Question: What are the (in characteristic 0 if needed) principal ideal domains that have finitely many units?
Can such rings be classified?
(This is a more specialised version of the question in Integral domains with finitely many units and got split off of this thread)
The well known examples are the imaginary quadratic integer rings of $\mathbb{Q} (\sqrt{d})$ for $d \in \{−1, −2, −3, −7, −11, −19, −43, −67, −163 \}$.
Is there a nice infinite family in characteristic 0?
There is not likely to be a good answer to this question, because of a very annoying Theorem due to Heinzer and Roitman:
If $D$ is any UFD, then there is a PID $R$ containing $D$ which has the same unit group as $D$ and such that every prime of $D$ remains prime in $R$.
Since UFD's with finite unit group are common, this will give lots of examples.
Here is a sketch of the proof of Heinzer and Roitman's result: If $D$ is a UFD, and $a$ and $b$ are relatively prime elements of $D$, set $D_{a,b} = D[x,y]/(ax+by-1)$. One can check that $D_{a,b}$ is a UFD and that the ideals $(a)$ and $(b)$ are comaximal in $D_{a,b}$; one can also check that $D^{\times} = D^{\times}_{a,b}$ and that prime elements of $D$ stay prime in $D_{a,b}$.
Using the $D_{a,b}$ construction and a transfinite induction procedure, we embed $D$ into a ring $R$ such that $R$ is a UFD and any two relatively prime elements of $R$ generate comaximal ideals. Such a UFD is necessarily a PID. One also has $R^{\times} = D^{\times}$, and every prime of $D$ stays prime in $R$. See Section 4 of Heinzer, William; Roitman, Moshe, Principal ideal domains and Euclidean domains having 1 as the only unit, Commun. Algebra 29, No. 11, 5197-5208 (2001). ZBL1094.13532. for the details.
As observed in the comments, if such a ring $A$ is finitely generated over $\mathbb{Z}$, it must be either be an order in the ring of integers in a number field, or else it must be the coordinate ring of an affine curve over $\mathbb{F}_q$.
In the first case, by Dirichlet's unit theorem, the number field $K$ must be an imaginary complkex field. Since PID's are integrally closed, $A$ must be the full ring of integers. The imaginary quadratic number fields with class number $1$ are well known, they are $\mathbb{Z}[\sqrt{-1}]$, $\mathbb{Z}[\sqrt{-2}]$ and $\mathbb{Z}[\tfrac{1+\sqrt{-D}}{2}]$ for $D \in \{ 3, 7, 11, 19, 43, 67, 163 \}$.
In the second case, let $A$ be the coordinate ring of an affine curve $U$ over $\mathbb{F}_q$, and let $X$ be the smooth projective curve completing $U$. By Dirichlet's unit theorem for function fields, $U \setminus X$ is a single closed point $x$; let that point $x$ have residue field $q^f$. Then $\mathrm{Pic}(U) = \mathrm{Pic}(X)/[x]$. We have a short exact sequence $$0 \to \mathrm{Pic}^0(X) \to \mathrm{Pic}(X) \to \mathbb{Z} \to 0$$ and the image of $[x]$ in $\mathbb{Z}$ is $f$ times the generator, so $\mathrm{Pic}(U)$ has a $\mathbb{Z}/f \mathbb{Z}$ quotient.
Our assumption that $A$ is a PID means that $\mathrm{Pic}(U)$ is trivial, so we deduce that $f=1$ and $x$ is an $\mathbb{F}_q$ point. In that case, we have $\mathrm{Pic}(U) = \mathrm{Pic}^0(X)$, so we must additionally have $\mathrm{Pic}^0(X)$ trivial. In other words, if $J$ is the Jacobian of $X$, we must have $J(\mathbb{F}_q)$ trivial.
Let's assume from now on that $X$ has genus $>0$, since genus $0$ gives us the case $\mathbb{F}_q[x]$, which we already know. Note that, with this assumption, requiring that $J(\mathbb{F}_q)$ be trivial imposes in particular that $X(\mathbb{F}_q)$ has at most one point.
Let $\lambda_1$, $\lambda_2$, ..., $\lambda_{2g}$ be the eigenvalues of $q$-power Frobenius on $X$. Then the size of $J(\mathbb{F}_q)$ is $\prod (\lambda_i-1)$. Each $\lambda_i$ is a complex number of norm $\sqrt{q}$, so $\lambda_i$ is a complex number of norm at least $\sqrt{q}-1$. Thus, all examples have $\sqrt{q}-1 \leq 1$, so $q \in \{ 2,3,4 \}$.
Heinzer and Roitman look at the $q=2$ case and give the examples of $y^2+y = x^3+x^2+1$ (genus one) and $y^2+y = x^5+x^3+1$ (genus two). For $q=3$, I noticed $y^2 = x^3-x+1$ (genus one) and for $q=4$ I noticed $y^2+y = x^3+\omega$ (genus one), where $\omega$ is a root of $\omega^2+\omega+1=0$ in $\mathbb{F}_4$. The eigenvalues of Frobenius in the genus $1$ cases are $\sqrt{2} e^{\pm(2 \pi i)/8} =1\pm i$, $\sqrt{3} e^{\pm(2 \pi i)/12} = \tfrac{3}{2} \pm \tfrac{\sqrt{-3}}{2}$ and $2$ (with multiplicity two). The eigenvalues of Frobenius in the genus $2$ case are $\sqrt{2} e^{\pm(2 \pi i)/24}$ and $\sqrt{2} e^{\pm 7(2 \pi i)/24}$, also known as $(1\pm i)\left( \tfrac{1\pm \sqrt{-3}}{2} \right)$.
