# 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:

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

2. 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?

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Why don't like other equivalent conditions (from the same source) such as: "R is Noetherian on one side and self-injective on one side"? –  user26857 Nov 27 '12 at 0:05
@YACP, yes, that would be easy by using the version you mentioned, so I asked the question because I do not know how to prove the above equivalence, and I google the "QF" because I heard a lecture on "Categorification of small quantum group", where the speaker mentioned the "QF" which I am not familar with. Really a long motivation, but thanks for your tip. –  ougao Nov 27 '12 at 1:15
Not an exact duplicate by any means, but there is some overlap with this question and its answer. –  Matt E Nov 27 '12 at 3:18
@ YACP, I was hesitant about choosing which answer as the most favorate, as you have said, the three ones are all good. And I completely forgot this thing in the past days, now I decided to accept uncookedfalcon's, for its detailed computation. –  ougao Dec 2 '12 at 2:57

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.

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Looks like you are proving "QF iff projectives and injectives coincide." I realize is equivalent to the two conditions listed, but the author's question is a slightly harder version. I'm going to upvote just because I appreciate your effort :) –  rschwieb Nov 27 '12 at 1:26
Hey man thanks for the reponse! My original layout was a bit confusing (I'm super low on sleep), but hopefully now it's clearer: the first half is showing property 1. holds for $\mathbb{Z}_n$, which is what the original poster wanted? The second half is an attempt at independently showing 2., as I'm ignorant of the proof of their equivalence. –  uncookedfalcon Nov 27 '12 at 1:39
in fact for clarity, let me just throw out the second half, I'll put it in a separate answer –  uncookedfalcon Nov 27 '12 at 1:58

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!

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It is a theorem of Bass that non-finitely generated projectives over an indecomposable commutative rings are in fact free, so this simplifies matters in that case; $R$ above is not indecomposable but is a finite direct product of indecomposable commutative rings, so a little work lets us use Bass' theorem to dispose of the non-finitely generated case. –  Mariano Suárez-Alvarez Nov 27 '12 at 1:52
when $n$ is not prime, the ring $R=\mathbb{Z}/n\mathbb{Z}$ is not a domain. –  ougao Nov 27 '12 at 2:13
so, I guess you mean that we first consider a module over $R$ as a module over $\mathbb{Z}$, just as the above answer given by uncookedfalcon, then determine which is injectve. –  ougao Nov 27 '12 at 2:32
Well, thanks for your step 1 and uncookedfalcon's detailed illustration of your step 2. –  ougao Nov 27 '12 at 2:37
@ougao, I meant: $R$ is a quotient of a principal ideal domain, so we can classify of the f.g. modules over the latter and then see which are really $R$-modules. –  Mariano Suárez-Alvarez Nov 27 '12 at 2:41

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.

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I think you have given all the points that I have failed to see before. Thanks! –  ougao Nov 27 '12 at 2:33
For the general case $M$, follow the below sketch step 1 given by Mariano Suárez-Alvarez. –  ougao Nov 27 '12 at 2:35
you're most welcome :) ! –  uncookedfalcon Nov 27 '12 at 3:04