Let $R$ be a commutative ring (for the sake of easiness) and $P$ a projective $R$-module. We recall that the character module of $P$ is defined by $$ P^*=\mathrm{Hom}_\mathbb{Z}\left( P, \mathbb{Q}/ \mathbb{Z} \right) $$

How can I show that $P^*$ is an injective $R$-module?

I do know that $M$ is a flat $R$-module if and only if its character module $M^*=\mathrm{Hom}_\mathbb{Z}\left( M, \mathbb{Q}/ \mathbb{Z} \right)$ is injective, and also that every projective module is flat, from which the result follows trivially.

However, how can I solve the problem without appealing to flat modules nor tensor products? (It is supposedly meant to be solved without even knowing that such things exist)

I've tried to show that the (contravariant) functor $ F=\mathrm{Hom}_R \big( -, \mathrm{Hom}_\mathbb{Z}\left( M, \mathbb{Q}/ \mathbb{Z} \right)\big) $ is exact, but I've had no luck so far (maybe I'm missing something, or there is an easier way to proceed)

Can you help me please? Thanks in advance!

There are some facts which may be useful:

  • For every abelian group (i.e., $\mathbb Z$-module) $G$ and $R$-module $M$, there is an isomorphism $\mathrm{Hom}_\mathbb{Z}\left( M, G \right)\cong \mathrm{Hom}_R\big( M, \mathrm{Hom}_\mathbb{Z}\left( R, G \right) \big)$
  • If $L\to M\to N$ is exact, then $N^*\to M^*\to L^*$ is exact

1 Answer 1


The module $P$ is a direct summand of a free module $F$, so $P^*$ is a direct summand of $\def\H#1{\operatorname{Hom}_{\mathbb{Z}}(#1,\mathbb{Q}/\mathbb{Z})}\H{F}$. Since $F$ is a direct sum of copies of $R$ and $\H{-}$ transforms direct sums into products, you just need to prove the result for the special case when $P=R$, because products of injective modules are injective.

Now you can use Baer's criterion. Suppose you have a homomorphism $f\colon I\to R^*=\H{R}$. Applying $\H{-}$, we get $$ f^*\colon R^{**}\to I^* $$ which we can compose with the canonical homomorphism $\omega_R\colon R\to R^{**}$, so we have $f^*\circ\omega_R\colon R\to I^*$. Let $\xi=f^*\circ\omega_R(1)$, which is a $\mathbb{Z}$-module homomorphism $I\to\mathbb{Q}/\mathbb{Z}$. Let $\omega_R(1)=\tilde{1}$ for simplicity, so $\xi=f^*(\tilde{1})=\tilde{1}\circ f$.

Since $\mathbb{Q}/\mathbb{Z}$ is divisibile, $\xi$ extends to a $\mathbb{Z}$-homomorphism $\eta\colon R\to\mathbb{Q}/\mathbb{Z}$, that is, to an element of $R^*$.

Now it is a matter of computations to verify that, for every $x\in I$, $f(x)=x\eta$.

Indeed, $x\eta\colon R\to\mathbb{Q}/\mathbb{Z}$ is defined by $$ x\eta(r)=\eta(rx)=\xi(rx) $$ (because $rx\in I$); thus $$ x\eta(r)=\xi(rx)=\tilde{1}\circ f(rx)=\tilde{1}(f(rx))=f(rx)(1)=f(x)(r) $$ Therefore $x\eta=f(x)$ as desired.


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