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How to show that direct product of $\mathbb Z\times\mathbb Z\times\mathbb Z\times...$ is not projective as a $\mathbb Z$ module?

I know that $\mathbb Z$ is a free $\mathbb Z$ module since it has $\{1\}$ as a basis.

An $R-$ module $P$ is said to projective if $P$ is a direct summand of a free $R-$ module .Also an $R-$ module is free if it is isomorphic to a direct sum of copies of the underlying ring.

However I cant use these facts to arrive at my proof.How to do it?

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  • The only proof I know is by showing that $\hom(\prod\Bbb Z,\Bbb Z)=\bigoplus\Bbb Z$, but for a free abelian group, the $\hom$ with $\Bbb Z$ must be much larger. – Asaf Karagila Jun 23 '15 at 16:31
  • This is essentially a duplicate of this question as well as this question. – Jim Belk Jun 23 '15 at 16:35
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    Using the definition, the identity map of $\Bbb Z/2\Bbb Z$ doesn't lift to a map $\Bbb Z/2\Bbb Z\rightarrow\Bbb Z/4\Bbb Z$ under the quotient map of the latter to $\Bbb Z/2\Bbb Z$ itself. This can be generalized to any torsion abelian group. – AdLibitum Jun 23 '15 at 16:58

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