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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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    @ 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.2012-12-02

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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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    in fact for clarity, let me just throw out the second half, I'll put it in a separate answer2012-11-27
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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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    Yes, I see. Thanks!2012-11-27
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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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    you're most welcome :) !2012-11-27