# how to prove an element is non-zero in a tensor-product

I was studying the following example from Atiyah & MacDonald's Introduction to Commutative Algebra:

Let $x$ be the non-zero element in $N := \mathbf{Z}/ 2\mathbf{Z}$, $M := \mathbf{Z}$, and $M' := 2 \mathbf{Z}$. The element $2 \otimes x$ is zero in $M \otimes N$, but non-zero in $M' \otimes N$. I suppose this can be seen by the fact that $2 \otimes x$ generates $M' \otimes N$, and the tensor-product $M' \otimes N$ is non-zero.

However, I was wondering, what one would do if that element weren't a generator of $M' \otimes N$. So my question is:

What are the methods to prove that an element is non-zero in a tensor-product of modules?

Thanks a lot!

• Well, how do you know that $M'\otimes N$ is non-zero in your example? – Eric Wofsey Feb 2 '16 at 9:58
• I thought first one could exploit the universal property of the tensor-product by considering the identity $M' \times N \rightarrow M' \times N$. but this is not a bilinear map. so following the answer of MooS: any non-zero bilinear map $f: M' \times N \rightarrow P$ will do: if $(M' \otimes N, g)$ were zero, then we get a contradiction, as $f$ induces a linear map $f' : M' \otimes N \rightarrow P$ such that $f' \circ g = f$ (and $f'$ would be the zero map). – Steven Feb 2 '16 at 10:21

## 1 Answer

If you dont have any further knowledge on $M \otimes N$, this is the very first step and an instant consequence of the universal property of the tensor-product:

$m \otimes n \in M \otimes N$ is non-zero iff there exists some $R$-Module $T$ and a bilinear map $M \times N \to T$, which maps $(m,n)$ to non-zero.

Proof:

If $m \otimes n \neq 0$, let $T = M \otimes N$ and consider the map $M \times N \to M \otimes N$ from the universal property.

On the other hand, if such a $T$ and a map exists, by the universal property, we get a map $M \otimes N \to T$, which maps $m \otimes n$ to non-zero. In particular $m \otimes n$ is non-zero.

In our case, choose $T=\mathbb Z/2\mathbb Z$. The map

$2\mathbb Z \times \mathbb Z/2\mathbb Z, (a,b) \mapsto \frac{ab}{2}$ is bilinear and maps $(2,1)$ to $1 \neq 0$, hence $0 \neq 2 \otimes 1 \in 2\mathbb Z \otimes \mathbb Z/2\mathbb Z$.