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Let $A$ be a ring and $M$ an $A$-module. Then $M$ is faithfully flat over $A$ $\Leftrightarrow$ $M$ is flat over $A$ and $M \otimes N=0 \Rightarrow N=0$. This is part of theorem 7.2, p. 47 in Matsumura's Commutative Ring Theory.

Let's consider the direction $\Leftarrow$. Matsumura says, "let $N'\stackrel{f}{\rightarrow} N \stackrel{g} \rightarrow N''$ be a sequence of $A$-modules and suppose $M \otimes N'\stackrel{f_M}{\rightarrow} M \otimes N \stackrel{g_M} \rightarrow M \otimes N''$ is exact. Then $g_M \circ f_M=(g \circ f)_M=0$ so that BY FLATNESS, $Im(g \circ f) \otimes M= Im(g_M \circ f_M)=0$."

Question: where was it that flatness was used in Matsumura's last deduction? Is it not immdediate by definition (independently of flatness) that $Im(g \circ f) \otimes M= Im(g_M \circ f_M)$?

Michael Hardy
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Manos
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1 Answers1

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Flatness is needed to ensure that $-\otimes M$ commutes with taking images. If $M$ isn't flat, there are a module $B$ and a submodule $A$ such that the inclusion map $i:A\to B$ fails to remain monic when tensored with $M$. Then the image of $i$ is $A$ (or, more precisely, $i$) itself, and tensoring with $M$ gives $A\otimes M$. But the image of $i_M$ is a proper quotient of $A\otimes M$, because $i_M:A\otimes M\to B\otimes M$ isn't monic.

Andreas Blass
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  • A higher-level view of the situation is that the notion of image (of a map $f$) is defined by factoring $f$ as an epimorphism followed by a monomorphism. So for a functor to "obviously" preserve images, you'd want it to preserve both epimorphisms (which $-\otimes M$ does in any case) and monomorphisms (which needs flatness). – Andreas Blass Sep 13 '13 at 16:03