Why is ℤ₁₀ ⊗_ℤ ℤ₁₂ isomorphic to ℤ₂?

[Kindayr]

Homework Statement

Show that $\mathbb{Z}_{10}\otimes_{\mathbb{Z}}\mathbb{Z}_{12} \cong \mathbb{Z}_{2}$

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!

 

[micromass]

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

 

[Kindayr]

yepp

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

 

[micromass]

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

 

[Kindayr]

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

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?

 

[Kindayr]

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

 

[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.

 

[Kindayr]

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.

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

 

[Kindayr]

Thanks for the help!

I got it.