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For $X,Y$ sets, let's denote $Y^X$ the set of all mappings $X\rightarrow Y$. If $Y(=R)$ is a ring, $R^X$ is a $R$-module (well, a bi-module but my question is - at first - concerning commutative rings). The arrow $$ R^X\otimes_R R^Y\rightarrow R^{X\times Y} $$
is given by the product $f\otimes g\rightarrow ((x,y)\rightarrow f(x)g(y))$ we know that, in case $R$ is a field, it is into (strictly if $X,Y$ are infinite).

What happens if $R$ is a general ring ? ($X,Y$ being infinite).

This question is related to that one Necessary and sufficient condition for $can : A^X\otimes_A A^Y\rightarrow A^{X\times Y}$ to be an embedding

For $X,Y$ sets, let's denote $Y^X$ the set of all mappings $X\rightarrow Y$. If $Y(=R)$ is a ring, $R^X$ is a $R$-module (well, a bi-module but my question is - at first - concerning commutative rings). The arrow $$ R^X\otimes_R R^Y\rightarrow R^{X\times Y} $$
is given by the product $f\otimes g\rightarrow ((x,y)\rightarrow f(x)g(y))$ we know that, in case $R$ is a field, it is into (strictly if $X,Y$ are infinite).

What happens if $R$ is a general ring ? ($X,Y$ being infinite)

For $X,Y$ sets, let's denote $Y^X$ the set of all mappings $X\rightarrow Y$. If $Y(=R)$ is a ring, $R^X$ is a $R$-module (well, a bi-module but my question is - at first - concerning commutative rings). The arrow $$ R^X\otimes_R R^Y\rightarrow R^{X\times Y} $$
is given by the product $f\otimes g\rightarrow ((x,y)\rightarrow f(x)g(y))$ we know that, in case $R$ is a field, it is into (strictly if $X,Y$ are infinite).

What happens if $R$ is a general ring ? ($X,Y$ being infinite).

This question is related to that one Necessary and sufficient condition for $can : A^X\otimes_A A^Y\rightarrow A^{X\times Y}$ to be an embedding

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For $X,Y$ sets, let's denote $Y^X$ the set of all mappings $X\rightarrow Y$. If $Y(=R)$ is a ring, $R^X$ is a $R$-module (well, a bi-module but my question is - at first - concerning commutative rings). The arrow $$ R^X\otimes R^Y\rightarrow R^{X\times Y} $$$$ R^X\otimes_R R^Y\rightarrow R^{X\times Y} $$
is given by the product $f\otimes g\rightarrow ((x,y)\rightarrow f(x)g(y))$ we know that, in case $R$ is a field, it is into (strictly if $X,Y$ are infinite).

What happens if $R$ is a general ring ? ($X,Y$ being infinite)

For $X,Y$ sets let's denote $Y^X$ the set of all mappings $X\rightarrow Y$. If $Y(=R)$ is a ring, $R^X$ is a $R$-module (well bi-module but my question is - at first - concerning commutative rings). The arrow $$ R^X\otimes R^Y\rightarrow R^{X\times Y} $$
is given by the product $f\otimes g\rightarrow ((x,y)\rightarrow f(x)g(y))$ we know that, in case $R$ is a field, it is into (strictly if $X,Y$ are infinite).

What happens if $R$ is a general ring ? ($X,Y$ being infinite)

For $X,Y$ sets, let's denote $Y^X$ the set of all mappings $X\rightarrow Y$. If $Y(=R)$ is a ring, $R^X$ is a $R$-module (well, a bi-module but my question is - at first - concerning commutative rings). The arrow $$ R^X\otimes_R R^Y\rightarrow R^{X\times Y} $$
is given by the product $f\otimes g\rightarrow ((x,y)\rightarrow f(x)g(y))$ we know that, in case $R$ is a field, it is into (strictly if $X,Y$ are infinite).

What happens if $R$ is a general ring ? ($X,Y$ being infinite)

Source Link
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