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I am trying to find an example of rings $A\subset B$ and module $N$ over $B$ such that $N \otimes_A B$ over $A$ (by restriction of scalars on $N$) is not same as $N$ as $B$-modules. I think I have found it: $$A=Q,~B=\Bbb Q[x],~N=\Bbb Q .$$

Treat $\Bbb Q$ as $\Bbb Q[x]$-module. We can consider any linear map for the action of $x$. Then $\Bbb Q\otimes_{\Bbb Q}\Bbb Q[x]$ over $\Bbb Q$ is not isomorphic to $\Bbb Q$. I argue as follows: $\Bbb Q\otimes \Bbb Q[x]$ is isomorphic to $\Bbb Q\otimes \Bbb Q^N = (\Bbb Q\otimes \Bbb Q)^N$ which is not same as $\Bbb Q$. Is my argument correct?

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What do you mean by «you can consider any linear map for the action of x»? –  Mariano Suárez-Alvarez Jul 7 '12 at 7:16
    
In any case, for any $\mathbb Q$-module $M$ there is a canonical isomorphism $\mathbb Q\otimes_{\mathbb Q}M\cong M$. –  Mariano Suárez-Alvarez Jul 7 '12 at 7:17
    
treat Q as Q[x] module, for that you need a linear map. –  Rose Jul 7 '12 at 7:29
    
in your comment you write «treat Q as Q[x] module» yet your question begins with «treat Q[x] as Q module».... –  Mariano Suárez-Alvarez Jul 7 '12 at 7:30
    
treat Q as Q[x] module, for that you need a linear map. okay so Q\otimes Q[x] is isomorphic to Q[x]. actually i am trying to find an example of rings A and B and module N over B such that N \otimes B over A (by restriction of scalars on N) is not same as N as B modules. I think i have found it. A=Q,B=Q[x],N=Q. –  Rose Jul 7 '12 at 7:34

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