TITLE: Kernel of a bilinear map and tensor product specificially
QUESTION [1 upvotes]: I am cementing my understanding of tensors and a book I am reading handwaves and simply says "of course we may denote $0\otimes 0$ as the $0$ of $U\otimes V$"
I have proved that the kernel is larger than that, but the proof is a little bit "backwards" - I'd like someone to tell me if I can write it in a more 'from definitions' way.
Let $f$ be a bilinear map on $U\times V$ to $W$
First off: $\text{Ker}(f)=\{(u,v)\in U\times V|f(u,v)=0\}$
I'd like a proof that says "Let $(u,v)\in\text{Ker}(f)$" and shows what is required. What I have is:

$\forall u\in U[0u=0]$
So using this logic $\forall (u,v)\in U\times V[0f(u,v)=0]$
Then $0f(u,v)=f(0,v)=f(0,u)=0$
So $\text{Ker}(f)\subseteq\{(u,v)\in U\times V|u=0\vee v=0\}$

I have proved that $[u=0\vee v=0]\implies f(u,v)=0$, I would like a "cleaner" proof of this. 

The second question I have is I'd like prove that $u\otimes v=0\iff[u=0\vee v=0]$ and I'm not sure how I'd go about doing that.
If you can only answer the first question that's fine! Having a nicer proof will be instrumental in tackling the second part.

REPLY [4 votes]: Proof of the second part:
We assume that both $U$ and $V$ contain a non-zero vector.
Suppose that $u$ and $v$ are non-zero.  We may extend each to a basis of $U$ and $V$ respectively.  That is, we now have the bases
$\{u = u_1,u_2,\dots\}$ and $\{v = v_1,v_2,\dots\}$ of $U$ and $V$.  We define the bilinear map
$$
f_{u,v}:\left(\sum_{i \in I} a_i u_i,\sum_{j \in J} b_j v_j\right) \mapsto a_1b_1
$$
Since $f_{u,v}(u,v) \neq 0$, we may conclude $u \otimes v \neq 0$.
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