A Nonarithmetic Example of a Noneuclidean Principal Ideal Domain
12 Oct 2021  Tags: featured
This blog post changed dramatically over the course of writing it… I’m keeping the title, because I like the alliteration of “nonarithmetic noneuclidean”, but by the end I actually understood the arithmetic example better too. Perhaps a better title would be “In which algebraic geometry explains noneuclidean PIDs in a way that finally makes sense to me”.
Every first course on rings will prove the chain of inclusions
\[\text{Fields} \subset \text{Euclidean Domains} \subset \text{PIDs} \subset \text{UFDs} \subset \text{Domains}\]and most first courses will prove each of these inclusions are strict by giving examples of rings that are one but not the other. Most of these are fairly elementary, and are also memorable in the sense that, after seeing the example, it helps you understand why the inclusion is strict, and conversely, once you really understand the why, it’s quite easy to come up with counterexamples.
The exception, of course, is finding a PID which is not a euclidean domain. I’ve looked this up enough times to have the “standard” counterexample memorized: It’s $\mathbb{Z} \left [ \frac{1 + \sqrt{19}}{2} \right ]$… Obviously. Also, if you asked me right now to prove that this is a PID and isn’t euclidean, I’m fairly confident I couldn’t. One big reason for this, I think, is that I really know embarrassingly little number theory. This ring is obviously number theoretic, and so it makes perfect sense to me that knowing some heavier hitting number theory would probably clarify why this ring works (and also why it’s the “simplest” such ring). See here for what I’m sure would be a great explanation if I knew more number theory.
Now, with this in mind, I wanted to try and find examples of this phenomenon outside of number theory, but it took me a surprisingly long time to find one^{1}! In fact, the only examples I had seen were rings of integers of fields like $\mathbb{Q}(\sqrt{19})$, $\mathbb{Q}(\sqrt{43})$, $\mathbb{Q}(\sqrt{67})$, $\mathbb{Q}(\sqrt{163})$.
As an aside, if I understand correctly, these four are the only number theoretic examples if we assume the generalized riemann hypothesis, and this got me wondering if the “generic” PID is actually euclidean^{2}. I’m still interested in that question, but while writing this post I found an article which shows how to construct a whole slew of examples^{3}, so I suspect it’s not true (if it can even be made precise).
On to the main event, though:
is a noneuclidean PID.
This is obviously some kind of geometric object, but it’s quite hard for me (at time of writing) to visualize… It’s the coordinate ring of some curve in $\mathbb{R}$ that doesn’t really exist (since $x^2 + y^2 + 1$ has no real solutions).
Even though $x^2 + y^2 + 1$ has no solutions in $\mathbb{R}^2$, we can still argue about this ring geometrically. This is part of the magic of algebraic geometry!
For instance, we expect the surface carved out by this polynomial to be one dimensional. After all, it’s obviously some kind of curve in the plane. We also expect it to be smooth, in the sense that the partial derivatives in the $x$ and $y$ direction never simultaneously vanish (note $(0,0)$ is not on our curve).
Now, I’m willing to believe that this is a UFD, since oftentimes a failure of factorization is associated with a singular point^{4} and we’ve just checked that our variety is smooth. Since we’ve also checked $R$ is one dimensional, it must actually be a PID^{5}.
I don’t know of a good geometric interpretation of euclideanness, but thankfully the direct proof that $R$ isn’t euclidean^{6} is much easier (to my sensibilities at least) than the proof for the number theoretic examples:
$\ulcorner$ Let \(\cdot\) be a euclidean function on $R$, and let $p$ be a \(\cdot\)minimal nonzero nonunit. Then $p$ is prime, thus maximal (in each case, do you see why?), so $R / (p)$ is a field algebraic over $\mathbb{R}$ and properly containing it. The only choice is $\mathbb{C}$.
Now, what’s the plan? We know $R^\times \cong \mathbb{R}^\times$, and $(\mathbb{R} / p)^\times \cong \mathbb{C}^\times$. But we have a surjection $\pi : R \twoheadrightarrow R/p$! At this point we should start smelling blood in the water, since there’s no obvious surjection from $\mathbb{R}^\times \twoheadrightarrow \mathbb{C}^\times$.
We can simplify things by noticing the projection map $\pi$ is injective on $R^\times = \mathbb{R}^\times$ (do you see why? Remember $\pi$ is a $\mathbb{R}$algebra homomorphism), so if we can show it’s also surjective, we’ll be done, since certainly $\mathbb{R}^\times \not \cong \mathbb{C}^\times$ (again, do you see why?).
How do we do it? Well, let’s take a unit $\overline{u} \in R / p$. Fix a lift $u \in R$, and (using the euclidean algorithm) write $u = pq + v$ with \(v \lt p\). Notice $\overline{u} = \overline{v}$, so since $\overline{u} \neq 0$ (it’s a unit) we must have $\overline{v} \neq 0$, and $v \neq 0$ too (do you see why?). But since $p$ was \( \cdot \)minimal amongst the nonzero nonunits, $v$ must be a unit. Then $\pi(v) = \overline{u}$, and $\pi$ is surjective (as needed). $\lrcorner$
Actually, now that I’ve typed this up, there is a morethansuperficial similarity between this and the $S \triangleq \mathbb{Z} \left [ \frac{1 + \sqrt{19}}{2} \right ]$ version…
First, we might look at $\mathbb{Z}[\sqrt{19}]$. This is not a UFD, but we can fix this by moving to $S$. This move to the integral closure “removes the singularities”, and gives us a UFD^{7}. Now this UFD is one dimensional^{8}, so it’s a PID, as before. As for why it isn’t euclidean:
$\ulcorner$ Again, let \(\cdot\) be a euclidean norm, and pick a \(\cdot\)minimal nonzero nonunit $p$. As before, $(p)$ is maximal. Next, notice that $S^\times = { \pm 1 }$. So by definition, there are only $3$ elements of norm less than \(p\): $0$, $1$, and $1$.
By the euclidean algorithm, each $x \in S$ can be written as $x = pq + r$ where $r \lt p$. So there are at most $3$ residue classes mod $p$, and thus $S / p$ (which must be a field) is either $\mathbb{F}_2$ or $\mathbb{F}_3$.
Finally, the polynomial $x^2 + x + 5$ has a root in $S$ (indeed, $\theta = \frac{1 + \sqrt{19}}{2}$ works), but this polynomial does not have a root in $\mathbb{F}_2$ or $\mathbb{F}_3$.
Now if $\pi : S \twoheadrightarrow S/p$ were as described, then $\pi(\theta)$ would have to satisfy this polynomial. A contradiction. $\lrcorner$
Notice in both cases, we were able to wave our hands past the PID proof by arguing that we’re a one dimensional UFD. Similarly, in both cases we took a minimal nonzero nonunit $p$ of a purported euclidean function. Then by understanding the field we get by quotienting by $p$^{9}, we were able to show that something bad must happen. For $R$, this was an isomorphism $\mathbb{R}^\times \cong \mathbb{C}^\times$. For $S$, this was a solution to an insoluble equation.
I definitely learned a lot writing this post! Hopefully you all enjoyed too ^_^. If people have ideas about more ways to make this analogy precise, I would love to hear about it. In particular, the handwaving about smoothness implying UFDness feels a bit aggressive to me, and I would love to know more about how to put that on solid ground.
If people have any references for “what fraction” of PIDs are actually Euclidean, I would also be super interested in hearing about that! I’m not sure if there’s a great way to measure that, though. Maybe there’s an informal way of thinking about it?
Anyways, that’s two posts in one night, so I’m feeling quite tired now, haha. Plus I (appropriately enough) have a number theory class tomorrow at $9.30$, so it’s ~ bed time ~. See you all soon ^_^.

