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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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You need infinitely many factors; otherwise, the direct product is isomorphic to the corresponding direct sum. – Arturo Magidin Mar 11 '12 at 4:57
up vote 19 down vote accepted

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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Thank you Arturo! I'm still easing my way into this sort of thing. Is there a pedestrian way to see from this example that the natural homomorphism is neither injective nor surjective? – Hailie Mathieson Mar 11 '12 at 8:57
Similar example: $G = \prod_{n = 1}^\infty \mathbf Z$, again tensoring with $\mathbf Q$ over $\mathbf Z$. The issue is that an element of $\prod \mathbf Q$ can involve infinitely many denominators. – Dylan Moreland Mar 11 '12 at 15:53
@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. – Arturo Magidin Mar 11 '12 at 22:14

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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Thanks Pierre-Yves! Do you mind saying a quick word why $\sum_{i\in\mathbb N}\ X^i\ Y^i$ is not in the image? Is it because there are only finitely many powers of $Y$ with nonzero coefficient for an element of $R[[X]][Y]$? – Hailie Mathieson Mar 11 '12 at 9:57
Dear hmIII: Yes! Every nonzero element of $R[[X]][Y]$ has a finite $Y$-degree. But, clearly, $$\sum_{i\in\mathbb N}\ X^i\ Y^i$$ has no such $Y$-degree. But I like your formulation better than mine... – Pierre-Yves Gaillard Mar 11 '12 at 10:16
Thanks again. I wish I could accept both answers, since each answered a different part of my question. – Hailie Mathieson Mar 12 '12 at 2:24
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. – Pierre-Yves Gaillard Mar 12 '12 at 3:27

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