10) Let \( R \) denote a commutative ring with \( 1 \neq 0. \) Let \( F \) denote a free \( R \)-module with
finite rank. Prove that the \( R \)-modules \( \text{Hom}_R(F, R) \) and \( F \) are isomorphic.
Proof. A free \( R \)-module of finite rank is much like a finite dimensional vector space — even the
term "basis" is used. To wit, a free \( R \)-module \( F \) of finite rank is an \( R \)-module with a finite set of
elements \( a_1, a_2, \ldots, a_n \in F \) such than every element \( f \in F \) can be represented uniquely as
\( f = r_1 a_1 + r_2 a_2 + \ldots + r_n a_n. \) That is, the \( r_i \) are unique for a given \( f, \) as if they
were vectors being represented in a given basis.
For each \( f \in F, \) consider the map \( \varphi_f \) from \( F \) to \( R \) defined as follows. Each \( f \in F \)
has a unique representation \( f = f_1 a_1 + f_2 a_2 + \ldots + f_n a_n, \) where \( f_i \in R. \) For each
\( x = r_1 a_1 + r_2 a_2 + \ldots + r_n a_n \in F, \) define \( \varphi_f(x) = \varphi_f(r_1 a_1 + r_2 a_2 + \ldots + r_n a_n) \)
\( = r_1 f_1 + r_2 f_2 + \ldots + r_n f_n. \)
The task is to show that the map \( \Psi : F \rightarrow \text{Hom}_R(F, R) \) with \( \Psi(f) = \varphi_f \) is well-defined and an \( R \)-module isomorphism.
1) \( \Psi(f) = \varphi_f \) is well-defined and an \( R \)-module homomorphism. This is straightforward, depending heavily
on unique representation.
2) \( \Psi \) is 1-1. For suppose \( \varphi_f = \varphi_g \) for \( f, g \in F .\) That is, \( \varphi_f(x) = \varphi_g(x) \)
for all \( x \in F. \) If \( x = r_1 a_1 + r_2 a_2 + \ldots + r_n a_n, \) this just says
\( r_1 f_1 + r_2 f_2 + \ldots + r_n f_n = \) \( r_1 g_1 + r_2 g_2 + \ldots + r_n g_n, \) and this will be true for all
combinations of the \( r_i \) as \( x \) varies. Putting \( r_1 = 1, r_i = 0, 2 \leq i \leq n \) results in \( 1 \cdot f_1 = 1 \cdot g_1; \)
that is, \( f_1 = g_1. \) Similarly for all the others — \( f_i = g_i \) for all \( i \). But these are the elements
uniquely defining \( f \) and \( g \) respectively in the basis \( a_1, a_2, \ldots, a_n, \) so \( f = g. \)
3) \( \Psi \) is onto. Given an \( R \)-module homomorphism \( \varphi \) from \( F \) to \( R, \) put \( f_i = \varphi(a_i). \)
Then for any \( x = r_1 a_1 + r_2 a_2 + \ldots + r_n a_n \in F, \) \( \varphi(x) = \varphi(r_1 a_1 + r_2 a_2 + \ldots + r_n a_n) = \)
\( r_1 \varphi(a_1) + r_2 \varphi(a_2) + \ldots + r_n \varphi(a_n) = \) \( r_1 f_1 + r_2 f_2 + \ldots + r_n f_n, \) so \( \varphi = \varphi_f. \)
It follows that \( F \) and \( \text{Hom}_R(F, R) \) are isomorphic, \( \Psi \) providing the isomorphism. QED.
Note: This is Ex. §10.3,#13 in Dummit & Foote.