Proving When a Tensor is 0: M \cong M \otimes K

[Site]

Let M be a module over the commutative ring K with unit 1. I want to prove that M \cong M \otimes K. Define \phi:M \rightarrow M \otimes K by \phi(m)=m \otimes 1. This is a morphism because the tensor product is K-linear in the first slot. It is also easy to show that the map is surjective. This is where I get stuck.

Suppose \phi(m)=\phi(n), so that 0 = m \otimes 1 - n \otimes 1 = (m-n) \otimes 1. How do I prove that this implies that m=n and thus the map is injective? More generally, how can you tell when a tensor is 0?

 

[micromass]

Use the universal property to find an explicit inverse.

 

[Site]

Thanks, micromass! I think I've got it. If you wouldn't mind, please let me know if this looks alright.

Define h: M \times K \rightarrow M by h(m,k)=mk. This map is bilinear, so it factors through the tensor map M \times K \rightarrow M \otimes K. Thus there is a unique (irrelevant?) morphism f such that f(m \otimes k)=mk. This is the right and left inverse of the morphism \phi, implying that \phi is an isomorphism.

 

[micromass]

That looks alright! The uniqueness of the map is indeed irrelevant here, but it's nice to know.