I would be thankful for a reference to any result that says "how often" an equation of the form $$c_1x_1 + c_2x_2 + ... + c_nx_n = 0,$$

where $n$ is fixed, $c_1, ..., c_n \in \mathcal{O}_K$ are arbitrarily given algebraic integers of a fixed number field $K$ and $x_1, ..., x_n$ vary in the set of units of $\mathcal{O}_K$, has/does not have a solution.

Thank you!


2 Answers 2


There's a paper of Evertse, Gyory, Stewart, and Tijdeman in which they prove that for "most" (in an appropriate sense) $a,b\in K$, the equation $ax+by=1$ has at most 2 soutions in units $x,y\in\mathcal{O}_K^*$. The proof is to assume that there are 3 solutions $(x_i,y_i)$, $1\le i\le 3$, use those three equations to eliminate $a$ and $b$, yielding a sum of units that are products of the $x_i$ and $y_i$. Then results of Schlickewei, Evertse, ... (which are based on Schmidt's subspace theorem) imply that there are only finitely many solutions in which no subsum vanishes. It remains to untangle what happens if some subsum does vanish, which isn't so bad for $ax+by=1$, but becomes combinatorially very complicated when there are more variables. In any case, this gives you an idea of how one attacks this sort of problem, where one wants to know that most choices of coefficients give few solutions in units.

Addendum: First, as GH noted, Gyory has 100+ papers, not 1000+ (162 at current MathSciNet count). Second, I tracked down the references:

MR0939471 Evertse, J.-H.; Győry, K.; Stewart, C. L.; Tijdeman, R., On S-unit equations in two unknowns., Invent. Math. 92 (1988), no. 3, 461–477.

The generalization to $n$ variables is the following:

MR2093164 Evertse, Jan-Hendrik, Linear equations with unknowns from a multiplicative group whose solutions lie in a small number of subspaces. Indag. Math. (N.S.) 15 (2004), no. 3, 347–355.

The conclusion of the paper is that for most choices of coefficients, the solutions in units lie in at most $2^n$ proper linear subspaces of $K^n$.

  • $\begingroup$ Probably you meant more than 100 papers instead of 1000 (cf. MathSciNet). $\endgroup$
    – GH from MO
    Mar 26, 2014 at 17:38
  • $\begingroup$ Thank you! It would be of interest to know if the question whether, given a number field, one can always find such $c_1, ..., c_n$ (with at least three of the $c_j$'s nonzero), that the equation above had no solutions, is open. In fact, any example for a number field with the unit group of rank at least $2$ would be nice to see. $\endgroup$
    – Albertas
    Mar 27, 2014 at 15:54
  • 1
    $\begingroup$ @Albertas You might try the analogous (but probably very similar) question using $S$-units in $\mathbb{Q}$. So for example, take $S=\{2,3\}$, so $\mathbb{Z}_S^*$ has rank 2. Now can you find $c_1,c_2,c_3$ so that $c_1x_1+c_2x_2+c_3x_3=0$ has no solutions $x_1,x_2,x_3\in\mathbb{Z}_S^*$. This seems more tractable, and might give ideas for the number field case. $\endgroup$ Mar 27, 2014 at 19:07
  • $\begingroup$ I forgot to mention that the principal ideals generated by $c_j$'s should be coprime (else it is easy to construct a polynomial that is never zero on the set of units, but this does not help me). One may expect to find a prime ideal $\mathfrak{p}$ such that all the fundamental units as well as units of finite order were, say, squares in the corresponding residue field - then one could choose $c_1 = p \in \mathfrak{p}$ a rational prime, $c_2$ quadratic residue and $-c_3$ quadratic nonresidue mod $\mathfrak{p}$, respectively. But it does not seem obvious there must exist such $\mathfrak{p}$. $\endgroup$
    – Albertas
    Apr 3, 2014 at 15:27

As a partial answer, one can show, that, given a number field $K$, there exist infinitely many $c_1, c_2, c_3 \in \mathcal{O}_K$, such that the principal ideals $c_1\mathcal{O}_K, c_2\mathcal{O}_K, c_3\mathcal{O}_K$ are pairwise coprime and the equation

$$c_1x_1 + c_2x_2 + c_3x_3 = 0$$

has no solutions with $x_j \in \mathcal{O}_K^*$.

It suffices to show that there exists an ideal $\mathfrak{m} \subset \mathcal{O}_K$ such that the image of $\mathcal{O}_K^*$ in the residue ring $\mathcal{O}_K/\mathfrak{m}$ is a proper subgroup of the multiplicative group of the latter (then one can choose $c_1$ to be any number in the image, $-c_2$ - any number in the complement of the image and $c_3$ to be the rational prime number $p \in \mathfrak{m}$ to conclude that the equation obtained in this way does not have solutions modulo $\mathfrak{m}$).

That one can always find (many) such $\mathfrak{m}$ seems very natural to expect, however the following is a quite indirect argument that probably should be (much?) simpler: the order of the ray class group of a chosen modulus $\mathfrak{m}$ is a product of the class number $h_K$ and the index $[(\mathcal{O}_K/\mathfrak{m})^*: (\mathcal{O}_K^*/\mathfrak{m})]$ (as explained, for instance, here: https://math.stackexchange.com/questions/74206/ray-class-group ). The orders of the ray class groups of $K$, on the other hand, are arbitrarily large, since the Galois group of each abelian extension of $K$ is isomorphic to some ray class group (and there exist abelian extensions of $K$ of arbitrarily large degree).

  • $\begingroup$ The argument for the existence of such $\mathfrak{m}$ is indeed much simpler: it suffices to know that, given a generator $\zeta$ of the group of roots of unity in $K$ and a system of fundamental units $u_1, ..., u_s$, there exist many prime ideals $\mathfrak{m}$ that split completely in, e.g., $K(\sqrt{\zeta}, \sqrt{u_1}, ..., \sqrt{u_s})$ - then any unit is a square modulo $\mathfrak{m}$. $\endgroup$
    – Albertas
    Jun 16, 2014 at 12:19

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