TITLE: Tensor product of two vector spaces
QUESTION [5 upvotes]: There is something which always intrigue me.
Let $U$ and $V$ be vector spaces over $k$ a field. Is it true that if $U\otimes_{k}V=0$, then $U=0$ or $V=0$.
Note that $U$ and $V$ are not necessarily finite dimensional.
I know that if $\{u\}$ and $\{v\}$ are basis vectors of $U$ and $V$, then $\{u\otimes v\}$ forms a basis for $U\otimes_{k}V$ but
1) where can I find the proof of this?
2) How does this implies conclusion?
3) What is the shortest way of proving this?
Thank you for your help!

REPLY [2 votes]: I was looking at the flatness of the modules. So hope I am right here!
Here we use the fact that a vector space is a flat module.
So phrasing differently, if $U\otimes_{k}V=0$, and $U\neq 0$, then I must show that $V=0$.
Thus let $u$ be a non-zero element of $u$ and we look at the exact sequence
$0\rightarrow ku\rightarrow U$.
Then by flatness we have
$0\rightarrow ku\otimes_{k}V\rightarrow U\otimes_{k}V$.
By hypothesis, this means that $ku\otimes_{k}V=0$.
Now it remains to show that we have an injection of vector space $V\hookrightarrow ku\otimes_{k}V$ via $v\mapsto u\otimes v$. So suppose $u\otimes v=0$. Then using the universal property 
\begin{array}{ccc}
 ku \times V &  &  & \\
\downarrow & \searrow & &\\
ku\otimes_{k}V &\rightarrow& V
\end{array}
With $f:ku\times V\rightarrow V$ via $f(ku,v)=kv$. This can be checked that it is bilinear. Hence there exists a unique homomorphism $\phi:ku\otimes_{k}V\rightarrow V$ that makes the diagram commutes.
Now, we have $\phi(u\otimes v)=\phi(0)=0=v$. Hence we have an exact sequence
$0\rightarrow V\rightarrow ku\otimes_{k}V$.
But $ku\otimes_{k}V=0$. Hence so is $V$.
Hope I am right here!