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 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 ?

  • $\begingroup$ So $A^X$ and $A^Y$ mean direct powers of $A$ (that is, $A^X = \prod_{x \in X} A$ and $A^Y = \prod_{y \in Y} A$), not invariant rings? And you ask for the canonical map to be an embedding, not just for the existence of some embedding? (Just making sure that I understand correctly.) $\endgroup$ Apr 2 '15 at 19:59
  • $\begingroup$ Also, why is the family $\left(g_i\right)_i$ free in your argument? $\endgroup$ Apr 2 '15 at 20:03
  • $\begingroup$ [And you ask for the canonical map to be an embedding]--> Yes, I need (if possible) ness and suff conditions on $A$ so that the canonical map be an embedding (if it is NOT clear from my text, I'll amend). $\endgroup$ Apr 3 '15 at 1:43
  • $\begingroup$ @Darij [Also, why is the family $(g_i)_i$ free in your argument?]-->because $n$ is minimal. If it were not free, some $\alpha_0g_{i_0}$ would be a linear combination of others for some $\alpha_0\not=0$ and then, changing $\alpha$ to $\alpha_0\alpha\not=0$ (there are no zero divisor), we could reduce $n$. $\endgroup$ Apr 3 '15 at 1:49
  • $\begingroup$ Oh, thanks -- I missed the condition that $A$ be a domain. $\endgroup$ Apr 3 '15 at 1:52

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 $$x_s\left((tx_i)_{i\in I}\otimes(t^{-1}y_j)_{j\in J}-(x_i)_{i\in I}\otimes(y_j)_{j\in J}\right)=0.$$

  • $\begingroup$ Nice example of torsion (+1) ! By the way, I "feel" that $(tx_i)_{i\in I}\otimes(t^{-1}y_j)_{j\in J}-(x_i)_{i\in I}\otimes(y_j)_{j\in J}$ might not be zero, but I see no way of proving it formally ... $\endgroup$ Apr 6 '15 at 22:49
  • 1
    $\begingroup$ @DuchampGérardH.E. I'll add a proof later today when I have time. But essentially: if it's zero in $A^I\otimes_AA^J$ then it's zero in $A^I\otimes_BA^J$ for some finitely generated subalgebra $B$. And then it's possible to find a quotient $\bar{A}$ such that $\bar{B}=k$ and the image of the element in $\bar{A}^I\otimes_{\bar{B}}\bar{A}^J$ is evidently non-zero. $\endgroup$ Apr 7 '15 at 8:05
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    $\begingroup$ @DuchampGérardH.E. I've added the proof. $\endgroup$ Apr 7 '15 at 9:29
  • $\begingroup$ .@Jeremy I find your proof convincing (I must go into details however). As it answers my subquestion and nobody answered my main question (it seems however out of reach to have conditions, I will refine my needs), I am glad to accept it as an answer. $\endgroup$ Apr 7 '15 at 20:11

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