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So this is what I have for the injectivity:

Knowing $\mathbb{Z}$ is a principal domain, $\mathbb{Q}/\mathbb{Z}$ is an injective module if and only if $\mathbb{Q}/\mathbb{Z}$ is divisible.

Checking the divisibility: Since $\mathbb{Q}/\mathbb{Z}$ is abelian, we want to see if it is divisible as a $\mathbb{Z}-module$. Let $a+\mathbb{Z} \in \mathbb{Q}/\mathbb{Z}$ and $k \in \mathbb{Z}$ with $k \neq 0$. Since $\mathbb{Q}$ is divisible, there exists $x \in \mathbb{Q}$ such that $kx = a$; thus $k(x+\mathbb{Z}) = a+\mathbb{Z}$. Hence $\mathbb{Q}/\mathbb{Z}$ is divisible and therefore INJECTIVE.

Is this correct? For the not projective bit, would it be enough to show that every element in $\mathbb{Q}/\mathbb{Z}$ has finite order?

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    $\begingroup$ Yes and yes. (But for the second question only if you know why.) $\endgroup$
    – Qi Zhu
    Apr 6, 2021 at 18:08
  • $\begingroup$ So I have this written down: a finitely generated module is projective if and only if it is the direct summand of a finitely generated free module. But Q/Z can't be a direct summand of Z^(\oplusX) since every element of Z^(\oplusX) has infinite order. Is this why? $\endgroup$ Apr 6, 2021 at 18:19
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    $\begingroup$ Yes. (But there is no need for the finiteness assumption, a projective module is a direct summand of a free module.) $\endgroup$
    – Qi Zhu
    Apr 6, 2021 at 18:25

2 Answers 2

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The divisible -> injective approach is correct in this case.

Here's a general purpose observation: consider an exact sequence of modules $0\to B\to A\to A/B\to 0$. We know that if $A/B$ is projective the sequence has to split, so that $B$ is a summand of $A$.

But it is easy to check that $\mathbb Z$ is an essential $\mathbb Z$ submodule of $\mathbb Q$: there's no way you can make it into a summand because it intersects all nonzero submodules nontrivially.

Therefore $\mathbb Q/\mathbb Z$ can't be projective.

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It is enough to check that $\Bbb Q/\Bbb Z$ is an injective $\Bbb Z$-module. Then it cannot be projective , see here:

Prove that no $\mathbb{Z}$-module is both projective and injective

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