I wonder if it is possible to show, without using the Schmidt subspace/Roth theorem/Baker's bounds on linear forms in logarithms or other very deep results, that, in a number field, not all integral elements are sums of two units/most of them are not.

The known results that state that there are few elements that are sums of two units in number fields (see e.g. Fuchs, Tichy, Ziegler, "On quantitative aspects of the unit sum number problem", Arch. Math. 93 (2009)) seem to use either Thue-Siegel-Roth theorem or Baker's bounds on linear forms in logarithms. Both I find very difficult to reach. Since I rely on a similar result in a statement I have made [1], [2], [3], the motivation behind the question is a wish to find a simpler argument (or to convince oneself that existence of (many) elements that are not sums of two units is a claim that is itself of strength that is seemingly out of reach to significantly easier means than Thue-Siegel-Roth/Baker theorems).


[1] Zinevičius A., On the congruent number problem over integers of real number fields, Albanian J. Math. 8 (2014), 49-53.

[2] Zinevičius A., On the congruent number problem over integers of cyclic extensions, Math. Slovaca 66 (2016), 561-564,

[3] Zinevičius A., Corrigendum to "On the congruent number problem over integers of cyclic extensions", Math. Slovaca 69 (2019), 1233-1233.

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    $\begingroup$ Units are meant to be multiplied, not added, LOL. $\endgroup$
    – Fan Zheng
    Oct 9, 2015 at 15:23
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    $\begingroup$ I included the text you gave in the bounty notice, with a minimal change, in the post, as I think it is relevant context that should be (permanently) (more) visible. $\endgroup$
    – user9072
    Oct 9, 2015 at 18:10
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    $\begingroup$ Just to be sure, are you aware that the finiteness of solutions to the two-variable unit equation has a proof based on explicit Pade approximations to $(1-x)^n$ (arising from a hypergeometric identity), together with a short geometry of numbers argument? This is given in Ch. 4 of Bombieri and Gubler. By its explicit nature (see Prop. 5.2.11 in [BG]), it avoids all the difficult non-vanishing problems from diophantine approximations [that arise in the Thue-Siegel-Roth method], and it is also probably much simpler than the proof of Gel'fond-Baker. $\endgroup$ Oct 10, 2015 at 1:40
  • $\begingroup$ I very probably miss something obvious, but suppose $u$ and $v$ are two distinct units the sum thereof is a non zero integral element of the considered field. Then $(u+v)^2-(u-v)^2=4uv$ is not the sum of two units. $\endgroup$ Oct 10, 2015 at 2:10
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    $\begingroup$ @SylvainJULIEN: I am not sure I am following this correctly, but you are not saying that $4 = (2+\sqrt{5}) + (2-\sqrt{5})$ cannot be the sum of two units of a number field, are you? Every $\alpha \in \bar{\mathbb{Z}}$ is the sum of two units in some number field: just take the two solutions of $X^2 = \alpha X + 1$. (By the way, there is a typo in my comment above: the chapter of Bombieri and Gubler given the Pade approximations solution is the 5th one, not the 4th.) $\endgroup$ Oct 10, 2015 at 2:27

3 Answers 3


$\newcommand\p{\mathfrak{p}}$ $\newcommand\OL{\mathcal{O}}$ $\newcommand\P{\mathfrak{P}}$

Here is a solution which is essentially an elaboration on Felipe's answer. Instead of working with squares, consider working with $m$th powers instead.

Lemma: If $(1 - v u^m)$ is exactly divisible by a prime $\p$ of $\OL_K$, then, assuming $K(v^{1/m}) \ne K$, the prime $\p$ is not inert in $L = K(v^{1/m})$.

Proof: In $\OL_L$, the ideal $\P = (\p,1 - v^{1/m} u)$ has norm $\p$.

By Cebotarev, the density of primes $\p$ which remain inert in $\OL_L$ is non-zero. Hence by the analytic arguments Felipe alluded to, the set of principal ideals of the form $(1 - v u^m)$ with $v$ ranging over a the (finite) set of non-zero representatives in $\OL^{\times}_K/\OL^{\times m}_K$ has density zero. So it remains to deal with ideals of the form $(1 - u^m)$, where we now have flexibility in choosing $m$.

