Here is an apparent gap in a discreteness result of Lang that is a preliminary step in his proof of Dirichlet’s $S$-unit theorem. I have been working on a Minkowski-free approach to algebraic number theory, the goal being to rewrite Neukirch’s chapter on the Rieman-Roch theorem in his book Algebraic Number Theory in terms of adeles and ideles. I have been working through Tom Weston’s paper “The Idelic Approach to Number Theory and ran into a problem with his proof of the discreteness of the logarithmic representation of the $S$-units. He follows Lang’s book Algebraic Number Theory very closely in this regard.

Let $K$ be a number field, i.e. a finite algebraic extension of degree $n$ of the rational numbers, $\mathbb{Q}$. Let $B$ be the ring of integral elements of $K$. Let $S$ be a finite set, of cardinality $s$, of non-trivial absolute values of $K$, including all the Archimedean absolute values, taking every real Archimedean absolute value and a representative of each conjugate pair of complex Archimedean absolute values. An $S$-unit is a non-zero member of $K$ whose absolute value is 1 for every absolute value not in $S$. Write $s$-dimensional Euclidean space as $R’$. Define a map $\operatorname{Log}: \{ S\text{-units}\} → R’$ by $\operatorname{Log}(x)=(\log(|x|_1), … , \log(|x|_s))$.

The assertion is made, by Lang on p.144 of his book Algebraic Number Theory and in Tom Weston’s paper that any bounded region of $R’$ contains only finitely many members of $\operatorname{Log}(S\text{-units})$. Lang’s “proof” is “This is clear since prescribing a bounded region of $R’$ in effect defines bounds on the absolute value of an element of $K$ and hence bounds on the coefficients of the equation which this element satisfies over $\mathbb{Z}$.”

If $x$ is an $S$-unit, it is not clear which equation is involved but I take it to mean the field polynomial or the minimal polynomial of $x$ being $0$ at $x$. The statement about bounds presents no problem, since the absolute values of the complex roots of the field polynomial are the Archimedean absolute values of $x$ and the and the coefficient of degree $n-I$ is, up to parity, the sum of the products of the roots taken $I$ at a time, since the field polynomial is monic. Lang’s argument about finiteness appears to be that there are only finitely many possibilities for each coefficient, because there are only finitely many integers with ordinary absolute value less than some bound and thus there are only finitely many possible field or minimal polynomials and hence only finitely many possible $x$.

The problem is that an $S$-unit is not necessarily an algebraic integer-the coefficients of the field or minimal polynomials of $x$ are rational but they are not necessarily integers. There are infinitely many rational numbers with absolute value less than a bound. Of course we can multiply by the product of the denominators to obtain a polynomial with integral coefficients but then the polynomial is no longer monic and the bounds on the coefficients no longer apply.

The only way I can see to rescue the proof is to find some sort of granularity result on the coefficients of the field or minimal polynomial, such as showing that each coefficient of all the $S$-units has the form $c/d$ where $c$ and $d$ are integers and $d$ does not depend on $x$ but I cannot see how to obtain such a result. The non-Archimdean absolute values in $S$ correspond to prime ideals in $B$, each of which lies over a prime ideal in $\mathbb{Z}$ generated by a prime number. Could we do something with the set of all such prime numbers? Or am I barking up the wrong tree?

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    $\begingroup$ Probably lots of readers are more dedicated than I, but I gave up at the wall of text. Please format at least with line breaks, and preferably also with TeX where appropriate. $\endgroup$
    – LSpice
    Dec 24, 2019 at 2:31
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    $\begingroup$ Are you able to show the $S$-integers $\mathcal O_{K,S}$ form a discrete (co-compact) subset of $\prod_{v \in S} K_v$? For example, taking $K = \mathbf Q$ and $S = \{2,\infty\}$, can you show $\mathcal O_{\mathbf Q,S} = \mathbf Z[1/2]$ is discrete (and co-compact) in $\mathbf R \times \mathbf Q_2$? $\endgroup$
    – KConrad
    Dec 24, 2019 at 5:05

1 Answer 1


I think what is meant here is indeed not a monic polynomial over $\mathbb Z$, but rather one whose coefficients are (jointly) coprime. This, of course, is uniquely determined by the minimal polynomial of an algebraic number $x$.

Now, if $x$ satisfies the equation $$ a_nx^n + \dots + a_1 x + a_0 = 0 $$ with $a_i\in\mathbb Z$ jointly coprime, and is an $S$-unit, then for $v\not\in S$ we have $\lVert{x}\rVert_v\leqslant 1$. Say, we have $\lVert{x}\rVert_v\leqslant R$ for $v\in S$; then, analogously to your argument in the post, we get $\lVert{a_k/a_n}\rVert_p \leqslant CR^{2n}$ for all $p$-adic valuations on $\mathbb Q$. But this together with joint coprimality means that the denominator $a_n$ is bounded: the inequality $$ \lVert{a_n}\rVert_p\geqslant \frac{\lVert{a_k}\rVert_p}{CR^{2n}},\quad k=0,\dots,n-1 $$ together with the fact that $a_0,\dots,a_n$ cannot all be divisible by $p$ implies $$ \lVert{a_n}\rVert_p\geqslant \frac{1}{CR^{2n}}, $$ so $a_n$ is indeed bounded as long as we have a bound on norms of $x$.

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    $\begingroup$ An ingenious solution to to a problem that's bothered me for several months. I wonder if that is what Lang meant but it doesn't matter. Vadim, you have given me a beautiful Christmas present.. $\endgroup$ Dec 24, 2019 at 17:10
  • $\begingroup$ Glad I could help! Merry Christmas! :) $\endgroup$ Dec 24, 2019 at 17:12
  • $\begingroup$ @P.Lawrence: Please accept this answer officially (so that it turns green). $\endgroup$
    – GH from MO
    Jan 4, 2020 at 17:48

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