# When is a tensor 0?

1. Aug 3, 2013

### 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?

2. Aug 3, 2013

### micromass

Staff Emeritus
Use the universal property to find an explicit inverse.

3. Aug 3, 2013

### 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.

4. Aug 3, 2013

### micromass

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