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Let $R$ be a PID. Every proper ideal is a finite product of maximal ideals, which are uniquely determined up to order.

Here is a partial solution: Since $R$ is a PID, it must also be a UFD. If $I \subseteq R$ is a nontrivial proper ideal, then there exists $x \in R \setminus \{0\}$ that is not a unit, otherwise $I=R$. Since $x$ is a nonzero, nonunit, there exists irreducible elements $c_1,..,c_n$ such that $x=c_1...c_n$ and therefore $(x) = (c_1...c_n) = (c_1)...(c_n)$. Recall that $c_i$ is irreducible iff $(c_i)$ is maximal with respect to all principal ideals; but as they are the only ideals, $(c_i)$ is maximal in $R$.

As one may notice, I haven't dealt with uniqueness yet. I found this and this, but unfortunately uniqueness isn't discussed in either of these MSE posts. I have tried quite few different things (e.g., showing this element divides that element, and that element divides this element, etc.) but I haven't had much luck. I could use a hint.

EDIT:

Here is the best I can do: Suppose that $(x) = (p_1)...(p_m) = (p_1...p_m)$ Then $x \mid (p_1 ... p_m)$ and $(p_1 ... p_m) \mid x$ and therefore $p_i \mid x$ for every $i$. But I don't know how to conclude from this $x = p_1...p_m$.

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    $\begingroup$ The uniqueness of the composition comes from the U in UFD. $\endgroup$ – Arthur Nov 14 '17 at 22:46
  • $\begingroup$ @Arthur Yes. I figured as much. $\endgroup$ – user193319 Nov 14 '17 at 22:52
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    $\begingroup$ Well it's not true that $(x)=(p_1 \dotsm p_m)$ would imply $x = p_1 \dotsm p_m$. There could be a unit factor there. $\endgroup$ – Zach Teitler Nov 15 '17 at 2:49
  • $\begingroup$ @ZachTeitler Ah! You are right. And isn't it true that $(ru)=(r)$ if $u$ is a unit? If so, then $(x) = (p_1....p_m)$ implies $x = u p_1 ... p_m$ or $c_1 ... c_n = u p_1 ... p_m$ and therefore $(p_1)...(p_m) = (p_1...p_m) = (u p_1... p_m) = (c_1 ...c_n) = (c_1) ... (c_n)$. Does this sound right? $\endgroup$ – user193319 Nov 15 '17 at 11:57
  • $\begingroup$ @user193319 Yes, that sounds right to me. Well, it sounds true. I'm not sure if I can see a proof in your comment (but maybe you weren't trying to write a proof in that comment). You want to prove that the factorization is unique, right? If $(x)$ factors as both $(p_1)\dotsm(p_m)$ and also as $(c_1)\dotsm(c_n)$, then $x = u p_1 \dotsm p_m = u' c_1 \dotsm c_n$, and then unique factorization shows the $p$'s and $c$'s are equal up to permutation. Probably that's what you were getting at but I wanted to try to be clear. $\endgroup$ – Zach Teitler Nov 15 '17 at 21:25

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