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.
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