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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?

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    What about the function $f(x,y) = 1$ if $x=y$ but zero otherwise? Can you factorize this? – Pedro May 19 '21 at 08:57
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    @PedroTamaroff But how do you show such factorisation does not exist? Keep in mind that you can also take finite sums. –  May 19 '21 at 09:25
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    Yes, write a generic finite sum like in your post, and try to decide if it can equal the function $f(x,y)$ described in the first comment. – Pedro May 19 '21 at 09:26
  • This is basically a duplicate of https://math.stackexchange.com/questions/1686215/in-kunneth-formula-for-cohomology-the-finitely-generated-condition-is-necessary though there may be a better duplicate target out there. – Eric Wofsey May 24 '21 at 20:04
  • @EricWofsey Hard to see how that is a duplicate! –  May 25 '21 at 18:45
  • @PedroTamaroff Also, I'd like another example: $X$ and $Y$ need not be the same. –  May 25 '21 at 18:55
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    That question is asking why the canonical map $\prod_I\mathbb{Z}\otimes_{\mathbb{Z}}\prod_J\mathbb{Z}\to\prod_{I\times J}\mathbb{Z}$ is not surjective when $I$ and $J$ are infinite. If you replace $I$ and $J$ with $X$ and $Y$ and all the $\mathbb{Z}$s with $\mathbb{C}$s, this becomes exactly the same as your question, and my answer there works just as well over $\mathbb{C}$. – Eric Wofsey May 25 '21 at 20:11
  • Every example with $X=Y$ also leads to an example with $X \neq Y$ because you can of course $Y$ replace by a set isomorphic to $Y$. – Martin Brandenburg May 26 '21 at 12:57
  • Fun fact: There is an isomorphism of $\mathbb{C}$-vector spaces between $\mathbb{C}(X) \otimes \mathbb{C}(Y)$ and $\mathbb{C}(X \times Y)$ (by dimension considerations), but it is not natural. – Martin Brandenburg May 26 '21 at 12:59

2 Answers2

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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.)

Martin Brandenburg
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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.

Jack Schmidt
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  • Thanks for your answer! I will carefully read through it. Why community wiki though? –  May 25 '21 at 19:59
  • Well, I'm standing on the shoulders of Pedro Tamaroff :-) I tried to write my answer so it was useful to you, but I mostly just gave a more explicit reason why you can't factor like that – Jack Schmidt May 25 '21 at 20:04
  • This is essentially the same as what you do with matrix ranks, but here is how I prefer to think about it. Given $h:X\times Y\to\mathbb{C}$, consider the functions $h_x(y)=h(x,y)$ for each $x\in X$. If $h$ has the form $\sum f_k(x)g_k(y)$, then every $h_x$ is a linear combination of the functions $g_k$, and so the span of all the $h_x$'s must be a finite dimensional subspace of $\mathbb{C}(X)$. – Eric Wofsey May 25 '21 at 20:14
  • (I like the argument using ranks, I recall meeting it in a "deeper context" in a lecture by Gowers about additive number theory!) – Pedro May 25 '21 at 20:21