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 $v \otimes w$?

Lets assume that $\dim V=m$ and $\dim W=n$ with $m\geq 2$ and $n\geq2$.

Suppose that $\{v_1,v_2\}$ are linearly independent in $V$ and that $\{w_1,w_2\}$ are linearly independent in $W$. Seeking a contradiction, suppose that $$ v_1\otimes w_1+v_2\otimes w_2=v\otimes w\tag{1} $$ Then extend $\{v_1,v_2\}$ to a basis $\{v_1,v_2,v_3,\dotsc,v_m\}$ of $V$ and extend $\{w_1,w_2\}$ to a basis $\{w_1,w_2,\dotsc,w_n\}$ of $W$. Write \begin{align*} v &= \alpha_1\,v_1+\dotsb+\alpha_m\,v_m \\ w &= \beta_1\,w_2+\dotsb+\beta_n\,w_m \end{align*} so that $$ v\otimes w=\sum_{i=1}^m\sum_{j=1}^n\alpha_i\beta_j\,v_i\otimes w_j\tag{2} $$ But $\mathcal B=\{v_i\otimes v_j:1\leq i\leq m,1\leq j\leq n\}$ is a basis for $V\otimes W$ (check this!) so (1) and (2) imply $$ \alpha_i\beta_j= \begin{cases} 1 & i=j=1 \\ 1 & i=j=2 \\ 0 & \text{otherwise} \end{cases} $$ which clearly is equivalent to \begin{align*} \alpha_i &= \begin{cases} 1 & i =1,2 \\ 0 & i\neq 1,2 \end{cases} & \beta_j &= \begin{cases} 1 & j=1,2 \\ 0 & j\neq 1,2 \end{cases} \end{align*} Now, we may rewrite (1) as \begin{align*} v_1\otimes w_1+v_2\otimes w_2 &= v\otimes w \\ &= (v_1+v_2)\otimes (w_1+w_2) \\ &= v_1\otimes w_1+v_1\otimes w_2+v_2\otimes w_1+v_2\otimes w_2 \end{align*} which contradicts that $\mathcal B$ is a basis for $V\otimes W$.

This is in general not true. One easy way to see this, is the following: Assume $V$, $W$ are $n$ and $m$ dimensional vector fields over the complex numbers, then it is fairly easy to show that $V\otimes W$ is isomorphic to $\mathbb{C}^{n\times m}$ with the following isomorphism $\phi: V\otimes W \rightarrow \mathbb{C}^{m\times n}$, which is defined as

$\phi(u\otimes w) := uw^{H}$ for elementary tensors and extended by linearity for all $x \in V\otimes W$.

Now due to ismorphism between both spaces, we can study the same question in $\mathbb{C}^{n\times m}$: Notice that elementary tensors $u\otimes w$ correspond to matrices $uw^{H}$, hence only of rank one. A general matrix $M\in\mathbb{C}^{n\times m}$ can be decomposed into a linear combination of rank one matrices (using SVD for example) but obviously is not corresponding to a rank one matrix. Hence transferring this property to $V\otimes W$ using the isomorphism, we see that not every element in $V\otimes W$ is an elementary tensor.

It is necessarily possible to write all tensors as a simple tensor when either $V$ or $W$ has dimension $1$ (or $0$). I will leave this to you to confirm, the proof is fairly trivial.

On the other hand, suppose that $V$ and $W$ have dimensions $2$ or greater, and that we are given a basis of vectors $v_i$ and $w_i$ for the respective spaces. Then I am fairly certain that we can say: $$ v_1 \otimes w_2 + v_2 \otimes w_1 \neq v \otimes w \qquad \forall v \in V, w \in W $$ In the context of quantum mechanics, this amounts to the statement "the pure state $v_1 \otimes w_2 + v_2 \otimes w_1$ is entangled" (QM assumes, however, that $V$ and $W$ have an inner product).

In fact, the states $v_1 \otimes w_1 \pm v_2 \otimes w_2$ and $v_1 \otimes w_2 \pm v_2 \otimes w_1$ are referred to as the Bell States, since they are the "canonical example" of entanglement (i.e. none of these vectors can be written as simple tensors).

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.

As has been pointed out, $V\otimes W$ is isomorphic to $Hom(V, W)$ once you pick a basis, with simple elements going to rank one matrices. Since (over any coefficients) as soon as $V$ and $W$ both have dimension at least 2 there is always a linear map from $V$ to $W$ with at least 2D image, we know that not every matrix is rank 1 and so not every tensor in $V\otimes W$ is simple.

I want to note that there is another proof that works for real or complex coefficients, or coefficients in a finite field. It works simply by comparing dimensions.

In the real and complex cases, the canonical map $V\times W/scalars \to V\otimes W$ is not linear, but it is smooth. And there is no surjective smooth map from a smaller dimensional space to a larger dimensional space, by say, Sard's theorem. This means there are non-simple elements in $V\otimes W$.

If $V$ and $W$ are real vector spaces the dimensions or direct sum and tensor product are $\dim V+\dim W$ and $\dim V\dim W$, respectively. (For complex vector spaces the real dimensions are $2\dim V+2\dim W$ and $2(\dim V\dim W)$.) So as long as $\dim V+\dim W-1< \dim V \dim W$ i.e. $(1-\dim V)(1-\dim W)<0$ i.e. one of $V$ and $W$ is at least 2-dimensional, we are done.

For coefficients in finite fields, we can see that there is no surjective map from a smaller dimensional space to a larger dimensional space simply by counting elements, and then the proof is the same.