Is it true that every element of $V \otimes W$ is a simple tensor $v \otimes w$? I know that every vector in a tensor product $V \otimes W$ is a sum of simple tensors $v \otimes w$ with $v \in V$ and $w \in W$. In other words, any $u \in V \otimes W$ can be expressed in the form$$u = \sum_{i=1}^r v_i \otimes w_i$$for some vectors $v_i \in V$ and $w_i \in W$. This follows from the proof of the existence of $V \otimes W$, where one shows that $V \otimes W$ is spanned by the simple tensors $v \otimes w$; the assertion now follows from the fact that, in forming linear combinations, the scales can be absorbed in the vectors: $c(v \otimes w) = (cv) \otimes w = v\otimes (cw)$.
My question is, is it true in general that every element of $V \otimes W$ is a simple tensor $v \otimes w$?

This is equivalent to asking, does every multivariable polynomial factor into polynomials of one variable? No.
Consider the polynomial $x^2 + y$. If it could be factored into $P(x)Q(y)$ Then for some value of $y$, $Q(y) = 0$, and thus $x^2 + y = 0$ no matter the value of $x$. This is obviously false.

To be precise, let $\{1, x, x^2\}$ be a basis of $X$ and $\{1, y\}$ be a basis of $Y$. Then the basis of $X \otimes Y$ is just
$$\{x^2 \otimes y, x^2 \otimes 1, x \otimes y, x \otimes 1, 1 \otimes y, 1 \otimes 1 \}$$
The element $x^2 \otimes 1 + 1 \otimes y$ cannot be factored into a pure tensor.
