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I am self-learning some algebraic number theory and my question is regarding the advantages to studying PIDs. I have seen that Euclidean Domains $\subseteq$ Principal Idea Domains $\subseteq$ Unique Factorization Domains. What are some important properties we gain from requiring that all ideals are generated by one element? In what context were PIDs first introduced? What properties or questions do they help us understand (other than uniqueness of factorization)?

For example, $\mathbb{Z}[i]$ is a Euclidean domain with the norm $a + bi \mapsto a^2 + b^2$, so $\mathbb{Z}[i]$ is a PID and a UFD. However, I don't see what we pick up or lose by having ideals be principal. Perhaps examples of UFD that are not PID or PID that are not Euclidean would be enlightening.

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    $\begingroup$ For a UFD that is not a PID take polynomials over a field in more than one variable $\mathbb{F}[x_1,...x_n]$ with $n > 1$. You can compare this to the case where $n=1$ which gives a PID. $\endgroup$ Nov 5, 2015 at 15:50
  • $\begingroup$ For a PID that is not a UFD look at this paper. $\endgroup$ Nov 5, 2015 at 16:31
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    $\begingroup$ @rVitale you mean a PID that is not a Euclidean domain :) $\endgroup$
    – Mathmo123
    Nov 5, 2015 at 18:26
  • $\begingroup$ @Mathmo123 oops yes I did, thanks! $\endgroup$ Nov 5, 2015 at 18:35
  • $\begingroup$ The most famous example of a PID that is not Euclidean is $\mathcal{O}_{\mathbb{Q}(\sqrt{-19})}$: math.stackexchange.com/questions/1040330/… $\endgroup$
    – Mr. Brooks
    Nov 5, 2015 at 22:19

2 Answers 2

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One thing you get is a notion of gcd. If you want to find $\gcd(x,y)$ in a PID, $R$, you can use the fact that $(x)+(y)=(z)$ for some $z \in R$. Then a gcd of $x$ and $y$ is any generator of $(z)$. You don't need a PID for this though, you can take a look here

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    $\begingroup$ For rings of integers of number fields, UFD is equivalent to PID. Euclidian is definitely stronger. As for the "advantage" of PID, just think: if Z[w_p], where w_p is a primitive p-th root of 1, were a PID, Fermat's last theorem wouldn't be so famous ! $\endgroup$ Apr 19, 2016 at 5:53
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There are plenty of examples of UFDs which aren't PIDs. The most basic example is the polynomial ring $\mathbb Z[X]$, which is certainly a UFD, but the ideal $(2,X)$ is not principal.

However, in algebraic number theory, one of the key objects of study is the ring of integers of a number field, namely, the ring $\mathcal O_K$ of all the algebraic integers in a field $K$ which is a finite field extension of $\mathbb Q$ (so $K$ is a finite dimensional $\mathbb Q$-vector space). In these rings, which are examples of Dedekind domains, being a UFD is equivalent to being a PID, so to find out which rings have unique factorisation, it suffices to find out which rings are PIDs. $\mathbb Z[i]$ is an example of such a ring, and is the ring of integers of $\mathbb Q(i)$. An example of such a ring which is not a UFD (and hence is not a PID) is $\mathbb Z[\sqrt{-5}]$.


If you're looking for a result that applies specifically to PIDs:

You may have seen the structure theorem for finitely generated abelian groups, which says that if $G$ is a finitely generated abelian group, then $G$ is isomorphic to a direct product of cyclic groups.

This theorem is a special case of the following more general theorem (using the fact that abelian groups are exactly $\mathbb Z$-modules):

Let $M$ be a finitely generated $R$-module, where $R$ is a PID. Then there are uniquely determined ideals $$I_1 \supset I_2\supset\cdots\supset I_n$$such that $$M\cong \bigoplus_{i=1}^nR/I_i$$

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