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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 ness and suff 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 ?

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 ?

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 ness and suff 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 Then, 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 ?

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 ness and suff 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 ?

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 torsionless

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 ness and suff 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 Then, 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 ?

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 ness and suff 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 Then, 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 ?