Why is finite rank necessary here? - Dummit & Foote 10.3.13

Question

Problem statement: "Let $R$ be a commutative ring and let $F$ be a free $R$-module of finite rank. Prove the following isomorphism of $R$-modules: Hom$_R(F,R) \simeq F.$"

My proof relies on the fact that for any element $\phi \in \text{Hom}_R(F,R)$ we can map it to $\sum\limits_{i=1}^n \phi(a_i)a_i$ for $\{a_1,\ a_2,\ \dots,\ a_n\}$ a basis for $F$. But shouldn't you also be able to use infinite sums here instead, if the basis were infinite? Or does something break in the infinite case?

(And for any given element of $F$, all but finitely many basis elements should have a coefficient of 0 according to the definition of free modules in the text, but that shouldn't be an issue.)

Answer 1

No, with the given hypotheses you cannot use infinite sums here. No object in this construction has a topology so there's no way to define convergence of such a sum. And in fact the statement is false in the infinite rank case: for example if $R = \mathbb{Q}$ and $F = \bigoplus_{\mathbb{N}} \mathbb{Q}$ is a countable-dimensional $\mathbb{Q}$-vector space, then $\text{Hom}_R(F, R) \cong \mathbb{Q}^{\mathbb{N}}$ is uncountable-dimensional.

Answer 2

I think finite-dimensionality is needed, even for the simple case that $R$ is a field and $F$ is a vector space over $R$. This is because $\operatorname{hom}_{R}(F,R)$ is the dual space to $F$, and an infinite-dimensional vector space is never isomorphic to its dual.