# $C \otimes A \cong C \otimes B$ does not imply $A \cong B$

Let $R$ be a commutative unital ring and let $M$ be an $R$-module and let $S$ be a multiplicative subset of $R$.

Today I proved both of the following: $$S^{-1} R\otimes_R S^{-1}M \cong S^{-1} M$$ and $$S^{-1} R \otimes M \cong S^{-1} M$$

Now I'm slightly confused.

Either my proofs are wrong or $C \otimes A \cong C \otimes B$ does not imply $A \cong B$. But I can't come up with an example. Can someone give me an example? (Or tell me that my proofs are wrong.)

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Well, if nothing else, $C\otimes A$ and $C\otimes B$ can both be trivial. For example, tensoring $\mathbb{Z}_2$ with $\mathbb{Q}$ or $\mathbb{R}$ will kill it, but $\mathbb{Q}$ and $\mathbb{R}$ certainly aren't isomorphic. –  Miha Habič Jun 13 '12 at 19:24
If you're interested in cases where this does work, look up examples of faithful flatness. –  Dylan Moreland Jun 13 '12 at 19:30
I don't think your first isomorphism can be right. When $R$ is a domain and $S^{-1}R=K$ is its field of fractions, $S^{-1}M$ should be the extension of scalars of $M$ to $K$; in your first isomorphism you claim that restricting scalars to $R$ then extending again to $K$ gives the same thing; however, even comparing dimensions should show this is false. Maybe I don't understand what's hidden in your notation, or maybe I'm confusing myself. –  vgty6h7uij Jun 13 '12 at 19:30
@vgty6h7uij: $R\to S^{-1}R$ is epic, so tensoring over $R$ and $S^{-1}R$ is the same for $S^{-1}R$ modules. Try it for $R=\mathbb{Z}$ and $S^{-1}R=\mathbb{Q}$ to see how the fractions just slide around once you already assume you have $\mathbb{Q}$-modules. –  Jack Schmidt Jun 13 '12 at 19:45
This isn't constructive at all, but I just love the terminology: $R\to S^{-1}R$ is EPIC!! –  Miha Habič Jun 13 '12 at 19:48

You are correct in noticing that tensoring with a fixed module isn't an injective operation. Various things can go wrong; probably the simplest thing to notice is that tensoring can kill torsion. To repeat the example given in the comments, both $\mathbb{Z}_2\otimes \mathbb{Q}$ and $\mathbb{Z}_2\otimes\mathbb{R}$ (everything in sight is a $\mathbb{Z}$-module) are trivial, while $\mathbb{Q}$ and $\mathbb{R}$ aren't isomorphic. There is a multitude of examples in the same vein, e.g. tensoring any two finitely generated $\mathbb{Z}$-modules of the same rank with $\mathbb{Q}$ will produce isomorphic modules.
$$(\mathbb Z/5\mathbb Z) \otimes (\mathbb Z/3\mathbb Z) = (\mathbb Z/7\mathbb Z )\otimes (\mathbb Z/3\mathbb Z ) =0$$ but right sides are not isomorphic (everything as $\mathbb Z$-module).