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Question:

I want to determine when the ideals of $\mathbb{Z}/n\mathbb Z$ as $\mathbb{Z}/n\mathbb Z$-modules are projective/injective modules.

Here are some of the cases:

Proj:

(0) $\mathbb{Z/nZ}$ itself is projective $\mathbb{Z/nZ}$-module;

(1) $\mathbb{Z/nZ}$ is local ring iff $n=p^r$. (See Show that $\mathbb{Z}_n$ is local ring iff $n$ is a power of a prime number).

Since projective is equivalent to free over a local ring, and ideal $I=\mathbb{{p^sZ}\over{p^rZ}}$ is not free ($p^{r-s}\in ann(I)$), all nontrivial ideals of $\mathbb{Z/p^rZ}$ are not projective;

(2) $\mathbb{Z/nZ}\simeq \mathbb{Z/p_1Z\times\dots\times Z/p_rZ}$ is a semisimple ring, every module is projective and injective.

Inj:

(0) $\mathbb{Z/nZ}$ itself is injective $\mathbb{Z/nZ}$-module (Baer's criterion);

(1) $\mathbb{Z/p^rZ}$:also apply Baer's criterion. If ideal $I=\mathbb{{p^sZ}\over{p^rZ}}$ is injective, then we can extend $1_I$ to $f:R\to I$. Suppose $f(1+\mathbb{p^rZ})=kp^s+\mathbb{p^rZ})$, then $kp^{2s}=p^s+tp^r$, a contradiction.

Therefore, all nontrivial ideals of $\mathbb{Z/p^rZ}$ are not injective;

(2) $\mathbb{Z/nZ}\simeq \mathbb{Z/p_1Z\times\dots\times Z/p_rZ}$ is a semisimple ring, every module is projective and injective.

But the general cases are unsolved. Can you help me with that? Thanks in advance!

user26857
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shdvt
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  • Localize $\hspace{0cm}$ – math54321 Mar 21 '23 at 16:43
  • @math54321 sorry,i'm not familiar with localization...could you please explain it? – shdvt Mar 22 '23 at 02:20
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    If you're not familiar with localization then any explanation here probably won't be useful. But anyways, you worked out the local case which is $\mathbb{Z}/p^r\mathbb{Z}$, and both projectivity and injectivity are preserved by localization, so you're actually done (once you understand the localization process) – math54321 Mar 22 '23 at 18:59
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    To be precise, you've shown that if an ideal in $\mathbb{Z}/n\mathbb{Z}$ is projective/injective, then locally it must be all of $\mathbb{Z}/p^r\mathbb{Z}$ or $0$. Conversely all such ideals are easily seen to be projective and injective. This is equivalent to being generated by idempotents as in the answer below. – math54321 Mar 22 '23 at 19:02

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It's known that $\mathbb Z/n\mathbb Z$ is always a quasi-Frobenius ring, and since the injective modules coincide with the projective modules, solving one problem solves the other.

Considering that $R$ is self-injective, we can say that an ideal is injective iff it is a summand of $R$. So I would be comfortable with characterizing the injective/projective ideals of such rings as "the ones generated by idempotents."

The idempotents of $\mathbb Z/n\mathbb Z$ are not difficult to enumerate after expressing $\mathbb Z/n\mathbb Z=\prod \mathbb Z/p_i^{e_i}$ for various primes $p_i$ with exponents $e_i$.

rschwieb
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    I worked out the projective case myself with an ad hoc method, and found out that $d\mathbb{Z}/n\mathbb{Z}$ is projective if and only if $d$ and $n/d$ are coprime. I never realized it was equivalent to the existence of an idempotent generating this ideal (which is straightforward to prove once you expect it). Fun !! – GreginGre Mar 22 '23 at 14:00
  • @rschwieb I understand this is an old question, but I just came up with it. How do you characterize the projective/injective ideals hear by the ideals generated by idempotents. Which theorem are you using? I know that Re and R(1-e) are the summands of R but don't think you using this fact. – Tim Dec 31 '23 at 22:05
  • The part about projective-injective coincidence is a core theorem about quasi-Frobenius rings. You can find it by searching for the Faith-Walker Theorem. That injective submodules split out of their super-module, and that summands are projective (hence injective) is just bread and butter module theory. – rschwieb Jan 01 '24 at 04:14