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After looking back over some finite field theory, I've been thinking about the ring $\mathbb{Z}/p^k\mathbb{Z}$ for some prime $p$.

I'm just curious, are the $\mathbb{Z}/p^k\mathbb{Z}$ simple modules all isomorphic to each other? I suppose a better question to ask is: does $\mathbb{Z}/p^k\mathbb{Z}$ have a simple module, and if so, is it unique up to isomorphism?

I'd even appreciate a reference if this idea is collected somewhere. Thank you.

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Hace you tried to find all simple modules? (By the way, a unitary ring always has simple modules: just quotient the ring by any maximal left ideal) –  Mariano Suárez-Alvarez Apr 29 '12 at 5:38

3 Answers 3

up vote 3 down vote accepted

Let $A$ be a commutative ring with 1. Let $M$ is a simple $A$-module. Then $M\cong A/\mathfrak{m}$ for some maximal ideal $\mathfrak{m}$. The ring $\mathbb{Z}/p^k$ has only one maximal ideal with residue field $\mathbb{Z}/p\mathbb{Z}$.

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They are all isomorphic to $\mathbb{Z}/p\mathbb{Z} = p^{k-1}\mathbb{Z}/p^k\mathbb{Z}$. Note that any $\mathbb{Z}/p^k\mathbb{Z}$-module is a union of its finitely generated submodules, so we may apply the fundamental theorem of finitely generated abelian groups to deduce the structure of the module. The only finitely generated abelian group in which $p^k$ acts as 0 and which has no nontrivial submodules is $\mathbb{Z}/p$.

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A $\mathbb{Z}/p^k\mathbb{Z}$ module is the same thing as an abelian group of exponent $p^k$. Since an abelian group is simple if and only if it is cyclic of prime order, all simple $\mathbb{Z}/p^k\mathbb{Z}$ modules are isomorphic to $\mathbb{Z}/p\mathbb{Z}$.

Going a bit further, we can describe all modules:

By a theorem of Baer and Prufer, an abelian group of bounded exponent is a direct sum of cyclic groups, whether the group is finitely generated or not; so any $\mathbb{Z}/p^k\mathbb{Z}$-module is a direct sum of cyclic groups of order $p^i$, $1\leq i \leq k$; the indecomposable ones are cyclic of order $p^i$, $1\leq i \leq k$, and the only simple ones are the ones of prime order.

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You need to invoke the theorem only if what you want is to find indecomposable modules. But here we can just take the submodule generated by any non-zero element! –  Mariano Suárez-Alvarez Apr 29 '12 at 5:47
@Mariano: If you are looking for the simple modules and nothing more (which I had already described); I was describing all modules... –  Arturo Magidin Apr 29 '12 at 5:49
@Arturo : I corrected a trivial typo that looked a little weird. $1 \leq \leq k$ –  Patrick Da Silva Apr 29 '12 at 6:00
@Patrick: Thanks! –  Arturo Magidin Apr 29 '12 at 19:16

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