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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?

user26857
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Hailie Mathieson
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2 Answers2

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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 $$\left(\prod_{n=1}^{\infty}\mathbb{Z}/p^n\mathbb{Z}\right)\otimes \mathbb{Q}\not\cong \prod_{n=1}^{\infty}(\mathbb{Z}/p^n\mathbb{Z}\otimes\mathbb{Q}).$$

Daniel Fischer
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Arturo Magidin
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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 08:57
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    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
  • 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. – Ben123 Sep 21 '23 at 00:15
  • @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. – Arturo Magidin Sep 21 '23 at 01:52
  • Thanks for the answer, yes, my bad, it should be $$G = \prod_{i = 1}^{\infty} \mathbb{Z}/p^i\mathbb{Z}$$ – Ben123 Sep 21 '23 at 02:00
  • @Ben123 And like I said, isn't the argument you propose essentially the same as the I gave 11 and a half years ago? Note that you still have to argue that $A\otimes\mathbb{Q}$ is trivial if and only if $A$ is a torsion group. Tensor products are weird, after all. It isn't necessarily immediately clear that if $x$ and $y$ both have infinite order then so does $x\otimes y$, you need to prove it. – Arturo Magidin Sep 21 '23 at 02:04
  • I am not sure I understand your argument, but from what I can pick out, our arguments differ, well, maybe in detail, but not in idea (perhaps). You seem to pick elements on the form $$((1,1,\ldots) \otimes q) \quad (q \in \mathbb{Q})$$ which was not my choice.

    Edit: Although now when I think about it, it is perhaps needed, I am not sure.

     – Ben123 Sep 21 '23 at 02:16
  • @Ben123 I'm picking a nontorsion element, same as you. The next part shows that the resulting element in the tensor is not torsion, something you seem to just assume is true. Once you prove that, it's the same argument: you have a nontorsion element, you prove tensori g with $\mathbb{Q}$ will give a nontorsion element, and this shows the tensor with the product is not trivial. – Arturo Magidin Sep 21 '23 at 02:19
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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 $A$.

Pierre-Yves Gaillard
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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 09:57
  • 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 02:24
  • 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 03:27