Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
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
@MattE Same can be said about this. –  user26857 Nov 27 '12 at 10:59
@ 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

3 Answers 3

up vote 3 down vote accepted

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.

share|improve this answer
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!

share|improve this answer
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.

share|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.