Abstract Nonsense

PIDs (Pt. I)

Point of Post: In this post we discuss PIDs, and some of the pursuant theorem.

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Motivation

In this post we discuss a more general class of rings than Euclidean domains. The type of ring we are talking about is a natural one to consider. In particular, we have seen from experience that the ideal theory of a ring is a key-determinant in the rings complexity. For example, the nicest rings (fields) are precisely those with the nicest ideal theories (just the full ring and trivial ideal). Consequently, since our current goals (as laid out in the last post) is create a series of integral domains which are “nice” (so that, when encountered in practice, all our great theory applies to them) it makes sense to create some integral domains with nice ideal theory. So, we saw that having few ideals can make an ideal theory, but it is not the only way. In particular, besides having a small number of ideals its clear that we can make the ideal theory of a ring “nice” by requiring that each ideal is “nice”. Well, it’s pretty clear what the nicest type of ideals are, principal (singly generated). Thus, in this post we shall discuss rings which are integral domains for which every ideal is principal, or as we shall call them, PIDs.  These clearly contain Euclidean domains since, as we proved before, every ideal in a Euclidean domain is principle. In fact, for most basic uses we shall see that Euclidean Domains are just “practical” PIDs in the sense that the theoretical niceties of Euclidean domains are just the niceties from being a PID, but the applications of the theory is much more practical for Euclidean domains (e.g. actually finding the element which generates the ideal, finding greatest common divisors, etc.).

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PIDs

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Let $R$ be an integral domain, then we say that $R$ is a principal ideal domain, abbreviated PID, if every ideal of $R$ is principal. For example, we showed in our last post that every Euclidean domain is a PID so that things like $\mathbb{Z}$, $\mathbb{Z}[i]$, $k[x]$ (where $k$ is a field), etc. are all PIDs. Symbolically we have that $\left\{\text{PIDs}\right\}\supseteq\left\{\text{Euclidean domains}\right\}$. This inclusion is strict though, for $\mathbb{Z}\left[\frac{1}{2}(1+\sqrt{-19})\right]$ is a PID but not a Euclidean domain (this is a notorious annoying fact to prove, see here for an elementary proof). Ok, so fine, PIDs are cool, but what makes them so nice? Well, perhaps a first indication of this is the following very nice fact:

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Theorem: Let $R$ be a PID, then $\text{Spec}(R)=\text{MaxSpec}(R)$.

Proof: We have by first principles we have that $\text{MaxSpec}(R)\subseteq\text{Spec}(R)$ and so it suffices to prove the other inclusion. To do this let $\mathfrak{p}\in\text{Spec}(R)$, by assumption that $R$ is a PID we know that $\mathfrak{p}=(p)=Rp$ for some $p\in\mathfrak{p}$. Suppose now that $\mathfrak{a}\supseteq\mathfrak{p}$, with $\mathfrak{a}=(a)$. From the fact that $(a)\supseteq(p)$ we have that $p=ab$ for some $b\in R$. Now since $(p)$ is prime this implies that either $a\in(p)$ or $b(p)$. If the former is true then $\mathfrak{a}=(a)=(p)=\mathfrak{p}$ and if the latter is true we get, from $R$ being an integral domain, that $a\in R^\times$ so that $\mathfrak{a}=(a)=R$. From this we conclude that $\mathfrak{p}$ is maximal, by containment, among all proper ideals of $R$ and so, by definition, maximal. The conclusion follows. $\blacksquare$

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There is a very interesting consequence of this which, at first, is surprising. We have seen that the functor $R\mapsto R[x]$ is fairly well-behaved (property preserving in particular) as of yet, so there is no reason to believe that it won’t be well-behaved here. Said less cryptically, one might expect/hope that $R$ being a PID implies that $R[x]$ is a PID. This is, in fact, not true. Recalling that we stated (yet to be proven) that if $k$ is a field then $k[x]$ is a Euclidean domain, and thus a PID we can state the following:

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Theorem: Let $R$ be a ring. Then, $R[x]$ is a PID if and only if $R$ is a field.

Proof: If $R$ is a field we know (since it’s a Euclidean domain) $R[x]$ is a PID. Conversely, if $R[x]$ is a PID since $R$ embeds into $R[x]$ we have that $R$ is an integral domain. From this we know that $(x)\in\text{Spec}(R[x])$ and so, by the previous theorem, $(x)\in\text{MaxSpec}(R[x])$. Thus, $R[x]/(x)$ is a field, but $R[x]/(x)\cong R$ and so $R$ is a field. $\blacksquare$

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Now, it’s difficult to imagine proving, in general, that all the ideals of a given ring are principal. Luckily though, we don’t actually have to. In fact, we only have to check the prime ideals are principal. Indeed:

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Theorem: Let $R$ be an integral domain such that every prime ideal of $R$ is principal. Then, $R$ is a PID.

