Kernel of a bilinear map and tensor product specificially 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.
 A: 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 $u \otimes v \neq 0$.
A: I believe that the first thing you want to proof (in a "clean" way) is the following claim:

Let $U,V,W$ be vector spaces and $f:U\times V\rightarrow W$ be billinear. For every $u\in U$ and $v\in V$ $f(u,0_V)=f(0_U,v)=0$

This is true for $$f(0_U,v)=f(0\cdot 0_U,v)=0 \cdot f(0_U,v)=0=0\cdot f(u,0_V)=f(u,0\cdot0_V)=f(u,0_V)$$
The above is the sufficient condition for being in the kernel as for the necessary one we have:

Let $U,W$ be vector spaces. For every $u\in U$ and $v\in V$
$$u\otimes v=0_{U\otimes V}\implies \text{$u$ or $v$ equals 0 in its own space.}$$

Proof:
Assume that $u\otimes v=0_{U\otimes V}.$
If $u\neq 0_U,$ then we have a linear map $f:U\rightarrow\mathbb{R}$ for example dot product in U,such that $f(u)$ is a unit and let $g:V\rightarrow\mathbb{R}$ be arbitrary linear map. Consider now bilinear map $\phi:U\times V\rightarrow \mathbb{R}$ given by formula
$$\phi(x,y)=f(x)\cdot g(y)$$
By property of tensor product it gives rise to linear map $h:U\otimes V\rightarrow \mathbb{R}$ such that
$$h(x\otimes y)=f(x)\cdot g(y).$$
Since $u\otimes v=0_{U\otimes V}$ we get that
$$0=h(u\otimes v)=f(u)\cdot g(v)\leadsto g(v)=0.$$
We are now in the situation that for every linear map $g:V\rightarrow\mathbb{R}$, $g(v)=0$ holds. Hence $v=0_V.$(take norm if you like)
Similarly if $v\neq0_V,$ then $u=0_U.$
$\square$
