## The dual of a module

Let $R$ and $S$ be rings with 1. Let $M$ be a left $R$-module, and let $N$ be an $(R,S)$-bimodule.

1. Define an action of $S$ on $\mathsf{Hom}_R(M,N)$ by $(\varphi s)(m) = \varphi(m)s$. Prove that this action is well defined and makes $\mathsf{Hom}_R(M,N)$ into a right $S$-module. Deduce that, via this action, $\mathsf{Hom}_R(M,R)$ is a right $R$-module for any left $R$-module $M$; this is called the dual module to $M$.
2. Let $N = R$ be considered as an $(R,R)$-bimodule as usual. Show that, under the action defined in part (1), $R$ and $\mathsf{Hom}_R(R,R)$ are isomorphic as right $R$-modules. Deduce that if $M$ is a finitely generated free left $R$-module, then $\mathsf{Hom}_R(M,R)$ is a free right $R$-module of the same rank.
3. Show that if $M$ is a finitely generated projective left unital $R$-module then its dual module $\mathsf{Hom}_R(M,R)$ is projective as a right $R$-module.

1. Note that, for all $x,y \in M$ and $r \in R$, $(\varphi s)(x + ry) = \varphi(x + ry)s$ $= (\varphi(x) + r\varphi(y))s$ $= \varphi(x)s + r\varphi(y)s$ $= (\varphi s)(x) + r(\varphi s)(y)$. So this action is well defined. To see that $\mathsf{Hom}_R(M,N)$ is a right $S$-module, let $s,t \in S$ and let $\varphi,\psi \in \mathsf{Hom}_R(M,N)$.
1. Note that, for all $m \in M$, $(\varphi(s+t))(m) = \varphi(m)(s+t)$ $= \varphi(m)s + \varphi(m)t$ $= (\varphi s)(m) + (\varphi t)(m)$ $= (\varphi s + \varphi t)(m)$. Thus $\varphi(s+t) = \varphi s + \varphi t$.
2. Note that, for all $m \in M$, $(\varphi (st))(m) = \varphi(m)(st)$ $= (\varphi(m)s)t$ $= ((\varphi s)(m))t$ $= ((\varphi s)t)(m)$. Thus $\varphi(st) = (\varphi s)t$.
3. Note that, for all $m \in M$, $((\varphi + \psi)s)(m) = (\varphi + \psi)(m)s$ $= (\varphi(m)+\psi(m))s$ $= \varphi(m)s + \psi(m)s$ $= (\varphi s)(m) + (\psi s)(m)$ $= (\varphi s + \psi s)(m)$. Thus $(\varphi + \psi)s = \varphi s + \psi s$.
4. Note that, for all $m \in M$, $(\varphi 1)(m) = \varphi(m)1$ $= \varphi(m)$, so that $\varphi 1 = \varphi$.

Thus, under this action, $\mathsf{Hom}_R(M,N)$ is a right unital $S$-module. In particular, letting $S = R$ and $N = R$ (considering $R$ as an $(R,R)$-bimodule in the usual way), we have that $\mathsf{Hom}_R(M,R)$ is a right unital $R$-module.

2. We define a mapping $\Psi : R \rightarrow \mathsf{Hom}_R(R,R)$ as follows: $\Psi(t)(a) = at$. Note that, for all $x,y,r \in R$, $\Psi(t)(x + ry) = (x+ry)t$ $= xt + (ry)t$ $= xt + r(yt)$ $= \Psi(t)(x) + r\Psi(t)(y)$, so that $\Psi(t)$ is a homomorphism $R \rightarrow R$, and thus $\Psi$ is well-defined. Now let $t,u,r \in R$. Note that for all $a \in R$, $\Psi(t + ur)(a) = a(t+ur)$ $= at + a(ur)$ $= at + (au)r$ $= \Psi(t)(a) + \Psi(u)(a)r$ $= \Psi(t)(a) + (((\Psi(u))r)(a)$ $= (\Psi(t) + ((\Psi(u))r)(a)$. So $\Psi(t + ur) = \Psi(t) + \Psi(u)r$, so that $\Psi$ is a homomorphism of right $R$-modules. Now suppose $t \in \mathsf{ker}\ \Psi$. Then $0 = \Psi(t)(1) = 1t = t$. In particular, $\mathsf{ker}\ \Psi = 0$, so that $\Psi$ is injective. Now suppose $\chi : R \rightarrow R$ is a left $R$-module homomorphism. Note that, for all $a \in R$, $\Psi(\chi(1))(a) = a\chi(1) = \chi(a)$, so that $\Psi(\chi(1)) = \chi$. That is, $\Psi$ is surjective. Thus, as right $R$-modules, we have $R \cong_R \mathsf{Hom}_R(R,R)$. (Using this action on $\mathsf{Hom}_R(R,R)$.)

