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I recently read about the result that the tensor product distributes over direct sums. I was curious if it also distributes over direct products, but google tells me it doesn't.

What are some simple counterexamples to why this property isn't true? I know that there is a natural homomorphism $ \left(\prod M_i\right)\otimes N\to \prod (M_i\otimes N) $ given by $(\prod m_i)\otimes n\mapsto \prod (m_i\otimes n)$ when $M$ and $N$ are modules over some commutative ring $R$. Are there standard examples where this homomorphism is not injective/surjective and hence not an isomorphism?

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    Related: https://math.stackexchange.com/questions/1810042018-06-30

2 Answers 2

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We consider $\mathbb{Z}$-modules (i.e., abelian groups).

Since $\mathbb{Q}$ is divisible, if $A$ is a torsion abelian group, then $A\otimes\mathbb{Q}$ is trivial.

Let $G$ be the direct product of cyclic group of order $p^n$, with $p$ a prime, and $n$ increasing; that is: $G = \prod_{n=1}^{\infty}\mathbb{Z}/p^n\mathbb{Z}.$

Then $\prod_{n=1}^{\infty}\left(\mathbb{Z}/p^n\mathbb{Z}\otimes\mathbb{Q}\right) = 0.$

But $G\otimes\mathbb{Q}$ is not trivial: if we let $x$ be the element that corresponds to the class of $1$ in every coordinate, then $x$ has infinite order. Therefore, $\langle x\rangle \otimes\mathbb{Q}\cong \mathbb{Z}\otimes\mathbb{Q} \cong\mathbb{Q};$ but tensoring with $\mathbb{Q}$ over $\mathbb{Z}$ is exact; therefore, the embedding $\langle x\rangle \hookrightarrow G$ induces an embedding $\langle x\rangle\otimes \mathbb{Q}\hookrightarrow G\otimes \mathbb{Q}$. Therefore, $G\otimes\mathbb{Q}\neq 0$. Thus, we have $\left(\prod_{n=1}^{\infty}\mathbb{Z}/p^n\mathbb{Z}\right)\otimes \mathbb{Q}\not\cong \prod_{n=1}^{\infty}(\mathbb{Z}/p^n\mathbb{Z}\otimes\mathbb{Q}).$

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    @hmIII: Since, in this case, $\prod(M_i\otimes N)$ is the trivial module, the map is necessarily onto; since $(\prod M_i)\otimes N$ is not trivial, the map is necessarily not one-to-one.2012-03-11
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Let $X$ and $Y$ be indeterminates.

For an example where your map is not surjective, take $M_i:=R$ for all $i\in\mathbb N$, and $N:=R[Y]$.

Then you get the natural map $ R[[X]][Y]\to R[Y][[X]], $ and $ \sum_{i\in\mathbb N}\ X^i\ Y^i $ is not in the image.

EDIT. Same example with different notation: Put $ A:=\left(\prod M_i\right)\otimes N,\quad B:=\prod\ (M_i\otimes N), $ and, for all $i,j\in\mathbb N$, $ M_i=R_i=R_j=R_{ij}=R. $ Set also $N:=\bigoplus R_j$. Then we have canonical isomorphisms $ A=\bigoplus_j\ \prod_i\ R_{ij},\quad B=\prod_i\ \bigoplus_j\ R_{ij}. $ We also have the inclusions $ A\subset B\subset\prod_{i,j}\ R_{ij}, $ and your map becomes the first inclusion.

Note that the Kronecker symbol $(\delta_{ij})$ is in $B$ but not in $A$.

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    Dear @hmIII: You're welcome. I would also have accepted Arturo's answer: it was the first one, and it is outstanding. My goal is to be as helpful as possible.2012-03-12