Elementary tensor product of non-zero elements Let $C$ be a $\mathbb{K}$-vector space and let $x, y \in C$ be two non-zero elements. Is it true that, in this case, $x\otimes y$ is also nonzero?
 A: We have $x\otimes y = 0$ if and only if for any bilinear map $C\times C\to V$ for any $\Bbb K$-vector space $V$, $(x, y)\mapsto 0$. This is basically the universal property of the tensor product.
Let $f:C\to \Bbb K$ and $g: C\to\Bbb K$ be linear maps. Then $b:C\times C\to \Bbb K$ given by $b(a, b) =  f(a)g(b)$ is a bilinear map. If we make it1 so that $f(x), g(y)\neq 0$, then $b(x, y)\neq 0$. Thus $x\otimes y$ cannot be $0$ either.
1 This requires Zorn's lemma in the general case.
A: Yes. If $x$ is non-zero there is $f_x \in C^*$ with $f_x(x) \neq 0$. Similarly, if $y$ is non-zero there is $f_y \in C^*$ with $f_y(y) \neq 0$. So consider the bilinear map $C\times C \ni (v,w) \mapsto f_x(v)f_y(w) \in \Bbb K$. This is bilinear and so induces a map $f_x\otimes f_y \colon C\otimes C \to \Bbb K$. Since $$f_x\otimes f_y(x\otimes y) = f_x(x)f_y(y) \neq 0,$$we have $x\otimes y \neq 0$.
A: Yes. I will show the case when $\dim_{\mathbb{K}}(C) < \infty$. Assume that $C$ is a $\mathbb{K}$ vector space with basis $\{ e_{i} : i \in [n] \}$. Then $x,y \ne 0 \in C \implies x = \sum_{i=1}^{n} a_{i} e_{i}$ and $y = \sum_{i=1}^{n} b_{i} e_{i}$, with $a_{i} , b_{i} \in \mathbb{K}$ and the $a_{i}$'s, and $b_{i}$'s not all zero. Then $x \otimes y = \sum_{i,j = 1}^{n} (a_{i} \cdot b_{j}) e_{i} \otimes e_{j}$. Since $x \ne 0$, $\exists i_{1} \in [n] : a_{i_1} \ne 0$, similarly $\exists b_{j_{1}} \ne 0$. Hence, $a_{i_{1}} \cdot b_{j_{i}} \ne 0 \implies x \otimes y \ne 0 \in C \otimes C$ because the set $\{e_{i} \otimes e_{j} : i,j \in [n] \}$ is a basis for $C \otimes C$. 
