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If there is a ring homomorphism $A\rightarrow B$ and if $Q$ is an injective $A$-module, is it true that $Q\otimes_A B$ is an injective $B$-module? I don't think it's true but can't think of a counterexample.

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Question If $Q$ is an injective $A$-module and if $f:A\to B$ is a ring morphism, is the extended module $Q\otimes_A B$ injective over $B$ ?

Answer No!

Class of counterexamples Let $A=k$ be a field. Like any $k$-vector space, $Q=k$ is injective over $k$. However if you take for$B$ the polynomial ring $B=k[X]$ and for $f:k\to k[X]$ the inclusion, then the extended module $Q\otimes_A B= k[X]$ is not injective over $k[X]$.
Indeed, for a PID "injective" is equivalent to "divisible" and $k[X]$ is definitely not divisible: for example it is not divisible by $X$. This just means that multiplication by $X$
(the map $k[X]\to k[X]: P\to X.P$) is not surjective

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That is marvelous. Thanks! –  ashpool Jul 23 '11 at 19:17

Let $A=k$ be a field, let $B$ be a unital $k$ algebra, let the ring map be $1 \mapsto 1$, and let $Q=k$. Then $Q\otimes _A B \cong B$ as a right $B$-module. Therefore you can take any non-self-injective $k$-algebra for $B$ to get a counterexample.

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oh, I see someone else got there while I was writing my answer –  mt_ Jul 23 '11 at 19:18
    
Don't worry mt_, it has often happened to me too. –  Georges Elencwajg Jul 23 '11 at 19:23

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