tag/00HT part $(6)$. (Apply $(6)$ taking $R = A$, $A = B_{f(a)}$, $M = M_a$, where $R$, $A$, $M$ are the symbols used in part $(6)$)

Rankeya
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  • Dear Rankeya, how does the condition of $M$ being finite over $B$ come in your answer? Or is the hypothesis superfluous? –  Apr 23 '13 at 23:15
  • I believe it is superfluous. I had a comment in my answer for about 4 hours that my proof could be flawed because I don't use the finiteness of $M$. However, no one corrected me so far, so I assume my proof is okay. If not, please let me know :) – Rankeya Apr 23 '13 at 23:17
  • Dear Rankeya, thanks for your reply. I will read your proof more closely later (as I have to head to class now). Though whenever I see a question like this that has the words "finite as a module over (...)" and involves localization immediately I try to think of how I can apply Nakayama's Lemma. Regards, –  Apr 23 '13 at 23:28
  • Dear Rankeya，thanks for your detailed reply, and the finiteness of $M$ as a $B$-module is not necessary in this question. – nick Apr 24 '13 at 08:25
  • Dear Rankeya, How do we see that $A_a\otimes_A B\cong B_{f(a)}$. I can see it to be $B_a$. –  Sep 11 '15 at 21:35
  • It's an exercise that if $A \rightarrow B$ is a ring map, $S \subset A$ a multiplicative set, and $T$ is the image of $S$ in $B$, then $S^{-1}B \cong T^{-1}B$. Check Atiyah-Macdonal, Ch 3, Exercise 4. – Rankeya Sep 14 '15 at 14:48
  • @Rankeya : Why $M_{\mathfrak{q}} = (M_a){\mathfrak{q}B{f(a)}}$ ? By explicit construction of isomorphism? – Plantation Mar 10 '23 at 08:04