## Nontrivial divisible ZZ-modules are not projective

Prove that no nontrivial, divisible $\mathbb{Z}$-module is projective. Deduce that $\mathbb{Q}$ is not projective.

We begin with a lemma.

Lemma: Let $F$ be a free $\mathbb{Z}$-module. Then $\bigcap_{\mathbb{N}^+} nF = 0$. Proof: Suppose $B = \{e_i\}_I$ is a basis for $F$ indexed by a set $I$, and let $x \in \bigcap_{\mathbb{N}^+} nF$. For all $n$, there exist $k_{i,n} \in \mathbb{Z}$ such that $x = \sum nk_{i,n}e_i$ and all but finitely many $k_{i,n}$ are zero. Since $F$ is free on $B$, if $x = \sum \ell_i e_i$, we have $n|\ell_i$ for all $n \in \mathbb{N}^+$ so that $\ell_i = 0$. Thus $x = 0$. Hence $\bigcap_{\mathbb{N}^+} nF = 0$. $\square$

Now suppose the nontrivial, divisible $\mathbb{Z}$-module $Q$ is projective. Then there exists a module $N$ such that $Q \oplus N$ is free. By the lemma, $0 = \bigcap_{\mathbb{N}^+} n(Q \oplus N)$ $= \bigcap_{\mathbb{N}^+} nQ \oplus nN$. Since $Q$ is injective and $\mathbb{Z}$ is a principal ideal domain, $nQ = Q$ for all $n \in \mathbb{Z}$. So we have $0 = \bigcap_{\mathbb{N}^+} Q \oplus nN = Q \oplus \bigcap_{\mathbb{N}^+} nN$. Note, however, that this module contains $Q \oplus 0$, and since $Q$ is nontrivial, we have a contradiction. So $\mathbb{Q}$ is not projective.

In particular, since $\mathbb{Q}$ is nontrivial and divisible (as we showed here), it is not projective as a $\mathbb{Z}$-module.