$A\rightarrow B$ 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?
$\begingroup$
$\endgroup$
2
-
$\begingroup$ 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 $A$. $\endgroup$– ashpoolCommented Jun 29, 2010 at 18:11
-
3$\begingroup$ Matsumura Commutative Ring Theory, Theorem 7.1, page 46. $\endgroup$– Georges ElencwajgCommented Jun 29, 2010 at 18:39
Add a comment
|