I attempted to prove that the curves above were the only examples, but my proof had a gap. Fortunately, pregunton discovered a paper of Mercuri and Stirpe which lists all curves with $J(\mathbb{F}_q)$ trivial (both with $X(\mathbb{F}_q)$ singleton, like we want, and with $X(\mathbb{F}_q)$ empty). The curves listed above are cases (i), (ii), (vi) and (vii) on their list; I believe all the other curves on their list have $X(\mathbb{F}_q)$ empty.
Here is what I can salvage from my computation, which I still like:
There are no examples with $q=4$ and genus $>1$. If there were, then we would have $\lambda_1 = \lambda_2 = \cdots = \lambda_{2g}=2$. But then the number of points in $X(\mathbb{F}_4)$ would be $4-\sum \lambda_i + 1 = 5-4g<0$, a contradiction.
Now let $q$ be $2$ or $3$. We observe that $\sqrt{q}$ must have even multiplicity as an eigenvalue of Frobenius. Proof: We have $\#J(\mathbb{F}_q) = \prod (1-\lambda_j)$. Group together the terms with complex conjguate eigenvalues; then every factor except the one coming from $1-\sqrt{q}$ is a positive real, so $(1-\sqrt{q})$ must be raised to an even power as well. Using $\#J(\mathbb{F}_q) = \prod (\lambda_j-1)$, we see that $-\sqrt{q}$ likewise has even multiplicity.
Therefore, we can write the eigenvalues of Frobenius as $\sqrt{q} e^{\pm i \theta_j}$ for $1\leq \theta_j \leq g$, remembering to take both signs in the exponent even when $\theta_j$ is $0$ or $\pi$. Let $c_j = \cos \theta_j$.
The equation $X(\mathbb{F}_q)=1$ translates to $$q-2 \sqrt{q} \sum \cos \theta_j +1 =1 \ \implies \ \sum c_j = \tfrac{\sqrt{q}}{2} \quad (\ast)$$ The equation $J(\mathbb{F}_q)=1$ translates to $$\prod (q+1-2 \sqrt{q} c_j) = 1 \quad (\dagger).$$ We note that $(\ast)$ is a convex polytope whose vertices are $(1,1,\ldots,1, -1, -1, \ldots, -1, \tfrac{\sqrt{q}}{2})$ for $g$ odd and $(1,1,\ldots,1, -1, -1, \ldots, -1, \tfrac{\sqrt{q}}{2}-1)$ for $g$ even.
We will try to show that $\prod (q+1-2 \sqrt{q} c_j) \geq 1$ everywhere on this polytope. Note that $\log \prod (q+1-2 \sqrt{q} c_j)$ is concave, so it is enough to check the inequality at the vertices.
We compute that the value at the vertices is $$\begin{array}{c|cc} & g=2k+1 & g=2k+2 \\ \hline q=2 & 1 & (3-2 \sqrt{2}) (3 - 2 \sqrt{2} (\tfrac{\sqrt{2}}{2}-1)) \approx 0.657 \\ q=3 & 4^k & 4^k (4-2 \sqrt{3}) (4 - 2 \sqrt{3} (\tfrac{\sqrt{3}}{2}-1)) \approx 2.39 \times 4^k \\ \end{array}.$$
We see that the only solution with $q=3$ is $g=1$ with $c_1 = \tfrac{\sqrt{3}}{2}$.
If $q=2$ and $g=2k+1$, then $(\dagger)$ only occurs at the vertices of the polytope, corresponding to the eigenvalue sequence $\sqrt{2}$ (multiplicity $2k$), $-\sqrt{2}$ (multiplicity $2k$) and $1 \pm i$. However, this would lead to a curve with $1-4k$ points over $\mathbb{F}_4$, so this can only happen for $k=0$ (and thus $g=1$.)
However, I am unclear as to how to deal with the case of $q=2$, $g$ even. When $g=2$, the polytope is just a line segment and $(\dagger)$ occurs at two symmetrically placed points of the line segment, corresponding to $(c_1, c_2) = ( \tfrac{1 \pm \sqrt{3}}{\sqrt{2}}, \tfrac{1 \mp \sqrt{3}}{\sqrt{2}})$. However, for larger $k$, $(\dagger)$ occurs on a little $2k$-dimensional manifold in the corners of the polytope, and it isn't clear to me how to rule out more solutions here.
A fun fact is to note that none of these affine curve examples (except genus zero) will be Euclidean. This is due to the following nice exercise: If $A$ is a Euclidean domain and not a field, then there is some prime $\mathfrak{p}$ of $A$ such that $A^{\times} \to (A/\mathfrak{p})^{\times}$ is surjective. (Hint: Take a non-unit of minimal norm.) Since $U$ has no $\mathbb{F}_q$ points in any of these examples, all of the residue fields $A/\mathfrak{p}$ are proper extensions of $\mathbb{F}_q$, whereas $U^{\times} = \mathbb{F}_q^{\times}$.
