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.

 

[blue_raver22]

To prove that \phi is injective, we need to show that if \phi(m)=\phi(n), then m=n. So suppose \phi(m)=\phi(n), which means that m \otimes 1 = n \otimes 1. Since we are working in a commutative ring, we can commute the elements in the tensor product, so this is equivalent to 1 \otimes m = 1 \otimes n. Now, using the fact that the tensor product is K-linear in the first slot, we can pull the 1 out of the tensor product, giving us 1 \otimes (m-n) = 0. Since the tensor product is 0 if and only if one of the elements is 0, this implies that m-n = 0, and thus m=n. Therefore, \phi is injective and hence an isomorphism.

To determine when a tensor is 0, we can use the same logic as above. If we have an element x \otimes y = 0 in a tensor product, then either x=0 or y=0. This is because the tensor product is 0 if and only if one of the elements is 0. So if we can show that neither x nor y is 0, then we can conclude that x \otimes y \neq 0. In the case of \phi(m)=\phi(n), we showed that if the tensor product is 0, then m=n. So in general, to determine when a tensor is 0, we need to show that if the tensor product is 0, then the elements involved must also be 0.