The Kronecker--Hurwitz property for rings of integers in global function fields

Question

In Ireland and Rosen's book on number theory they give a proof of the finiteness of the class group of a number field which they attribute to Hurwitz, but which is essentially due to Kronecker (as I learnt from a comment by Franz Lemmermeyer to this answer by KConrad). The proof is based on showing that the ring of integers in a number field satisfies the following property which I here formulate for an arbitrary integral domain.

Let's call an integral domain $R$ a Kronecker--Hurwitz domain if there exists a function $f:R\rightarrow\mathbb{Z}_{\geq0}$ and a finite subset $T\subset R\setminus\{0\}$ such that

1. $f(\alpha)=0$ if and only if $\alpha=0$;
2. for any $\alpha,\beta\in R$, $\beta\neq0$, there is a $t\in T$ and an $\omega\in R$ such that $f(t\alpha-\omega\beta)<f(\beta)$.

Perhaps it is natural to also assume that $f$ is multiplicative so that the above inequality can be written as $f(t γ - ω ) < 1$, for any $γ$ in the field of fractions of $R$ (this is what is actually proved in Ireland--Rosen when $f$ is the absolute value of the field norm).

At first sight, this looks a bit like the conditions for a Dedekind--Hasse norm, but of course there is a massive difference because if $R$ has a Dedekind--Hasse norm then it must be a PID.

Question. Let $R$ be a ring of integers in a global function field (i.e., $R$ is the integral closure of $\mathbb{F}_q[t]$, for some $t$ in the fraction field, transcendent over $\mathbb{F}_q$). Is $R$ a Kronecker--Hurwitz domain?

I would also be interested to know whether the defining property of Kronecker--Hurwitz domains has appeared anywhere in the literature.

I have asked a couple of experts about this, but have so far not made any progress towards an answer.

Note. I have not been able to adapt the proof in Ireland--Rosen (p. 178) that rings of integers in number fields are Kronecker--Hurwitz to the function field case because it is based on embedding the coordinates of an element w.r.t. an integral basis into the Euclidean space $\mathbb{R}^n$. More precisely, let $N$ be the ideal norm on $\mathbb{F}_q[t]$ extended to $\mathbb{F}_q(t)$. Partition the unit cube $\{x\in\mathbb{F}_{q}(t)\mid0\leq N(x)\leq1\}^{n}$ into $m^{n}$ subcubes with side length $1/m$. At the end of the proof, we obtain a $\delta=\{h\gamma\}-\{l\gamma\}$ with $\{h\gamma\}$ and $\{l\gamma\}$ in the same subcube. This means that the $\mathbb{F}_{q}(t)$-coordinates of $\{h\gamma\}_{i}$ and $\{l\gamma\}_{i}$ of $\{h\gamma\}$ and $\{l\gamma\}$ satisfy $$ \frac{a_{i}}{m}\leq N(\{h\gamma\}_{i}),N(\{l\gamma\}_{i})\leq\frac{a_{i}+1}{m}, $$ for some integers $0\leq a_{i}\leq m-1$. For the proof to go through, however, we would need $N(\{h\gamma\}_{i}-\{l\gamma\}_{i})\leq1/m$, which does not follow from the above inequalities in the function field case (e.g., if $m=q^{2}$, $a_{1}=q-1$, $\{h\gamma\}_{1}=\frac{1}{t}$, $\{l\gamma\}_1=\frac{2}{t}$, assuming $\text{char }\mathbb{F}_{q}>2$; then $N(\{h\gamma\}_{1})=N(\{l\gamma\}_{1})=1/q=\frac{a_{i}+1}{m}$, but $N(\{h\gamma\}_{1}-\{l\gamma\}_{1})=N(-1/t)=1/q>1/m$).

Answer 1

Let us take $f$ to be the norm (i.e. view $R$ as a finite rank module over $\mathbb F_q[t]$. Each element of $R$ acts by multiplication on this module, so its determinant lies in $\mathbb F_q[t]$. Take $f$ to be $q$ to the degree.

Then it suffices to prove for $\gamma$ in the field of fractions of $R$ that there is $c \in T$ and $\omega \in R$ with $f( c\gamma - \omega)< 1$, i.e. with the degree of the determinant of $c \gamma -\omega$ negative.

We can generalize this to allow $\gamma \in R \otimes_{ \mathbb F_q[t] } \mathbb F_q((t^{-1}))$, where $\mathbb F_q((t^{-1}))$ is the field of formal Laurent series. The norm makes sense for elements of this ring for the same reason - it's a free module over $\mathbb F_q((t^{-1}))$, so we can take determinants, and then look at the degree in $t$ of the leading term.

The advantage of this generalization is that $(R \otimes_{ \mathbb F_q[t] } \mathbb F_q((t^{-1})) ) / R = ( \mathbb F_q((t^{-1}))/\mathbb F_q[t])^n$ is compact, and since for each $c\in R$, the set of $\gamma$ such that there exists $\omega \in R$ with $f( c\gamma -\omega)<1$ is open and invariant under translation by $R$, it suffices to check that for each $\gamma \in R \otimes_{ \mathbb F_q[t] } \mathbb F_q((t^{-1})) $ there exists $c,\omega \in R$ with $f( c\gamma - \omega)<1$, since this gives an open cover indexed by $c$ and we can find a finite subcover.

Given $\gamma$, express multiplication by $\gamma$ as an $n\times n$ matrix over $\mathbb F_q((t^{-1}))$ and let $d$ be the greatest degree in $t$ of an element of that matrix. Then for $c \in R$ whose coordinates are polynomials of degree $\leq N$, and $\omega \in R$ whose coordinates are polynomials of degree $\leq N+d$, $c\gamma-\omega$ has coordinates in $\mathbb F_q((t^{-1}))$ of degree $\leq N+d$.

The map $(c, \omega) \to c \gamma - \omega \mod t^{-N} \mathbb F_q[[t^{-1}]]$, where $c$ has coordinates of degree $\leq N$ and $\omega$ has coordinates of degree $\leq N+d$, is a map from an $n(N+1) + n (N+d+1)$-dimensional vector space to an $n ( 2N+d)$-dimensional vector space, and thus has nontrivial kernel, so there exist $c, \omega$ in $R$ with $c\gamma - \omega$ having all coordinates of degree $\leq -N$. Taking $N$ sufficiently large depending on the coefficients of the polynomial expressing of the determinant, we get $c \gamma -\omega$ of small norm.