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$R$ is a commutative ring with unity.

If $R$ is P.I.D. I want to show that each of its proper ideal is written as a product of prime ideals.

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Since $R$ is a P.I.D. every ideal is a prime ideal, right?

So, how can it be that each proper ideal is a product of prime ideals?

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    $\begingroup$ What about $\mathbb{Z}$ ? $\endgroup$ – Captain Lama Apr 26 '16 at 16:43
  • $\begingroup$ Are all its proper ideals a product of prime ideals? So, aren't all its ideals prime ideals? @CaptainLama $\endgroup$ – Mary Star Apr 26 '16 at 16:49
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    $\begingroup$ All ideals of $\mathbb{Z}$ are of the form $(n)$ for some $n\in \mathbb{Z}$. Is the ideal $(n)$ always prime ? $\endgroup$ – Captain Lama Apr 26 '16 at 16:50
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    $\begingroup$ Every ideal in a PID is principal ideal. $\endgroup$ – drhab Apr 26 '16 at 16:52
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In a PID, $R$, every ideal is principal. That is every ideal is of the form $(a)$ for some $a \in R$.

So let $(a)$ be a proper ideal in $R$. PIDs are unique factorization domains, so we can write $a = p_1p_2 \cdots p_n$ for some prime/irreducible elements $p_1,p_2,\dots, p_n \in R$. The ideals $P_i = (p_i)$ are prime ideals.

Claim: $(a) = P_1P_2\dots P_n$.

If $r \in (a)$, then $r = sa = s p_1p_2 \cdots p_n$ for some $s \in R$. But then $$r = s p_1p_2 \cdots p_n \in P_1P_2\dots P_n.$$ Therefore $(a) \subseteq P_1P_2\dots P_n$.

If $r \in P_1P_2\dots P_n$, $\exists s \in R$ and $\exists p_i' \in P_i$ for $i= 1,\dots,n$ such that $r = sp_1'p_2'\dots p_n'$. But since $p_i' \in P_i = (p_i)$, $\exists t_i \in R$ such that $p_i'=t_ip_i$. Hence $$ r = s(t_1\dots t_n)(p_1\dots p_n) = s(t_1\dots t_n)a \in (a).$$ Therefore $P_1P_2\dots P_n \subseteq (a)$.

And thus $(a) = P_1P_2\dots P_n$.

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