Tensor Product of \mathbb{Z}_{10} and \mathbb{Z}_{12} with a Surprising Result

[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.