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If $g: A\rightarrow B$ is a ring homomorphism and $M$ is a flat $A$-module. Then $M_{B}=B\otimes_{A}M$ is flat $B$-module.

We need to prove if $f:S\rightarrow T$ is injective, then $f\otimes 1: S_{B}\otimes( B\otimes_{A}M)\rightarrow T_{B}\otimes (B\otimes_{A}M)$ is injective. But this is the same as $$f\otimes 1: (S_{B}\otimes B)\otimes_{A}M=S\otimes_{A}M\rightarrow (T_{B}\otimes B)\otimes_{A}M=T\otimes_{A} M$$ is injective by canonical isomorphism. Then this would make us treat $S\otimes_{A}M$ as an $A$-module by having $a(s\otimes m)=g(a)s\otimes m=s\otimes am$. Since $M$ is flat, the above sequence must be injective.

Is this properly justified? I feel unclear because $S\otimes_{A}M$ should be treated as a $B$ module instead. Should I prove $f\otimes 1: S\otimes_{A}M\rightarrow T\otimes_{A}M$ is injective given $f:S\rightarrow T$ is injective directly instead? This can be done $$f(s)\otimes m=0\Leftrightarrow f(s)=0\wedge m=0\rightarrow s\otimes m=0$$ but feels somehow cumbersome to use. Notice there is a similar question asked in here:

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up vote 4 down vote accepted

We need to prove if $f:S\rightarrow T$ is an injective $B$-morphism, then $f\otimes 1: S\otimes_{B}( B\otimes_{A}M)\rightarrow T\otimes_{B} (B\otimes_{A}M)$ is injective.

Since $S\otimes_{B}( B\otimes_{A}M) = S\otimes_{A} M$ and $T\otimes_{B}( B\otimes_{A}M) = T\otimes_{A} M$ and $f:S\rightarrow T$ is an injective $A$-morphism, the assertion is clear.

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I see. Did you use $s\otimes b\otimes m=sb\otimes m$ already? – Bombyx mori Sep 18 '12 at 0:44
@user32240 I used $S\otimes_{B}( B\otimes_{A}M) = (S\otimes_{B} B)\otimes_{A}M$. – Makoto Kato Sep 18 '12 at 0:54
oh, I thought you would use something different; this is what I wrote in the question already. But thanks for the response. – Bombyx mori Sep 18 '12 at 0:55

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