Projective and Injective Modules

Question

Let $M$ be a free $\mathbb{Z}$-module. Is $\text{Hom}_{\mathbb{Z}}(M,\mathbb{Q})$ an injective or a projective $\mathbb{Z}$-module?

very thanks

Answer 1

First, notice that $\mathbb{Q}$ is an injective $\mathbb{Z}$-module. If $M$ is free (or just flat), then $\newcommand{\Hom}{\text{Hom}_{\mathbb{Z}}} \Hom(M, \mathbb{Q})$ is injective, since $\Hom(\_, \Hom(M, \mathbb{Q})) \cong \Hom(\_ \otimes_\mathbb{Z} M, \mathbb{Q}) = \Hom(\_, \mathbb{Q}) \circ (\_ \otimes_Z M)$ as functors (by hom-tensor adjointness), and both $\_ \otimes_{\mathbb{Z}} M$ and $\Hom(\_, \mathbb{Q})$ are exact functors.

If $M = \oplus_{i \in I} \mathbb{Z}$, then $\Hom(M, \mathbb{Q}) \cong \prod_{i \in I} \Hom(\mathbb{Z}, \mathbb{Q}) \cong \prod_{i \in I} \mathbb{Q}$. Thus even if $M = \mathbb{Z}$, then $\Hom(M, \mathbb{Q}) \cong \mathbb{Q}$ is not free, and hence not projective (since over $\mathbb{Z}$, projective = free). Note however that $\Hom(M, \mathbb{Q})$ is always flat over $\mathbb{Z}$, since $\mathbb{Q}$ is flat over $\mathbb{Z}$ and arbitrary products of flat $\mathbb{Z}$-modules remain flat.