Rather embarrassingly, while writing this post I learned the example which I’m about to give is actually on the wikipedia page for euclidean domains… So… I probably could have found an example sooner, haha. ↩

Interestingly, while many rings of integers are euclidean domains, the norm isn’t the norm you would expect (inherited from $\mathbb{C}$). See here for more information, as well as the claim I cited about the generalized riemann hypothesis. ↩

D. D. Anderson’s An Existence Theorem for Noneuclidean PIDs, which you can find here, for instance.
The author shows:
If $(R,\mathfrak{m})$ is any $2$dimensional local UFD, then $R[1/f]$ is a nonprincipal PID for any $0 \neq f \in \mathfrak{m}^2$ a principal prime.
For instance, if $k$ is a field, then $k[[x,y]]$ is a $2$ dimensional local UFD (with maximal ideal $(x,y)$). So taking $f \in (x,y)^2$ irreducible, say, $f = x^2 + y^3$, we find $k[[x,y]][1/f]$ is principal and noneuclidean.
Also, a quick clarifying note: the author uses “quasilocal”, to mean a nonnoetherian ring with a unique maximal ideal. It seems it used to be standard to assume noetherianness when discussing local rings, but this has fallen out of fashion. Nowadays we use the word “local” in a broader sense. You can see some discussion about this here.
I imagine that a “principal prime” element is a similarly archaic name for an irreducible element. Since in a UFD, the principal ideal generated by an irreducible is prime. I haven’t actually read through the proof closely enough to verify this, though. ↩

Though this argument doesn’t really work here, since you can also “accidentally” fail to be a UFD when your underlying field isn’t algebraically closed. For instance, the coordinate ring of a circle
\[\mathbb{R}[x,y] \big / \langle x^2 + y^2  1 \rangle\]is superficially very similar to our ring (in particular it’s smooth) but unique factorization fails (do you see why?) ↩

In a one dimensional UFD, there’s not enough room for a nonprincipal ideal to exist. Indeed, say we had a nonprincipal ideal $(f,g)$. Then
\[(0) \subset (f) \subset (f,g)\]is a chain of length $2$, which can’t fit in a ring of dimension $1$.
The UFD condition avoids certain pathologies (again, associated with singularities), for instance in $k[t^2, t^3] = k[x,y] \big / \langle x^3  y^2 \rangle$, which has a singularity, we find the ideal $(t^2, t^3)$ (corresponding to the singular point at the origin) is nonprincipal, even though $k[t^2,t^3]$ is one dimensional. ↩

Recall the passage to the integral closure is how we get the Noether Normalization, which removes the singular points of a curve. ↩

See here, for instance. We need to combine this with the fact that normalizing (that is, taking the integral closure) doesn’t change the dimension. ↩

Which is easy, since the elements of the field are exactly the elements of norm \(\lt p\), and we can get our hands on them by using the euclidean algorithm ↩