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$A\rightarrow B$ a ring homomorphism, $N$ a $B$-module which is flat over $A$. $\mathfrak{q}\subset B$ a prime ideal, $\mathfrak{p}\subset A$ its contraction in $A$. Then is it true that $N_{\mathfrak{q}}$ is flat over $A_{\mathfrak{p}}$? Bruns seems to be suggesting that, but I don't see how. I can see that $N_{\mathfrak{p}}$ is flat over $A_{\mathfrak{p}}$ and that $N_{\mathfrak{q}}$ is a localization of $N_{\mathfrak{p}}$ as a $B_{\mathfrak{p}}$-module, but I can't apply the usual argument which works for rings. Help?

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I just discovered that the argument works for modules as well, in the following form: $A\rightarrow B$ a ring homomorphism, $N$ a $B$-module which is flat over $A$, $T\subset B$ a multiplicative subset. Then $T^{-1}N$ is flat over $A$. –  ashpool Jun 29 '10 at 18:11
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Matsumura Commutative Ring Theory, Theorem 7.1, page 46. –  Georges Elencwajg Jun 29 '10 at 18:39
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