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Find modules $M,N$ and a ring $A$ s.t. $M\otimes_A N\not \cong M\otimes_{\mathbb{Z}}N$.

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Do I have to? I'm only auditing the course! – Arturo Magidin Nov 17 '10 at 17:39
up vote 4 down vote accepted

Well, $A\otimes_A A$ is certainly isomorphic to $A$. So all we need is a ring for which $A\otimes_{\mathbb{Z}} A$ isn't isomorphic to $A$. An obvious choice is a ring like the Gaussian integers $A=\mathbb{Z}[i]$ which is free of rank $2$ as a $\mathbb{Z}$-module, for then $A\otimes_{\mathbb{Z}} A$ is free of rank $4$ as a $\mathbb{Z}$-module.

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Let $A=\mathbb Z[X]$ the polynomial ring, let $M=\mathbb Z[X,X^{-1}]$ be the ring of Laurent polynomials views as an $A$ module, and let $N=A/(X)$. As $\mathbb Z$-modules, both $M$ and $N$ are free $\mathbb Z$-modules. You can easily check that $M\otimes_AN=0$ yet $M\otimes_\mathbb ZN\neq0$.

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OK. I'm relative new on tensor products. Why $M\otimes_AN=0\ne M\otimes_\mathbb{Z}N$? – Jaska Nov 17 '10 at 20:14

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