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> Question: What are the (in characteristic 0 if needed) principal ideal domains that have finitely many units?
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Can such rings be classified?


(This is a more specialised version of the question in [Integral domains with finitely many units](https://mathoverflow.net/questions/388936/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?



Principal ideal domains with finitely many units
There is not likely to be a good answer to this question, because of a very annoying Theorem due to Heinzer and Roitman:



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> 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 $R$.
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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**](http://dx.doi.org/10.1081/AGB-100106813), Commun. Algebra 29, No. 11, 5197-5208 (2001). [ZBL1094.13532](https://zbmath.org/?q=an:1094.13532). for the details.




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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 $D \in \{ 3, 7, 11, 19, 43, 67, 163 \}$.




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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](https://arxiv.org/abs/1406.5365) 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.




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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.




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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 $U^{\times} = \mathbb{F}\_q^{\times}$.


