Why is $\mathbb{Z}/n\mathbb{Z}$ quasi-Frobenius? When surfing the wiki, I found the definition of Quasi-Frobenius rings
$R$ is quasi-Frobenius if and only it satisfies the following equivalent conditions:


*

*All right (or all left) R modules which are projective are also injective.

*All right (or all left) R modules which are injective are also projective.
Then, it mentions that the quotient ring $\frac{\mathbb{Z}}{n\mathbb{Z}}$ is QF for any positive integer $n>1$. But how to prove this directly by using the above definition?
 A: Well, let's prove $M$ projective over $\mathbb{Z}_n$ $\Rightarrow$ $M$ injective. I claim it suffices to prove that $M$ free $\Rightarrow$ $M$ injective, since a direct summand in an injective module is injective (and projectives are summands in free modules).
I also use Baer's criterion: it suffices to check we can extend for injections $I \rightarrow R$ of an ideal of the ground ring into the ring itself. Such an ideal is given by $(b)$ for $b|n$. 
Suppose $b \mapsto \bar{a} \in \oplus_i \mathbb{Z}_n$; to extend the map we wish to find $\bar{a'}$ in $\oplus_i \mathbb{Z}_n$ such that $b \cdot \bar{a'} = \bar{a}$ (and send 1 to $\bar{A'}$). Clearly it suffices to work coordinate wise (since in coordinates where $a = 0$ we can take $a' = 0$), by the Chinese remainder theorem it suffices to take $n = p^k$, in which case $b = p^l$ for some $l < k$. To be a homomorphism, in particular we have that $a p^{k-l} \equiv 0 (p^k) \Rightarrow p^l | a$ as desired. 
A: A sketch:


*

*The ring $R=\mathbb Z/n\mathbb Z$ is artinian, so every projective module is a direct sum of indecomposable finitely generated projectives. Since every direct sum of injectives is injective because $R$ is noetherian, we need only consider finitely generated modules.

*The ring $R$ is a quotient of a principal ideal domain, and there is a well-known theorem giving us the classification of all finitiely generated modules. It is very easy to see which, exactly, are the projectives and which are the injectives.

*The two classes actually coincide.

*Profit!
A: Let's directly see that $M$ injective implies $M$ projective 
First consider the case that $M$ is finitely generated, by the structure theorem (say for $\mathbb{Z}$) we may write $$M \simeq \oplus_i \mathbb{Z}_{m_i}$$To be $\mathbb{Z}_n$ modules is the claim that each $m_i|n$. We need to analyze when a summand $\mathbb{Z}_m$ can be injective, writing $n = mm'$, I claim $\mathbb{Z}_m$ injective implies $(m,m') = 1$.
Indeed, we have a map $(m') \subset \mathbb{Z}_n \rightarrow \mathbb{Z}_m$ given by $m' \mapsto 1$. Being able to extend this would mean there exists an element $a$ of $\mathbb{Z}_m$ such that $a \cdot m' = 1$. That is, $(m, m') = 1$.
In this case, we split $\mathbb{Z}_n \simeq \mathbb{Z}_{m} \oplus \mathbb{Z}_{m'}$ so we see each $\mathbb{Z}_{m_i}$ is a summand in a free module (of rank 1), hence each is projective, hence so is $M$, as desired. 