Proof: Assume to the contrary that there exists non-principal ideals, then the set $P$ of such ideals is non-empty. We claim that $\left(P,\subseteq\right)$ has a maximal element. As one always does in these situations we will proceed by Zorn’s lemma. In particular, let $\mathcal{C}$ be a non-empty chain in $\left(P,\subseteq\right)$ and let $U$ be the union over all the elements of $\mathcal{C}$. We know, in general, that the union over the chain of ideals is an ideal, and so to show that $U\in P$ it suffices to prove that $U$ is non-principal. To see this suppose to the contrary that $U=(x)$ for some $x\in R$. We claim that $x\notin C$ for any $C\in\mathcal{C}$. Indeed, by assumption we have that $C\subseteq (x)$ and if $x\in C$ then $(x)\subseteq C$ so that $C=(x)$ contradictory to the assumption that each $C\in\mathcal{C}$ was non-principal. Thus, $x\notin C$ for every $C$ and so $x$ is not in the union over all the $C\in\mathcal{C}$, but this is ridiculous since by assumption this is equal to $(x)$.

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What we shall now show is that $\mathfrak{a}$ is, in fact, principal contradicting (since this was the only assumption) the non-emptiness of $P$. Indeed, suppose that $ab\in\mathfrak{a}$ but $a,b\notin\mathfrak{a}$. We can define $\mathfrak{a}_a=(\mathfrak{a},a)$ and $\mathfrak{a}_b=(\mathfrak{a},b)$ and then $\mathfrak{b}=\left\{r\in R:r\mathfrak{a}_a\subseteq\mathfrak{a}\right\}$. Now, note that since $\mathfrak{a}_a$ properly contain $\mathfrak{a}$ it cannot be non-principal, say $\mathfrak{a}_a=(\alpha)$. Furthermore, we define $\mathfrak{b}=\left\{r\in R:r\mathfrak{a}_a\subseteq\mathfrak{a}\right\}$ (the ideal quotient, if the reader is familiar with the lingo). We claim that $\mathfrak{a}\subsetneq\mathfrak{a}_b\subseteq\mathfrak{b}$. The first of these (strict) containments is clear because $b\notin\mathfrak{a}$ and the second follows since if $x+rb\in\mathfrak{a}_b$ with $x\in\mathfrak{a}$ and $z+sa\in\mathfrak{a}_a$ then $(x+rb)(z+sa)=xz+xsa+rbz+rsab$. Now, the first three terms are in $\mathfrak{a}$ since $x,z\in\mathfrak{a}$ and $\mathfrak{a}$ is an ideal, and the last term is in $\mathfrak{a}$ since $ab\in\mathfrak{a}$ and so their sum is in $\mathfrak{a}$. Since $x+rb,z+sa$ were arbitrary we may conclude that $\mathfrak{a}_b\subseteq\mathfrak{b}$. From this we can conclude that $\mathfrak{b}$ must be principal, say $\mathfrak{b}=(\beta)$. Consider then $\mathfrak{a}_b\mathfrak{b}=(\alpha\beta)$. Evidently (by definition of $\mathfrak{b}$) we have that $(\alpha\beta)\subseteq\mathfrak{a}$. That said, let $x\in\mathfrak{a}$ since $x\in\mathfrak{a}_a$ there exists $r\in R$ with $x=r\alpha$, but we see then that $r\mathfrak{a}_a=r(\alpha)=(r\alpha)=(x)\subseteq\mathfrak{a}$ so that $r\in\mathfrak{b}$. From this we gather that $\mathfrak{a}\subseteq\mathfrak{a}_a\mathfrak{b}=(\alpha\beta)$ and thus we conclude that $\mathfrak{a}=(\alpha\beta)$. As stated, this contradicts the previous paragraph, and so the initial assumption that $P$ was non-empty (which allowed us to apply Zorn’s lemma and create a paradoxical maximal element) must have been false. Thus, every ideal of $R$ is principal and so the theorem follows. $\blacksquare$

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This allows us to quickly prove the following useful fact:

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Theorem: Let $R$ be a PID and $D$ a (unitally) multiplicative subset of $R$ not containing $0$. Then, the localization $D^{-1}R$ is also a PID.

Proof: Note that since $R$ is an integral domain and $0\notin D$ (so that $D^{-1}R\ne \{0\}$) one can see that $D^{-1}R$ is an integral domain, and so it suffices to prove that every ideal of $D^{-1}R$ is principal. By the previous problem it suffices to prove that every prime ideal of $D^{-1}R$ is principal. That said, we know that every prime ideal of $D^{-1}R$ is of the form $D^{-1}\mathfrak{p}$ for some prime ideal $\mathfrak{p}$. But, since $R$ is a PID we know that $\mathfrak{p}$ is principal, say $\mathfrak{p}=(x)$. It’s easy to see then that $D^{-1}\mathfrak{p}=(x)$ (where we can recall that we are identifying $x$ with its image $\ell_D(x)$ under the localization map). $\blacksquare$

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References:

[1] Dummit, David Steven., and Richard M. Foote. Abstract Algebra. Hoboken, NJ: Wiley, 2004. Print.

[2] Rotman, Joseph J. Advanced Modern Algebra. Providence, RI: American Mathematical Society, 2010. Print.

[3] Bhattacharya, P. B., S. K. Jain, and S. R. Nagpaul. Basic Abstract Algebra. Cambridge [Cambridgeshire: Cambridge UP, 1986. Print.

October 14, 2011 -

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