Let $V$ be a vector space and $V'$ its dual, similarly for $W$ and $W'$. Given $v \in V$ and $w \in W$ we get an element of $V' \times W'$, written $v \otimes w$, as follows: for $\varphi \in V'$ and $\tau \in W'$, define $(v \otimes w)(\varphi,\tau) = \varphi(v)\tau(w)$. The space $V \otimes W$ is the set of all possible linear combinations of such objects $v \otimes w$, where $(v,w) \in V \times W$.
Let $B : V\times W \to V\otimes W$, where $B(v,w) = v\otimes w$. The linearity of $\varphi$ and $\tau$ imply that $B$ is bilinear.
If $\dim V > 1$ and $\dim W > 1$, then we can show that $B(V\times W) \subsetneq V\otimes W$, i.e. there are elements of $V \otimes W$ which cannot be written as $v \otimes w$ for any $v \in V$ and $w \in W$. I wanted to examine the hypotheses that $\dim V>1$ and $\dim W>1$.
Assume that $\dim V = 1$ and $\dim W = n$. Moreover, let $v$ be a basis for $V$ and let the vectors $w_1,\ldots,w_n$ be a basis for $W$. The set of vectors $v \otimes w_j$ form a basis for $V\otimes W$. So a general element of $V \otimes W$ looks like $$ c_1(v\otimes w_1)+\cdots+c_n(v\otimes w_n)$$ On the other hand, a general element of $B(V\times W)$ looks like $$(\lambda v) \otimes (\mu_1w_1+\cdots+\mu_nw_n) = \lambda\mu_1(v\otimes w_1)+\cdots+\lambda\mu_n(v\otimes w_n)$$ As sets: $B(V \times W) \subseteq V \otimes W$: just set $c_i = \lambda\mu_i$ for all $1 \le i \le n$. Conversely, $V \otimes W \subseteq B(V\times W)$: just set $\lambda = 1$ and $\mu_i = c_i$ for all $1 \le i \le n$.
It started me thinking: What are the "fibres" of $B : V \times W \to V \otimes W$ when $\dim V = 1$? It feels like there is some projectivity at play here? For a fixed, non-zero $c_1(v\otimes w_1)+\cdots+c_n(v\otimes w_n) \in V \otimes W$, we have a "fibre" of $$(\lambda v) \times \left(\frac{c_1}{\lambda}w_1+\cdots+\frac{c_n}{\lambda}w_n\right)$$ This is one dimensional and parametrised by $\lambda \neq 0$. In the case of zero, I see that $B(v,0_W) = 0_{V\otimes W}$ for all $v \in V$ and $B(0_V,w) = 0_{V\otimes W}$ for all $w \in W$. In fact, as a set, $B(v,w)=0_{V \otimes W}$ if, and only if, $v=0_V$ or $w=0_W$.
Edit: @William suggested I prove $V\otimes W \cong W$. Wll, we know two vector spaces , over the same field, are isomorphic if, and only if, they have the same dimension. We also know that $\dim(V \otimes W) = (\dim V)(\dim W)$. Hence, $V \otimes W \cong W$ if, and only if, $(\dim V)(\dim W) = \dim W$. So, either $\dim V = 1$ and $V \otimes W \cong W$ or $\dim W = 0$ and $V \otimes \{0\} \cong \{0\}$.
Hint Suppose that $v \otimes w = v' \otimes w' \neq 0$ for $v, v' \in V$ and $w, w' \in W$. If $\dim V = 1$, $\{v\}$ spans $V$, so there is some nonzero scalar $r$ such that $v' = r v$, hence $$v \otimes w = v' \otimes w' = (r v) \otimes w' = v \otimes (r w') .$$