Freeness of $R^I$ ($R$ ring and $I$ infinite)

Question

It is well known that the (full) dual of $R^{(I)}$ ($R$ a ring and $I$ arbitrary) is $R^{I}$, the pairing $\langle ? | ? \rangle$ being defined on $R^{I}\otimes R^{(I)}$ by $$ \langle f | g \rangle=\sum_{i\in I}f(i)g(i) $$ If $I$ is finite, as $R^{I}=R^{(I)}$, we get that $R^{I}$ is free. But what is known if $I$ is infinite ? Is it free for some classes of rings ? all ? (of course it is free for fields, skew or not)

Answer 1

I presume $R^{(I)}$ denotes the direct sum of copies of $R$ indexed by $I$, while $R^I$ denotes the direct product. Of course when $R$ is a field, then $R^I$ is free. It is well-known that $\Bbb Z^I$ is not free over $\Bbb Z$ for $I$ infinite though.

When $R$ is a slender ring then the natural pairing $R^I\otimes R^{(I)}\to R$ is a perfect pairing, and I think this implies that for $I$ infinite then $R^I$ isn't free. For example, $\Bbb Z$ is a slender ring.