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Let $A,B,C$ be finitely generated $\mathbf{Z}/p^n\mathbf{Z}$-modules, where $p$ is prime and $n\geq 1$, and suppose $A\oplus C\cong B\oplus C$. Prove that $A\cong B$.

This problem has frustrated me since I can't seem to figure out how $\mathbf{Z}/p^n\mathbf{Z}$ and the finitely generated hypothesis play a role. I would like to do something like the following, but I am skeptical of the middle isomorphism (hence the question mark), though I'm fairly certain the others are okay: $$ A\cong \frac{A\oplus C}{0\oplus C}\overset{\text{?}}{\cong}\frac{B\oplus C}{0\oplus C}\cong B $$ Can anyone give me some hints? I was trying to construct the middle isomorphism explicitly by looking at the homomorphism $A\oplus C\rightarrow \frac{B\oplus C}{0\oplus C}$ induced by the isomorphism $A\oplus C\rightarrow B\oplus C$, but I couldn't seem to figure out how to use the hypotheses, or why things might fail without them...

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    $\begingroup$ The middle step doesn't work. Let's take nonfinitely generated modules, say $M=\Bbb{Z}^{\oplus \Bbb{N}}$, observe that $\newcommand{\ZZ}{\Bbb{Z}}\ZZ\oplus M \cong 0\oplus M$, but you cannot quotient both out by $M$ and preserve the isomorphism, since the isomorphism does not send the factor of $M$ in one to the factor of $M$ in the other. $\endgroup$
    – jgon
    Aug 20, 2018 at 0:54
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    $\begingroup$ I think you could use the structure theorem for PIDs here (this uses both the hypotheses) $\endgroup$
    – Sheel Stueber
    Aug 20, 2018 at 1:01
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    $\begingroup$ @SheelStueber $\mathbb{Z}/p^n\mathbb{Z}$ is not a PID for $n>1$, though.... $\endgroup$
    – Zilliput
    Aug 20, 2018 at 1:03
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    $\begingroup$ But $\mathbf{Z}/p^n\mathbf{Z}$ isn't a PID. $\endgroup$
    – Arbutus
    Aug 20, 2018 at 1:03
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    $\begingroup$ $\ZZ/p^n\ZZ$ isn't a domain. $\endgroup$
    – jgon
    Aug 20, 2018 at 1:04

2 Answers 2

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Hint: Since $A$, $B$ and $C$ are finitely generated modules over a finite ring, they are finite and have finite length. Thus, we may apply the Krull-Schmidt theorem. In particular, the uniqueness of a decomposition into indecomposable modules will imply the result you wish to prove.

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    $\begingroup$ Ah, I didn't know that theorem, but yes, I think that does do the trick. So the main point of the hypothesis is that $\mathbb{Z}/p^n\mathbb{Z}$ is finite (would any finite ring work here?) and the modules are finitely generated, which allows us to say they have finite length. Letting $M=A\oplus C$ and decomposing $A$ and $B$ as finite sums $\oplus U_i$ and $\oplus V_i$, the isomorphism $\oplus U_i\oplus C\cong \oplus V_i\oplus C$ implies that (up to permutation) $U_i\cong V_i$ by Krull-Schmidt, hence $A\cong B$. Is that the idea? $\endgroup$
    – Arbutus
    Aug 20, 2018 at 1:30
  • $\begingroup$ Yep, that's right and the proof works for any finite ring, as you say. $\endgroup$
    – Zilliput
    Aug 20, 2018 at 3:41
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There's a more elementary version, without using the Krull-Schmidt theorem: as $A,B,C$ are finitely generated over a finite ring, they are finite. But there's a general argument proving that finite structures are always simplifiable.

First of all, notice that the hypothesis yields $|A|=|B|$, so it suffices to prove that there is an injective morphism $A\to B$ to conclude.

Now note that for any finite module $D$, $hom(D,A\times C)\simeq hom(D,A)\times hom(D,C)$ and $hom(D,A\times C)\simeq hom(D,B\times C)\simeq hom(D,B)\times hom(D,C)$, now taking cardinals (which are all finite and nonzero) and simplifying by $|hom(D,C)|$ we get that for any finite module $D$, we have $|hom(D,A)|=|hom(D,B)|$.

Denote by $i(D,A)$ the number of injective morphisms $D\to A$, and similarly for $i(D,B)$. We prove by induction on $|D|$ that for all $D$, $i(D,A)=i(D,B)$.

Indeed by the factorization theorem, $|hom(D,A)| = \displaystyle\sum_{N\leq D} i(D/N,A)$, and so by applying the induction hypothesis, we get $|hom(D,B)|=|hom(D,A)| = \displaystyle\sum_{N\leq D, N\neq 0} i(D/N,A) + i(D,A) = \displaystyle\sum_{N\leq D, N\neq 0} i(D/N,B)+ i(D,A) = |hom(D,B)|-i(D,B) + i(D,A)$ and so $i(D,A)=i(D,B)$.

Apply this to $D=A$, since $i(A,A)\geq 1$ we get that $i(A,B)\geq 1$, and we may conclude.

This is more elementary than Krull-Schmidt and generalizes to many situations (the same argument applies, almost verbatim, to all finite algebraic structures, which are therefore all simplifable)

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