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If you mod out $\mathbb{Z}$ be a nontrivial prime power $p^k$, $k>1$, then why can't $\mathbb{Z}/(p^k)=\mathbb{Z}/(n)\oplus\mathbb{Z}/(m)$ for some such submodules?

If that where the case, then $\mathbb{Z}/(n)\cap\mathbb{Z}/(m)=0$, but $mn$ is in both submodules, so $p^k\mid mn$. If say $p^k\mid m$, then $\mathbb{Z}/(m)=0$, so we must have at least $p\mid m$ and $p\mid n$. Is there a problem here? Perhaps they cannot generate $\mathbb{Z}/(p^k)$ any more?

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Replace $mn$ with the least common multiple of $m,n$ and you have my proof. –  Jack Schmidt Oct 27 '11 at 3:16

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The subgroups of $\mathbb{Z}/a\mathbb{Z}$ are exactly $d\mathbb{Z}/a\mathbb{Z}$ where $d$ is a divisor of $a$. In particular, if $a=p^k$ is a prime power, the subgroups of $\mathbb{Z}/p^k\mathbb{Z}$ are exactly the $p^i\mathbb{Z}/p^k\mathbb{Z}$ for $0 \leq i \leq k$.

If $i=k$, we get the zero submodule. If $i<k$ we get a nonzero submodule, but it must contain $p^{k-1}\mathbb{Z}/p^k\mathbb{Z}$.

In particular, $\mathbb{Z}/p^k\mathbb{Z}$ does not have two nonzero submodules that intersect in the zero submodule.

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Thanks! (extra char). –  Lux Interior Oct 27 '11 at 4:25

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