I think this is kind of true, since $\square\otimes C$ is a functor, so it preserves the isomorphism.

But what if we consider the example, $\mathbb{Z}\otimes\mathbb{Z}_2$ is not isomorphic to $2\mathbb{Z}\otimes\mathbb{Z}_2$. Because the latter is trivial while the first is not.

We can also consider the exact sequences: $$0\to\mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}_2\to0,$$ where the first map is simply multiple by 2, and $$0\to2\mathbb{Z}\to\mathbb{Z}\to\mathbb{Z}_2\to0,$$ where the first map is the inclusion. If we tensor by $\mathbb{Z}_2$ we will get $$\mathbb{Z}_2\to\mathbb{Z}_2\to\mathbb{Z}_2\to0$$ and

$$(0\to)X\to\mathbb{Z}_2\to\mathbb{Z}_2\to0$$ respectively.

Apparently, the first one is the famous counterexample of non-left-exactness of tensor product. But the second one we just inject a submodule (ideal) to the whole module (ring). If $r\otimes m$ is 0 in $\mathbb{Z}\otimes\mathbb{Z}_2$, then apparently it is 0 in $2\mathbb{Z}\otimes\mathbb{Z}_2$. So we have the left exactness of the tensor product. This is again weird, since the two exact sequences are isomorphic (by wiki), why do I have different result?

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    $\begingroup$ @5xum but tensoring by $C$ is a functor in the category of $R$-modules, so it must preserve the isomorphism. $\endgroup$ – Upc Nov 22 '18 at 14:53
  • $\begingroup$ Are ${\Bbb Z}$ and ${\Bbb Z}_2$ considered as $\Bbb Z$ modules? $\endgroup$ – Wuestenfux Nov 22 '18 at 14:54
  • $\begingroup$ @Wuestenfux Yes. So, you mean $2\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Z}_2$ is also $\mathbb{Z}_2$? $\endgroup$ – Upc Nov 22 '18 at 14:56
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    $\begingroup$ Why is the second trivial? Shouldn’t $2 \otimes 1$ be a nonzero element of $2 \mathbb{Z} \otimes \mathbb{Z}_2$? $\endgroup$ – user328442 Nov 22 '18 at 15:01
  • $\begingroup$ You've implicitly assumed that $(-) \otimes \mathbb{Z}_2$ preserves inclusions, but it doesn't; in general tensoring with a module preserves monomorphisms iff it's exact iff the module is flat. $\endgroup$ – Qiaochu Yuan Nov 22 '18 at 21:31

$2\mathbb{Z}\otimes\mathbb{Z}_2$ is not trivial. It is trivial inside of $\mathbb Z \otimes \mathbb{Z}_2$, a subtle but important difference. The "proof" you might have in mind that it is trivial would be

$$ 2n \otimes 0 = 0 \text{ and } 2n \otimes 1 = n \otimes 2 = 0 $$

but $n \otimes 2$ is only an element of $2\mathbb{Z}\otimes\mathbb{Z}_2$ if $n$ is even. Otherwise, you need the ambient module $\mathbb Z \otimes \mathbb{Z}_2$ to make sense of it.

Secondly, for your exact sequences. Do keep in mind that the isomorphism $2\mathbb{Z} \to \mathbb{Z}$ is a different map than the inclusion $2\mathbb{Z} \to \mathbb{Z}$.

So if we take the exact sequence

$$ 2\mathbb{Z} \to \mathbb{Z}\to\mathbb{Z}_2\to 0 $$

where the leftmost map is inclusion and tensor with $\mathbb{Z}_2$, we get the following commutative diagram $\require{AMScd}$ \begin{CD} 2\mathbb{Z} \otimes \mathbb{Z}_2 @>\text{inclusion}>> \mathbb{Z} \otimes \mathbb{Z}_2 @>>> \mathbb{Z}_2\otimes \mathbb{Z}_2 @>>> 0 \\ @VVV @VVV @VVV @VVV \\ \mathbb{Z}_2 @>\times 2>> \mathbb{Z}_2 @>>> \mathbb{Z}_2 @>>>0 \end{CD} where the rows are exact and the vertical arrows are isomorphisms.

That map $\mathbb{Z}_2 \xrightarrow{\times 2} \mathbb{Z}_2$ on the bottom is the zero map. To see that this must be the case, let us note that the isomorphisms $\to \mathbb{Z}_2$ are

$$ \begin{array}{ccc} 2\mathbb{Z} \otimes \mathbb{Z}_2 & \longrightarrow & \mathbb{Z}_2 \\ 2n \otimes 0 &\longmapsto & 0 \\ 2n \otimes 1 & \longmapsto & n \bmod 2 \end{array} \qquad \text{and} \qquad \begin{array}{ccc} \mathbb{Z} \otimes \mathbb{Z}_2 & \longrightarrow & \mathbb{Z}_2 \\ n \otimes 0 &\longmapsto & 0 \\ n \otimes 1 & \longmapsto & n \bmod 2 \end{array} $$

So the map on the bottom we can compute by looking at the composition $\mathbb{Z}_2 \to 2\mathbb{Z} \otimes \mathbb{Z}_2 \to \mathbb{Z} \otimes \mathbb{Z}_2 \to \mathbb{Z}_2$ and this composition takes $1 \mapsto 2 \otimes 1 \mapsto 2 \otimes 1 \mapsto 0$.


$2\mathbb{Z}$ and $\mathbb{Z}_2$ are not isomorphic as $\mathbb{Z}$-modules. Take a generator of each module and multiply it by 2.

  • $\begingroup$ Yes, this is clear. $\endgroup$ – Upc Nov 22 '18 at 17:16

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