The key point is to generalize the problem in order to make more effective the use of limits. Consider more generally for any $R$-module $R$ the natural map $$f_{M,Y}: M \otimes_R \prod_{y \in Y} R \rightarrow \prod_{y \in Y} M$$ given by $m \otimes (r_y) \mapsto (r_y m)$. In the special case $M = R^X$ this recovers the map in question, so it would suffice more generally to prove that such maps $f_{M,Y}$ are injective.

If some $\xi$ lies in the kernel then by writing it as a finite sum of elementary tensors we get a finitely generated $R$-submodule $N \subset M$ such that $\xi$ comes from some $\theta \in N \otimes R^Y$ and then $f_{N,X}(\theta) = 0$ since the target is left-exact in $M$. Hence, it suffices to treat the case when $M$ is finitely generated. If $M$ is finitely presented then $f_{R,X}$ is an isomorphism because right-exactness of source and target allows one to reduce to the case of finite free $M$ (which is easy). So this gives an affirmative answer when $R$ is noetherian.

Are you interested in non-noetherian $R$?

The key point is to generalize the problem in order to make more effective the use of limits. Consider more generally for any $R$-module $R$ the natural map $$f_{M,Y}: M \otimes_R \prod_{y \in Y} R \rightarrow \prod_{y \in Y} M$$ given by $m \otimes (r_y) \mapsto (r_y m)$. In the special case $M = R^X$ this recovers the map in question, so it would suffice more generally to prove that such maps $f_{M,Y}$ are injective.

If some $\xi$ lies in the kernel then by writing it as a finite sum of elementary tensors we get a finitely generated $R$-submodule $N \subset M$ such that $\xi$ comes from some $\theta \in N \otimes R^Y$ and then $f_{N,X}(\theta) = 0$ since the target is left-exact in $M$. Hence, it suffices to treat the case when $M$ is finitely generated. In case $M$ is finitely presented then $f_{M,Y}$ is an isomorphism because right-exactness of source and target allows one to reduce to the case of finite free $M$ (which is easy). So this gives an affirmative answer when $R$ is noetherian.

Are you interested in non-noetherian $R$?

The key point is to generalize the problem in order to make more effective use of direct limits. Consider more generally for any $R$-module $R$ the natural map $$f_{M,Y}: M \otimes_R \prod_{y \in Y} R \rightarrow \prod_{y \in Y} M$$ given by $m \otimes (r_y) \mapsto (r_y m)$. In the special case $M = R^X$ this recovers the map in question, so it would suffice more generally to prove that such maps $f_{M,Y}$ are injective.

If some $\xi$ lies in the kernel then by writing it as a finite sum of elementary tensors we get a finitely generated $R$-submodule $N \subset M$ such that $\xi$ comes from some $\theta \in N \otimes R^Y$ and then $f_{N,X}(\theta) = 0$ since the target is left-exact in $M$. Hence, it suffices to treat the case when $M$ is finitely generated. If $M$ is finitely presented then $f_{R,X}$ is an isomorphism because right-exactness of source and target allows one to reduce to the case of finite free $M$ (which is easy). So this gives an affirmative answer when $R$ is noetherian.

Are you interested in non-noetherian $R$?

The key point is to generalize the problem in order to make more effective the use of limits. Consider more generally for any $R$-module $R$ the natural map $$f_{M,Y}: M \otimes_R \prod_{y \in Y} R \rightarrow \prod_{y \in Y} M$$ given by $m \otimes (r_y) \mapsto (r_y m)$. In the special case $M = R^X$ this recovers the map in question, so it would suffice more generally to prove that such maps $f_{M,Y}$ are injective.

If some $\xi$ lies in the kernel then by writing it as a finite sum of elementary tensors we get a finitely generated $R$-submodule $N \subset M$ such that $\xi$ comes from some $\theta \in N \otimes R^Y$ and then $f_{N,X}(\theta) = 0$ since the target is left-exact in $M$. Hence, it suffices to treat the case when $M$ is finitely generated. If $M$ is finitely presented then $f_{R,X}$ is an isomorphism because right-exactness of source and target allows one to reduce to the case of finite free $M$ (which is easy). So this gives an affirmative answer when $R$ is noetherian.

Are you interested in non-noetherian $R$?