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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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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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  • $\begingroup$ 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? $\endgroup$ Mar 11, 2012 at 8:57
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    $\begingroup$ 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. $\endgroup$ Mar 11, 2012 at 15:53
  • $\begingroup$ @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. $\endgroup$ Mar 11, 2012 at 22:14
  • $\begingroup$ Question, is there a reason why you can´t use the fact that $$G = \prod_{i = 1}^{n} \mathbb{Z}/p^n \mathbb{Z}$$ has elements with infinitely many non-zero components, and hence, those elements can not have torsion, i.e. there does not exist a non-zero $z \in \mathbb{Z}$ such that $z \cdot \pi = 0$, for $\pi \in G$, where $\pi$ is such that there are infinitely many non-zero elements. Now, take $\pi$ as such, together with $q = 1 \in \mathbb{Q}$. $q \otimes \pi \in G \otimes \mathbb{Q}$ will not have torsion. $\endgroup$
    – Ben123
    Sep 21, 2023 at 0:15
  • $\begingroup$ @Ben123 Well, because it is false that the group you write has no torsion. That group is finite, hence every element is torsion. If you meant the product from $n=1$ to $\infty$ instead, what you describe is essentially what I did, so I don't understand the question. You would be asking why I can't use the fact that I essentially used...Btw, "those elements cannot have torsion" is misuse of the term. Elements are or are not torsion, but they don't "have" torsion. Groups are the ones that may "have" or "not have" toesion. $\endgroup$ Sep 21, 2023 at 1:52
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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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  • $\begingroup$ 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 $R[[X]][Y]$? $\endgroup$ Mar 11, 2012 at 9:57
  • $\begingroup$ 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... $\endgroup$ Mar 11, 2012 at 10:16
  • $\begingroup$ Thanks again. I wish I could accept both answers, since each answered a different part of my question. $\endgroup$ Mar 12, 2012 at 2:24
  • $\begingroup$ 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. $\endgroup$ Mar 12, 2012 at 3:27

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