# Homework Help: Simple Tensor Product

1. Oct 15, 2012

### Kindayr

1. The problem statement, all variables and given/known data
Show that $\mathbb{Z}_{10}\otimes_{\mathbb{Z}}\mathbb{Z}_{12} \cong \mathbb{Z}_{2}$

3. The attempt at a solution
Clearly, for any $0\neq m\in\mathbb{Z}_{10}$ and $0\neq n \in \mathbb{Z}_{12}$ we have that $m\otimes n = mn(1\otimes 1)$, and if either $m=0$ or $n=0$ we have that $m\otimes n = 0\otimes 0$.

I just don't know how to finish it.

I'm just working through Vakil's Algebraic Geometry monograph for fun, and this seemingly trivial question is bothering me.

Thank you for any help!

Last edited: Oct 15, 2012
2. Oct 15, 2012

### micromass

Can you prove that $1\otimes 10=0$?

3. Oct 15, 2012

### Kindayr

yepp

$1 \otimes 10 = 10(1\otimes 1)=10\otimes 1 = 0\otimes 1=0$.

4. Oct 15, 2012

### micromass

What about $m\otimes 1$. Can you prove that this is 0 for even m?

5. Oct 15, 2012

### Kindayr

Well it is trivial if $m=0$, so suppose $m\neq 0$ even. Then it follows that $m=2k$ hence $m \otimes 1 = 2k\otimes 1 = 2(k\otimes 1)$

Hence, for any morphism of $\mathbb{Z}$-modules $\phi : (\mathbb{Z}_{10}\otimes_{\mathbb{Z}}\mathbb{Z}_{12})\to \mathbb{Z}_{2}$, it follows that $\phi(m\otimes1)=2\phi(k\otimes 1)=0\in\mathbb{Z}_{2}$.

Also, another question, if we're dealing with $\mathbb{Z}$-modules, we can treat them as abelian groups. So what would the tensor product of $\mathbb{Z}$-modules translate to for abelian groups?

6. Oct 15, 2012

### Kindayr

Actually, I guess that doesn't really prove anything since it isn't assumed that $\phi$ is injective. Hrmm...

7. Oct 15, 2012

### micromass

Try the following for example:

$$2\otimes 1=12\otimes 1=0$$

As for your other question. The tensor product of abelian groups is exactly defined as the tensor product of $\mathbb{Z}$-modules.

8. Oct 16, 2012

### Kindayr

We have $0= 1\otimes 0 = 1\otimes 12= 12(1\otimes 1)=12\otimes 1=(2\otimes 1)+ (10\otimes 1)=(2\otimes 1)+0=2\otimes 1$.

Hence, $m\otimes 1 = k(2\otimes 1)=0$.

9. Oct 16, 2012

### Kindayr

Thanks for the help!

I got it.