Why is this map on the tensor product not surjective? Let $X,Y$ be sets and consider $\mathbb{C}(X)$ the set of all functions $X \to \mathbb{C}$ with the componentwise operations. I showed that there is an embedding of algebras
$$\mathbb{C}(X) \otimes \mathbb{C}(Y) \to \mathbb{C}(X \times Y): f \otimes g \mapsto [(x,y)\mapsto f(x)g(y)]$$
If $X,Y$ are infinite, I believe this mapping is not surjective. My intuition is that there are functions $g(x,y)$ which can not be written as $\sum_i f_i(x) g_i(y)$ (i.e. we can not always detach the variables). However, I struggle to formalise this idea. Any help?
 A: Let $R$ be any non-trivial commutative ring and let $X,Y$ be two infinite sets. I claim that the natural map of $R$-modules
$$R(X) \otimes_R R(Y) \to R(X \times Y)$$ is not surjective. Since we have injective maps $\mathbb{N} \to X$ and $\mathbb{N} \to Y$, inducing a commutative diagram
$$\begin{array}{ccc}
R(X) \otimes_R R(Y) & \to & R(X \times Y) \\
\downarrow && \downarrow \\
R(\mathbb{N}) \otimes_R R(\mathbb{N}) & \to & R(\mathbb{N} \times \mathbb{N})
\end{array}$$
with surjective vertical maps, it suffices to show that
$$\vartheta : R(\mathbb{N}) \otimes_R R(\mathbb{N}) \to R(\mathbb{N} \times \mathbb{N})$$
is not surjective. This reduction to the generic case is a proof method which is useful in many situations.
Now $R(\mathbb{N})$ consists of sequences in $R$, whereas $R(\mathbb{N} \times \mathbb{N})$ consists of infinite matrices in $R$, and $\vartheta$ maps a pure tensor $a \otimes b$ to the matrix
$$\begin{pmatrix} a_0 b_0 & a_0 b_1 & a_0 b_2 & \dotsc \\
a_1 b_0 & a_1 b_1 & a_1 b_2 & \dotsc \\
\vdots & \vdots& \vdots \\
\end{pmatrix}$$
A general tensor is mapped to a sum of such matrices. Observe that each row is a multiple of $(b_0,b_1,b_2,\dotsc)$. It follows that for every matrix in $\mathrm{im}(\vartheta)$ the rows are contained in a finitely generated $R$-submodule of $R(\mathbb{N})$. Since the unit vectors $(1,0,0,\dotsc),(0,1,0,\dotsc),\dotsc$ are not contained in a finitely generated $R$-submodule of $R(\mathbb{N})$, it follows that the identity matrix is not contained in $\mathrm{im}(\vartheta)$.
(Notice that this essentially the same as Jack Schmidt's answer, which is also essentially the same as Eric Wofsey's answer.)
A: Here is Pedro Tamaroff's answer updated for the request to let $X$ be different from $Y$.
Let $X$ and $Y$ be infinite. Choose sequences $x_1, \ldots$ and $y_1, \ldots$ of distinct elements of $X$ and $Y$. Define $$h(x,y)=\begin{cases} 1 & \text{if there is some } n \text{ with } x=x_n \text{ and } y = y_n \\
0 & \text{otherwise}\end{cases}$$
Suppose $\def\ds{\displaystyle}\ds h(x,y) = \sum_{k=1}^n f_k(x) g_k(y)$ for functions $f_i:X\to \mathbb{C}$ and $g_i:Y\to \mathbb{C}$.
Then consider the $(n+1) \times (n+1)$ identity matrix $I$ whose $i,j$th entry is $h(x_i,y_j)$.
Consider the matrices $A_{i,k} = f_k(x_i)$ and $B_{k,j} = g_k(y_j)$.
Then $\ds h(x_i,y_j) = \sum_{k=1}^n A_{i,k} B_{k,j}$ is the $i,j$th entry of the matrix product $AB$. Since $A,B$ have rank at most $n$, but $I$ has rank $n+1$ we get a contradiction.
