# If $A\cong B$, then $A\otimes C\cong B\otimes C$.

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?

• @5xum but tensoring by $C$ is a functor in the category of $R$-modules, so it must preserve the isomorphism. – Upc Nov 22 '18 at 14:53
• Are ${\Bbb Z}$ and ${\Bbb Z}_2$ considered as $\Bbb Z$ modules? – Wuestenfux Nov 22 '18 at 14:54
• @Wuestenfux Yes. So, you mean $2\mathbb{Z}\otimes_\mathbb{Z}\mathbb{Z}_2$ is also $\mathbb{Z}_2$? – Upc Nov 22 '18 at 14:56
• Why is the second trivial? Shouldn’t $2 \otimes 1$ be a nonzero element of $2 \mathbb{Z} \otimes \mathbb{Z}_2$? – user328442 Nov 22 '18 at 15:01
• 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. – Qiaochu Yuan Nov 22 '18 at 21:31

## 2 Answers

$$2\mathbb{Z}\otimes\mathbb{Z}_2$$ 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.

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