Lemma: Let $\ell$ be any prime. Suppose that $m = |(\OL_K/\ell)^{\times}|$. Then the density of principal ideals of the form $(1 - u^m)$ is at most $1/\ell$.

Proof: Since $u^m \equiv 1 \mod \ell$, this is the same as saying that the density of principal ideals divisible by $\ell$ is at most $1/\ell$.

Taken together, it follows that the density of principal ideals of the form $(1 - u)$ has density at most $1/\ell$ for any prime $\ell$, and hence has density zero.

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    $\begingroup$ Very nice, thanks. (Not sure about the username, though...) $\endgroup$ Oct 10, 2015 at 18:16
  • $\begingroup$ Thank you very much, Michael, Vesselin, Felipe, Felipe's shadow and whoever bothered themselves with the question. I am happy with this answer and will try to patch with it what I intended to (maybe one could add that one can avoid Cebotarev here as well, and use Frobenius thm. :). $\endgroup$
    – Albertas
    Oct 10, 2015 at 18:29

This is not really an answer, but it gives an indication where the difficulty lies.

We first observe that the property of an interger in a number field $K$ to be a sum of two units is invariant under multiplication by a unit. So the question is equivalent to

How many principal ideals of the ring of integers $R$ of $K$ can be generated by an element of the form $1-u$ with a unit $u$?

Now by standard results from analytic number theory, the number of principal idals of $R$ of norm $\le X$ grows linearly in $X$. So we would like to show that the number of units $u$ such that $N(1-u) \le X$ grows more slowly.

If none of the conjugates of $u$ gets close to $1$, then $\log N(1-u)$ is not too far away from the height of $u$, and the number of units up to height $B$ grows polynomially in $B$, so the number of units such that $N(1-u) \le X$ and no conjugate of $u$ is close to $1$ grows like $(\log X)^r$ (where $r$ is the unit rank). So the remaining problem is to deal with the `bad' units. This is where the deep results come in to show that the growth is still like $(\log X)^r$.

What one needs to show is that one cannot have (too many) units $u$ such that $|1 - \sigma(u)| \ll e^{-ch(u)}$ for some embdedding $\sigma$ of $K$ into $\mathbb C$. Consider the case of unit rank $2$, with fundamental units $u_1$ and $u_2$. The condition that $|1 - \sigma(u)|$ should be small translates into $$n_1 \log |\sigma(u_1)| + n_2 \log |\sigma(u_2)| \ll e^{-c(n_1+n_2)} .$$ Now if we forget about where this came from and allow ourselves to use arbitrary real numbers $a_1$ and $a_2$ in place of $\log |\sigma(u_1)|$ and $\log |\sigma(u_2)|$, then we could pick $a_1$ and $a_2$ in such a way that $a_1/a_2$ can be extremely well approximated by rational numbers in the sense that there are arbitrary large $n_1$ and $n_2$ such that $|a_1/a_2 + n_2/n_1| \ll e^{-c' n_1}$ (with $c' > 2 c$). This would give us infinitely many linear combinations $|n_1 a_1 + n_2 a_2| \ll e^{-c(n_1 + n_2)}$.

So we need to use in some way that the logarithms of $u_1$ and $u_2$ are not just any real numbers, but have some specific (but rather subtle) properties. This seems to indicate that one really has to use rather sophisticated means to prove the result you want.


Not a full solution yet, but I think it may be possible to give a proof without the advanced tools you list. But we need to use something!

I'll start with Michael's observation that it's enough to restrict the ideals generated by $1-u, u$ a unit. Since the group of units is finitely generated (can I use that?), we can look at $1-u_0u^2, u_0$ in a finite set of representatives of units modulo squares. If $u_0$ is not a square, the prime factors of $1-u_0u^2$ split (or ramify) in $K(\sqrt{u_0})$ and using some fairly easy analytic number theory (is that allowed?) it follows that the ideals generated by $1-u_0u^2$ have density zero.

We still have to deal with the ideals generated by $1-u^2, u$ a unit.

  • $\begingroup$ Yes, this is all "allowed" (i.e., I can understand it). As for "something" that is to be used, anything that is significantly more accessible than Thue-Siegel-Roth or Baker is "allowed" (I'd hope for an approach that I could expect to understand, yet any fair simplification should count as an answer to the question). $\endgroup$
    – Albertas
    Oct 10, 2015 at 15:05

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