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.

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