The two sets are, of course, supposed infinite.


This question is related to that one
[Commutation of tensor products with inverse limits in a specific case](https://mathoverflow.net/questions/200442/commutation-of-tensor-products-with-inverse-limits-in-a-specific-case/200443#200443)
where it received a (partial) answer ($A$ noetherian is suff).


I can complement a bit: in any case, $A$ is supposed a commutative ring with unit and $1\not= 0$.


When $A$ is a domain (no zero divisor) and if $A^X\otimes\_A A^Y$ is torsion-free (which I do not know
in general) then the natural arrow $f\otimes g\mapsto ((x,y)\mapsto f(x)g(y))$
$$
A^X\otimes\_A A^Y\rightarrow A^{X\times Y}
$$
is an embedding.



> 
> Let $\Phi$ be the natural arrow $A^X\otimes\_A A^Y\rightarrow A^{X\times Y}$ and $t\in ker(\Phi)$. Among all expressions
> $$
> \alpha\, t=\sum\_{i=1}^n f\_i\otimes g\_i\ , 
> $$
> with $\alpha\not=0$ choose one with $n$ minimal.
> 
> 
> 



> 
> **(H)** *We suppose that the tensor product $A^X\otimes\_A A^Y$ is **torsion-free***
> 
> 
> 



> 
> if $n=0$, one has $\alpha\, t=0$ and then $t=0$, we are done. If $n>0$, the family $(g\_i)\_{i=1}^n$ is free and for all $(x,y)\in X\times Y$
> $$
> \Phi(\alpha\, t)[x,y]=\sum\_{i=1}^n f\_i(x)g\_i(y)
> $$
> which means that $(\forall x\in X)(\sum\_{i=1}^n f\_i(x)g\_i=0)$ and then $(\forall x\in X)(\forall i\in [1..n])(f\_i(x)=0)$. This implies $\alpha\, t=0$ and then, under **(H)**, $t=0$.
> 
> 
> 


Who knows more ? (I am specially interested in necessary and sufficient conditions but, of course, any partial further result is welcome).


On the way I cannot prove **(H)**, any kind of hint will be appreciated. Hence



> 
> **Subquestion** We know that, if $A^X\otimes\_A A^Y$ is torsionless (as a $A$-module), the ring $A$ is a domain. Is the converse true ? otherwise can one provide a counterexample ?
> 
> 
> 



Necessary and sufficient condition for $can : A^X\otimes_A A^Y\rightarrow A^{X\times Y}$ to be an embedding
Let $k$ be a field, $I$ and $J$ infinite sets, and $A$ the $k$-subalgebra of $$k(t)[x\_i,y\_j: i\in I,j\in J]$$ 
generated by 
$$\{x\_i,y\_j,tx\_i,t^{-1}y\_j: i\in I, j\in J\}.$$


Then 
$$(tx\_i)\_{i\in I}\otimes(t^{-1}y\_j)\_{j\in J}-(x\_i)\_{i\in I}\otimes(y\_j)\_{j\in J}$$ 
is a non-zero element of the kernel of the natural map $A^I\otimes\_A A^J\to A^{I\times J}$. 


[**EDIT:** Here's a proof that this element is non-zero.


If it's zero, then it can be shown to be zero using only finitely many elements of $A$, and so $(tx\_i)\_{i\in I}\otimes(t^{-1}y\_j)\_{j\in J}-(x\_i)\_{i\in I}\otimes(y\_j)\_{j\in J}=0$ in $A^I\otimes\_BA^J$ for some finitely generated subalgebra $B$ of $A$. We can choose $r\in I$ and $s\in J$ so that
$$B\subseteq k(t)[x\_i,y\_j:i\neq r, j\neq s].$$


Now $A$ has a basis consisting of elements of the form $t^lm$ where $m$ is a monomial in $\{x\_i,y\_j:i\in I,j\in J\}$ and $l\geq 0$ if $m$ involves no $y\_j$s and $l\leq0$ if $m$ involves no $x\_i$s. The basis elements other than those for which $m$ is a power of $x\_r$ or of $y\_s$ span an ideal. Let $\bar{A}$ be the corresponding quotient algebra; then the image $\bar{B}$ of $B$ in $\bar{A}$ is just $k$.


Consider the image of $(tx\_i)\_{i\in I}\otimes(t^{-1}y\_j)\_{j\in J}-(x\_i)\_{i\in I}\otimes(y\_j)\_{j\in J}$ in $\bar{A}^I\otimes\_{\bar{B}}\bar{A}^J$. This is non-zero since $tx\_r\otimes t^{-1}y\_s-x\_ry\_s\neq0$ in $\bar{A}\otimes\_{\bar{B}}\bar{A}$, as the set $\{x\_r,tx\_r,y\_s,t^{-1}y\_s\}$ is linearly independent in $\bar{A}$, and the components for $i\neq r$ or $j\neq s$ are all zero.]


As your proof shows, this means that $A^I\otimes\_AA^J$ must have torsion, and indeed, for any $s\in I$,
$$\begin{align}x\_s\left((tx\_i)\_{i\in I}\otimes(t^{-1}y\_j)\_{j\in J}\right)
&=(tx\_sx\_i)\_{i\in I}\otimes(t^{-1}y\_j)\_{j\in J}\\
&=(x\_i)\_{i\in I}\otimes(x\_sy\_j)\_{j\in J}\\
&=x\_s\left((x\_i)\_{i\in I}\otimes(y\_j)\_{j\in J}\right),\end{align}$$
and so
$$x\_s\left((tx\_i)\_{i\in I}\otimes(t^{-1}y\_j)\_{j\in J}-(x\_i)\_{i\in I}\otimes(y\_j)\_{j\in J}\right)=0.$$


