# algebraic series with transcendental conjugate

Let $$a$$ be an algebraic number such that $$|a|<1$$ and $$(d_k)_{k\in\mathbb N}$$ be a sequence of positive integers. Assume that $$\sum_{k\ge0}\frac{a^k}{d_k}$$ is an algebraic number. Denote by $$a_1,\cdots,a_n$$ the conjugates of $$a$$ over $$\mathbb Q$$. Suppose that for every $$1\le i\le n$$ $$|a_i|<1$$. Is the series $$\sum_{k\ge0}\frac{a_i^k}{d_k}$$ an algebraic number for every $$1\le i\le n$$?

• After your edit, the question makes even less sense. It's now unclear to me what you exactly want to ask. So I deleted my answer, which says nothing but the sum is $\frac 1{1 - a_i}$ hence algebraic. Edit: alright this last edit finally makes sense. Dec 13, 2019 at 3:12
• I made edit to my answer and now it gives you a counterexample. Dec 13, 2019 at 4:09

No, not necessarily.

We first prove a lemma:

For every real number $$x$$ such that $$0 < x < 2$$, there exists a squence of positive integers $$(d_k)_{k\geq0}$$ such that $$\sum_{k\geq 0}\frac{2^{-k}}{d_k} = x$$.

Proof: we define sequences $$(x_k)$$ and $$(d_k)$$ recursively by: $$x_0 = x, d_k = \lfloor \frac 1 {x_k}\rfloor + 1, x_{k + 1} = 2(x_k - 1/d_k).$$ To verify that this is well-defined, we show that $$0< x_k < 2$$ implies $$0 < x_{k + 1} < 2$$.

Since we have $$\frac 1 {x_k} < \lfloor \frac 1 {x_k} \rfloor + 1 = d_k$$, it is clear that $$x_{k + 1} > 0$$.

To show that $$x_{k + 1} < 2$$, we separate into two cases.

• If $$x_k \leq 1$$, then we have $$x_{k + 1} < 2x_k \leq 2$$.

• If $$x_k > 1$$, then we have $$d_k = 1$$ and hence $$x_{k + 1} = 2(x_k - 1) < 2$$.

Thus both sequences are well-defined and every $$d_k$$ is a positive integer.

By induction on $$k$$, it is easy to show that $$x - \sum_{k = 0}^{n - 1}\frac{2^{-k}}{d_k} = 2^{-n}x_n$$.

Taking limit $$n\rightarrow \infty$$, we have $$x = \sum_{k\geq 0}\frac{2^{-k}}{d_k}$$ as desired.

Now it's easy to give counterexamples to your question.

We choose $$a = 1/\sqrt 2$$ so that its conjugate is $$-a$$.

Choose a transcendental number $$x$$ with $$0 < x < \sqrt 2$$.

By the lemma, there exist sequences of positive integers $$(u_k)$$ and $$(v_k)$$, such that $$\sum_{k\geq 0} \frac{2^{-k}}{u_k} = x$$ and $$\sum_{k\geq 0} \frac{2^{-k}}{v_k} = \sqrt 2 x$$.

We define a sequence $$(d_k)$$ such that $$d_{2k} = u_k$$ and $$d_{2k + 1} = v_k$$.

We then calculate: $$\sum_{k \geq 0}\frac{(\pm a)^k}{d_k} = \sum_{k \geq 0}\frac{2^{-k}}{u_k} \pm \frac{1}{\sqrt 2}\sum_{k \geq 0}\frac{2^{-k}}{v_k} = x \pm x.$$

Hence for $$a$$, the sum is $$0$$, which is algebraic, while for $$-a$$, the sum is $$2x$$, which is transcendental.

• Very impressive. Thanks!! Dec 13, 2019 at 4:50

Let $$S$$ be the set of sequences $$d_k\ge 4^k$$ such that $$\sum_k \frac{(-\sqrt{15})^k}{d_k}=0$$. It is uncountably infinite thus for most of those $$\sum_k \frac{(\sqrt{15})^k}{d_k}$$ is transcendental.

It is hard to give a concrete example because there is a natural norm on $$\overline{\Bbb{Q}}$$ which is $$\|x\|=\sup_{\sigma\in Gal(\overline{\Bbb{Q}}/\Bbb{Q})} |\sigma(x)|$$ If $$(x_n)$$ is Cauchy for $$\|.\|$$ then for all $$\sigma$$, $$(\sigma(x_n))$$ is Cauchy in $$\Bbb{C}$$. This way we can define $$\sigma(\exp(\sqrt{2})) = \exp(\sigma(\sqrt{2}))=\exp(-\sqrt{2})$$ even if $$\sigma$$ is not continuous $$\Bbb{Q}[\sqrt{2}]\to \Bbb{Q}[\sqrt{2}]$$. The completion of $$\overline{\Bbb{Q}}$$ for $$\|.\|$$ is weird, it is not the complex numbers, ie. in there $$\exp(\sqrt{2})$$ is not a complex number, it is the limit of the exponential defining series, not the same.

The common functions such as $$\exp,\log$$, the algebraic functions satisfy their properties in $$\overline{\Bbb{Q}},\|.\|$$ not only in $$\Bbb{C}$$, thus we can't obtain a counter-example to your problem from them. My dream is to use $$\overline{\Bbb{Q}},\|.\|$$ to connect the properties (the non-trivial zeros..) of $$L(s,\chi^\sigma)$$ to those of $$L(s,\chi)$$.

• This $\|\cdot\|$ is not a norm in the usual sense, as it is not multiplicative. The multiplicative norms are of course well classified. Dec 13, 2019 at 14:17
• In particular, you will not be able to show that the completion is a field. In fact it's easy to see what it looks like. Let's consider the restriction to $\Bbb Q[\sqrt 2]$. Then the completion of $\Bbb Q[\sqrt 2]$ w.r.t. this norm is nothing but $\Bbb Q[\sqrt 2]\otimes_{\Bbb Q }\Bbb R$, which as a ring is isomorphic to $\Bbb R\times \Bbb R$, hence not a field. This example of course generalizes to arbitrary number field $F$, whose completion becomes $F\otimes_{\Bbb Q }\Bbb R$, which is the product of $F_v$ for all archimedean places $v$ of $F$. Dec 13, 2019 at 18:21
• Therefore the completion of $\overline{ \Bbb Q}$ w.r.t. this norm is something like direct product (or sum?) of all $\overline {\Bbb Q}_v$, where $v$ ranges through all archimedean places of $\overline {\Bbb Q}$. The topology is perhaps something like weak topology, which I don't understand well. And the action of any $\sigma\in Aut(\overline {\Bbb Q})$ is simply permuting the components, hence not very interesting. I don't think this norm and the completion can be seriously useful. Dec 13, 2019 at 18:30