Next we prove a lemma.

Lemma: Let $R$ and $S$ be unital rings. Let $M$, $N$, be left unital $R$-modules, and let $T$ be an $(R,S)$-bimodule. Let $\chi : N \rightarrow M$ be a left $R$-module homomorphism. Note that $\Theta_\chi : \mathsf{Hom}_R(M,T) \rightarrow \mathsf{Hom}_R(N,T)$ given by $\Theta_\chi(\varphi) = \varphi \circ \chi$ is well-defined. Moreover, $\Theta_\chi$ is a right $R$-module homomorphism (using the action from part (1)) and if $\chi$ is surjective, then $\Theta_\chi$ is injective. Proof: Let $\varphi,\psi \in \mathsf{Hom}_R(M,T)$ and let $r \in R$. Note that, for all $n \in N$, $\Theta_\chi(\varphi + \psi r)(n) = ((\varphi + \psi r) \circ \chi)(n) = (\varphi + \psi r)(\chi(n))$ $= \varphi(\chi(n)) + (\psi r)(\chi(n))$ $= (\varphi \circ \chi)(n) + \psi(\chi(n))r$ $= (\varphi \circ \chi)(n) + (\psi \circ \chi)(n)r$ $= \Theta_\chi(\varphi)(n) + \Theta_\chi(\psi)(n)r$ $= \Theta_\chi(\varphi)(n) + (\Theta_\chi(\psi) r)(n)$ $= (\Theta_\chi(n) + \Theta_\chi(\psi)r)(n)$. Thus $\Theta_\chi(\varphi + \psi r) = \Theta_\chi(\varphi) + \Theta_\chi(\psi) r$, so that $\Theta_\chi$ is a right $R$-module homomorphism. Now suppose $\chi$ is surjective. If $\Theta_\chi(\alpha) = \Theta_\chi(\beta)$, then $\alpha \circ \chi = \beta \circ \chi$. Because $\chi$ is surjective, it is right cancellable, and we have $\alpha = \beta$. Thus $\Theta_\chi$ is injective. $\square$

Now let $I$ be a finite index set, and let $\{M_i\}_I$ be a family of left unital $R$-modules. Note that the projections $\pi_k : \bigoplus_I M_i \rightarrow M_k$ are surjective. Using the lemma, for each $k$ we have an injective right unital $R$-module homomorphism $\Theta_k : \mathsf{Hom}_R(M_k,R) \rightarrow \mathsf{Hom}_R(\bigoplus_I M_i, R)$. By the universal property of direct sums, we have an injective module homomorphism $\Theta : \bigoplus_I \mathsf{Hom}_R(M_i,R) \rightarrow \mathsf{Hom}_R(\bigoplus_I M_i, R)$. Now let $\chi \in \mathsf{Hom}_R(\bigoplus_I M_i, R)$. For each $k \in I$, we have the natural injection $\iota_k : M_k \rightarrow \bigoplus_I M_i$. Let $\chi_k = \chi \circ \iota_k$. Certainly then $\Theta((\chi_i)) = \sum \chi_i \circ \iota_i \circ \pi_i = \chi$. That is, $\Theta$ is an isomorphism of right $R$-modules, so that $\bigoplus_I \mathsf{Hom}_R(M_i,R) \cong_R \mathsf{Hom}_R(\bigoplus_I M_i, R)$. (Using the action defined in (1)).

In particular, if $M = \bigoplus_I R$ is a free left $R$-module of finite rank, then \$latex $\mathsf{Hom}_R(M,R) = \mathsf{Hom}_R(\bigoplus_I R, R)$ $\cong_R \bigoplus_I \mathsf{Hom}_R(R,R)$ $\cong_R \bigoplus_I R$. Note that this is a chain of right $R$-module isomorphisms. That is, the dual module of $M$ is isomorphic to the free right $R$-module having the same rank as $M$.

3. Suppose $M$ is a finitely generated, projective, left unital $R$-module. Then $M \oplus N$ is free for some $N$, and moreover, we may choose $N$ so that $M \oplus N$ is finitely generated. Now as right $R$-modules, and using the action described in part (1), we have that $\mathsf{Hom}_R(M,R) \oplus \mathsf{Hom}_R(N,R) \cong_R \mathsf{Hom}_R(M \oplus N, R) \cong_R M \oplus N$ is free. Thus the dual module $\mathsf{Hom}_R(M,R)$ is projective as a right $R$-module.