Let $A$ be a ring and $I$ an infinite set. ¿Does $A^{(I)}\cong A^{(J)}$ implies $|I|=|J|$?

Question

Im learning aboute range on module and Im not sure about many things since my lectures didnt go deep enough on the subject. Is my question true? If so, why? If not, is it true under some additional hypothesis?

Answer 1

Yes, this is true (assuming $A$ is not the zero ring). More generally, the following theorem is true.

Theorem: Let $A$ be a ring, $M$ is an $A$-module, and $T\subseteq M$ is an infinite subset that generates $M$. Then for any generating set $S\subseteq M$, there is a subset $S_0\subseteq S$ which also generates $M$ such that $|S_0|\leq|T|$. (If $T$ is finite, we can instead conclude that there exists a generating set $S_0\subseteq S$ which is finite.)

In particular, if $A$ is a nonzero ring and $M=A^{(I)}$ for some infinite $I$, the standard basis of $M$ is a generating set $S$ of cardinality $|I|$ which has no proper subset which generates. If we had $M\cong A^{(J)}$ for some other infinite $J$, then there would also be a generating set $T$ of cardinality $|J|$, and so $|I|\leq |J|$ by the Theorem. (Or if $J$ is finite, we would conclude that $I$ is finite and immediately reach a contradiction.) But by swapping the roles of $I$ and $J$, we also have $|J|\leq |I|$. Thus $|I|=|J|$.

The proof of the Theorem is quite straightforward. Since $S$ generates $M$, each element $t\in T$ can be written as a linear combination of finitely many elements of $S$. Choosing finitely many such elements for each $t\in T$ and forming a set $S_0\subseteq S$ out of them, we have $|S_0|\leq \aleph_0\cdot |T|=|T|$. (If $T$ is finite, we can instead conclude that $S_0$ is finite since it has finitely many elements for each element of $T$.) Since every element of $T$ is in the submodule generated by $S_0$ and $T$ generates $M$, $S_0$ also generates $M$.