# As a $\mathbb{Z}$-module, is $\mathbb{R}\otimes_\mathbb{Z} \mathbb{R}$ isomorphic to $\mathbb{R}$?

Is it true that $\mathbb{R}\otimes_\mathbb{Z} \mathbb{R}$ (the tensor product of $\mathbb{R}$ and $\mathbb{R}$ over $\mathbb{Z}$) is not isomorphic to $\mathbb{R}$ as a $\mathbb{Z}$-module? Please give proof.

It is easy to prove that the tensor product of $\mathbb{Q}$ and $\mathbb{Q}$ over $\mathbb{Z}$ is isomorphic to $\mathbb{Q}$ as $\mathbb{Z}$-modules.

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As an abelian group, there is an isomorphism $\mathbb R\cong\mathbb Q^{(I)}$ with $I$ a set of cardinality equal to that of $\mathbb R$; here $\mathbb Q^{(I)}$ denotes a direct sum of $|I|$ copies of $\mathbb Q$, as usual. This is a consequence of the fact that $\mathbb R$ is a $\mathbb Q$-vector space, and it has a basis as such.
Since the tensor product distributes over arbitrary direct sums, $$\mathbb R\otimes_{\mathbb Z}\mathbb R\cong \mathbb Q^{(I)}\otimes_{\mathbb Z}\mathbb Q^{(I)}\cong(\mathbb Q\otimes \mathbb Q)^{(I\times I)}.$$ Now, since $I\times I$ has the cardinality of $I$ and you know that $\mathbb Q\otimes \mathbb Q\cong\mathbb Q$, we have $$(\mathbb Q\otimes \mathbb Q)^{(I\times I)}\cong\mathbb Q^{(I)}\cong\mathbb R$$.
 treat Q[x] as Q module. you can consider any linear map for the action of x. Then Q\otimes Q[x] over Q is not isomorphic to Q. I argue as following: Q\otimes Q[x] is isomorphic to Q\otimes Q^N = (Q\otimes Q)^N which is not same as Q. Is my argument correct?? – Rose Jul 7 '12 at 7:07 Why would $\mathbb Q\otimes_{\mathbb Q}\mathbb Q[x]$ be isomorphic to $\mathbb Q$? – Mariano Suárez-Alvarez♦ Jul 7 '12 at 7:12 How is it known that $\mathbb{Q}\otimes_\mathbb{Z}\mathbb{Q}\cong\mathbb{Q}$? – Leon Lampret Dec 31 '12 at 17:55 @LeonLampret, there is a more or less obvious map from one to the other; show it is an isomorphism. More generally, this follows from the fact that for an abelian group $M$, the tensor product $\mathbb Q\otimes_ZM$ is the localization of $M$ at the non-zero elements of $\mathbb Z$. – Mariano Suárez-Alvarez♦ Dec 31 '12 at 18:15 Aha, we have $a\otimes\frac{b}{k} = (\sum_{i=1}^k\frac{a}{k})\otimes\frac{b}{k} = \sum_{i=1}^k\frac{a}{k}\otimes\frac{b}{k} = \frac{a}{k}\otimes(\sum_{i=1}^k\frac{b}{k}) = \frac{a}{k}\otimes b$, hence $\mathbb{Q}\otimes_\mathbb{Z}\mathbb{Q} = \{1\otimes q; q\!\in\!\mathbb{Q}\}$. Did you mean $\mathbb{Q}\otimes_\mathbb{Z}M\cong M_{\langle\langle0\rangle\rangle}$? Is there a generalization? Perhaps $Q(R)\otimes_RM\cong M_{\langle\langle0\rangle\rangle}$ when $R$ is a domain? – Leon Lampret Dec 31 '12 at 18:38