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 $N(\{h\gamma\}_{1}-\{l\gamma\}_{1})=N(-1/t)=1/q>1/m$).

  • 3
    $\begingroup$ Surely we can follow the same proof by embedding the coordinates of an element with respect to an integral basis into the space $\mathbb F_q((t^{-1}))^n$? $\endgroup$
    – Will Sawin
    Feb 16, 2021 at 15:21
  • $\begingroup$ @WillSawin: I don't think the proof in Ireland--Rosen goes through in any straightforward way by embedding into $\mathbb{F}_q((t^{-1}))^n$. In particular, in their proof, the completion is irrelevant as all the elements lie in the number field. I have added some details on what it is that goes wrong. $\endgroup$ Feb 19, 2021 at 13:27
  • $\begingroup$ Why are you placing the norms of the coordinates in short intervals? Place the coordinates themselves in short intervals, where short intervals are defined by fixing the leading few coefficients (in $\mathbb F_q((t^{-1}))$) and letting the others vary. Equivalently, rational functions $a$ and $b$ are in the same short interval of length $q^{-n}$ if and only if $ a-b$ has norm at most $q^{-n}$. This is an equivalence relation, and there are exactly $q^n$ equivalence classes. $\endgroup$
    – Will Sawin
    Feb 19, 2021 at 14:11
  • $\begingroup$ Yes, that would work (no need for the completion though) because the norm is non-Archimedean in the function field case. $\endgroup$ Feb 23, 2021 at 8:53

1 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.

  • $\begingroup$ Very nice. I've corrected a few tiny typos. Note that this argument is different from that in Ireland--Rosen as the latter doesn't use any completion or compactness (the embedding into $\mathbb{R}^n$ can be replaced by an embedding into $\mathbb{Q}^n$). $\endgroup$ Feb 18, 2021 at 15:48
  • $\begingroup$ @AStasinski Thanks! I guess one can remove compactness from this argument by explicitly checking uniformity in the size of $N$ needed (they depend only on the coefficients of the polynomial expressing the determinant). Maybe it's the same (or more similar) after doing that. $\endgroup$
    – Will Sawin
    Feb 18, 2021 at 16:54

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.