The motivation for asking this question is to test how much of the module theory that I have learned so far I can apply it to the simple rings I know in algebra. I have decided to start with the two of the simplest available rings: $\mathbb{Z}$ and $\mathbb{Z_n}$. Later, I'm planning to ask the same questions for other rings.

Categorize all finitely generated free/injective/project/flat modules over $\mathbb{Z}$ and $\mathbb{Z_n}$.

Here is my attempt to classify finitely generated modules over $\mathbb{Z}$:

Free: $M = \oplus_{i=1}^n \mathbb{Z}$.

Because if $M$ over $\mathbb{Z}$ is free then it's a sum of copies of $\mathbb{Z}$ and since it is finitely generated the number of copies must be finite.

Projective: $M = \oplus_{i=1}^n \mathbb{Z}$.

Because $\mathbb{Z}$ is a PID and the sets of f.g. free modules and f.g. projective modules over a PID coincide.


$\mathbb{Z}$ is a PID, a module over integers is injective iff it is divisible. Moreover, a product of modules is injective if and only if each component module is injective. Therefore, I guess the problem of classifying all injective modules over the ring of integers is equivalent to the classification of all divisible Abelian groups. Is it right?

Flat: I have no idea.

Here is my attempt to classify finitely generated modules over $\mathbb{Z_n}$:

Free: The answer is similar. In fact, what I said about integers previously is true for any arbitrary ring in general, I guess.

Projective: I assume it has to do something with the Chinese remainder theorem, but I don't know how to go forward beyond that.

Injective: No idea

Free: No idea.


1 Answer 1


I think you have an OK answer for $\mathbb Z$, except for flatness, which I'll address below. Perhaps the thing holding you back on $\mathbb Z_n$ is that you have not seen much on quasi-Frobenius rings?

Well, the long and short of it is that $\mathbb Z/(n)$ is quasi-Frobenius whenever $(n)$ is a nontrivial ideal.

There is a theorem that says the classes of projective and injective modules over a quasi-Frobenius rings coincide. Furthermore such a ring is Noetherian, so f.g. flat modules are also projective. (This applies to $\mathbb Z$ as well, completing your solution to the first half a little more.)

So for finitely generated modules of $\mathbb Z_n$, flat, projective and injective modules coincide. Classifying one is classifying them all. You could say that they are direct summands of $R^n$ for some $n\in\mathbb N^+$.

  • $\begingroup$ No. It's the first time I read the term "quasi-Frobenius". How to show that $\mathbb{Z}/n$ is a quasi-Frobenius ring? (without falling into a circular logic, in our particular case) $\endgroup$
    – user375190
    Dec 11, 2016 at 14:40
  • $\begingroup$ @user375190 you could work to show the ring is self injective, since it is already artinian $\endgroup$
    – rschwieb
    Dec 11, 2016 at 18:57
  • $\begingroup$ Oh, I see. The thing is chain conditions on modules will be discussed in the second half of the course after midterm. The first half covers basic module theory only including category theory, modules, Hom, tensors, projective, injective and flat modules. Is there any other way to see $\mathbb{Z}/n$ is quasi-Frobenius? $\endgroup$
    – user375190
    Dec 11, 2016 at 21:20
  • $\begingroup$ @user375190 You could show that it has the dual-annihilator property. Or that it factors into a product of rings of the form $\mathbb Z/(p^k)$, and study why those are QF, since a finite product of QF rings is QF $\endgroup$
    – rschwieb
    Dec 12, 2016 at 1:31
  • $\begingroup$ The first sentence is fine, but nothing after that is correct. $\mathbb Z/(p^k)$ is definitely not a quotient of $\mathbb Z/(p)$. I don't know what you mean by "divisible ring." I think it would be better for you to prove the ring is self-injective by Baer's criterion, given that the ideals are linearly ordered principal ideals. $\endgroup$
    – rschwieb
    Dec 12, 2016 at 21:31

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .