A Strange Tensor Product

I'm taking a class on quantum computation at the moment, so I've been thinking a lot about tensor products (which I've written about before here). The tensor product is a weird operation. On vector spaces, it has a fairly straightforward definition, although it has some very strange, unintuitive properties. But the tensor product of general modules is even weirder.

Here's an example. What is the tensor product of $\mathbb{Z}/2\mathbb{Z}$ and $\mathbb{Z}/3\mathbb{Z}$ as $\mathbb{Z}$-modules? That is to say, what is $\mathbb{Z}/2\mathbb{Z} \otimes_{\mathbb{Z}} \mathbb{Z}/3\mathbb{Z}$? For convenience, I'll write $M$ for this tensor product from now on. Cyclic groups are not free $\mathbb{Z}$ modules, so we can't think about $M$ in the same way as we think about the tensor product of vector spaces. And the answer turns out to be very strange.

\[M = 0\]

This answer doesn't make very much sense to me. The tensor product of nonzero vector spaces $V \otimes W$ always contains copies of $V$ and $W$. How can two nonzero $\mathbb Z$-modules tensor to 0?

Although it is very counterintuitive, this fact is actually not very hard to prove. Suppose $a \otimes b \in M$. It's always true that $1 \cdot a \otimes b = a \otimes b$. But because $2$ and $3$ are relatively prime, we can express 1 as a linear combination of 2 and 3. In fact, $1 = 2 \cdot 2 - 3$. Then \[\begin{aligned} a \otimes b &= 1 \cdot (a \otimes b)\\ &= (2 \cdot 2 - 3) (a \otimes b)\\ &= 2 \cdot 2 (a \otimes b)- 3 (a \otimes b)\\ &= 2 [(2a) \otimes b] - a \otimes (3b)\\ &= 0 \end{aligned}\] Since we can use this trick to show that every element of $M$ is 0, $M$ must be the 0 module.

However, I don't find this proof very satisfying. To me, it doesn't feel like it explains why the tensor product should be 0. It just demonstrates that it is 0. Of course, the division between 'should' and 'is' is vague and subjective, but I would prefer a different perspective on the problem. Today, I came up with an alternative justification using tensor-hom adjunction. This justification isn't fully formal at the moment, but it gives me a sense of why I might think this tensor product should be 0.

Tensor-Hom Adjunction

To me, the tensor-hom adjunction is all about curried functions (which I've written about before here). The idea is essentially that if you have a bilinear function from $X \times Y$ to $Z$, you can express this as a linear function on $X$ that returns a linear function that takes in an element of $Y$ and returns an element of $Z$. This decomposition of a multi-argument function into single-variable functions that return other functions is precisely the idea of currying.

More formally, the tensor-hom adjunction gives us a natural isomorphism \[\text{Hom}(X \otimes Y, Z) \simeq \text{Hom}(X, \text{Hom}(Y, Z))\]

Using the Tensor-Hom Adjunction to Study Our Tensor Product

We can use this to study our module $M$. Let $Z$ be an arbitrary $\mathbb Z$-module. Then tensor-hom adjunction tells us that\[\text{Hom}(M, Z) = \text{Hom}(\mathbb Z /2 \mathbb Z \otimes \mathbb Z / 3 \mathbb Z, Z) \simeq \text{Hom}(\mathbb Z / 2 \mathbb Z, \text{Hom}(\mathbb Z / 3 \mathbb Z, Z))\]

The nonzero of $\mathbb Z / 3 \mathbb Z$ all have order 3. So their images under any $\mathbb Z$-linear map must have order 1 or order 3. Therefore, every element of $\text{Hom}(\mathbb Z / 3 \mathbb Z, Z)$ must have order 1 or 3 as an element of the group of $\mathbb Z$-linear maps. But the image of an element of $\mathbb Z / 2 \mathbb Z$ under a linear map must have order 1 or order 2. Since every element of the codomain has order 1 or order 3, we must map every element of $\mathbb Z / 2 \mathbb Z$ to the identity. So the only elements of $\text{Hom}(\mathbb Z / 2 \mathbb Z, \text{Hom}(\mathbb Z / 3 \mathbb Z, Z))$ are maps which send $\mathbb Z / 2 \mathbb Z$ to the zero map $\mathbb Z / 3 \mathbb Z \to Z$. This means that the only element of $\text{Hom}(M, Z)$ is the zero map, no matter what $Z$ is.

This is a pretty convincing argument to me that maybe $M$ should be 0. In particular, if $Z = M$, then we see that the only linear map from $M$ to itself is the zero map. So it looks like $M$ has to be 0. Furthermore, this argument generalizes naturally in the same way that the previous calculation does. The argument should work for $\mathbb Z / m \mathbb Z \otimes \mathbb Z / n \mathbb Z$ whenever $n$ and $m$ are relatively prime. So maybe it is a good perspective to keep in mind. We can always find nontrivial maps between nonzero vector spaces over a fixed field. But because modules can have restrictions on element orders, we can have modules which don't have nontrivial maps to each other. And this can result in their tensor products being 0.

I'm still not totally satisfied with my answer to this problem. The tensor product still seems to be a little spooky. If you have a different way of looking at the tensor product in general, or at this problem in particular, I'd love to hear it in the comments.