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In the usual definition, a family $\{x_i\}_{i\in I}$ of real (or complex) numbers is algebraically independent if every non-zero polynomial relation involving only finitely many variables is impossible.

This “finite character” is essential in the definition.
My question is about whether this finiteness already has consequences for certain infinite expressions.

Question.
Suppose $\{x_i\}_{i\in I}$ is algebraically independent in the usual (finite-support) sense. Consider an absolutely convergent series $$ S=\sum_{i\in I} c_i x_i , \qquad \sum_i |c_i x_i| < \infty,\qquad c_i\in\mathbb{Q}. $$ Is it necessarily true that $S$ cannot be algebraic unless only finitely many $c_i$ are non-zero?

Equivalently: does algebraic independence (defined via finite polynomial relations) already forbid the existence of an algebraic value of an absolutely convergent infinite linear form with infinitely many non-zero coefficients?

Any reference or known result about the interaction between algebraic independence and infinite-support linear forms would be very helpful.

KELVIN LAM
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    For any non-vanishing sequence $x_i$, any number $a$ and any dense set $C$, there are $c_i \in C$, s.t. $\sum c_i x_i$ converges absolutely, and converges to $a$. I don't have reference at hand, but it's easy to construct by just choosing each $c_i$ reducing remaining difference $|S_n - a|$ by about constant factor. – mihaild Nov 17 '25 at 13:12
  • Thanks! That makes sense.

    Just to follow up:
    if the family ${x_i}$ is already known to be algebraically independent as an infinite set (in the usual sense that no non-trivial polynomial relation among any finite subfamily exists), is it still true that an absolutely convergent infinite linear combination $$ \sum_i c_i x_i $$ cannot be algebraic unless only finitely many $c_i$ are non-zero?

    I'm mainly trying to confirm whether algebraic independence already rules out the possibility of an algebraic value for such an infinite sum. Any standard reference would be very helpful.

     – KELVIN LAM Nov 17 '25 at 13:40
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    Either I misread, or entirely opposite is true: if family is algebraicaly independent, then finite linear combinations can't be algebraic. But infinite can, and for any family there are infinite linear combinations that converge to, for example, $1$. – mihaild Nov 17 '25 at 16:22
  • In general, I think the following is true: if $x=\sum_{k\geq 0} x_k$ is convergent (absolutely or not) and $x_k$ are algebraically independent, then there is a $k_0$ such that $x_{k_0}-x$ and $x_k$ with $k\neq k_0$ are (jointly) algebraically independent. So we can modify just one term and get such a sequence with sum zero. I'm not quite sure how to prove this, but I feel that it should follow from the exchange lemma. – tomasz Nov 17 '25 at 23:50
  • @tomasz True. Will delete the earlier comment, and this one soon. – lulu Nov 18 '25 at 04:38
  • Kelvin, the question in the body of your post was (strongly) answered below. Please edit your title, which is misleading as it asks something different. – Anne Bauval Nov 18 '25 at 06:19
  • Your (new) title is still asking a different question. Please replace "can ... be algebraic" by "is ... necessarily transcendental". – Anne Bauval Nov 18 '25 at 15:02
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    By definition of polynomials, your "involving only finitely many variables" is redundant, and "in the usual (finite-support) sense" is useless. On the other hand, "with rational coefficients" is missing. – Anne Bauval Nov 18 '25 at 15:06
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    "infinite linear form" sounds weird. Your prior "infinite linear combination" was more usual. You obviously forgot to require the $c_i$s to be rational, or at least algebraic. Note that you could formulate your question more simply (though equivalently) by replacing "infinite linear form with infinitely many non-zero coefficients" by just "infinite sum" (or series), i.e. take all your $c_i$s equal to $1$. Also, better replace your index set $I$ by $\Bbb N$. And remove the redundant "$0<$". – Anne Bauval Nov 18 '25 at 15:27

1 Answers1

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Given any sequence of real numbers $x_n \ne 0$, one can find $c_n \in \mathbb Q, c_n \ne 0$ st $\sum c_nx_n=0$ with absolute convergence, $\sum |c_n||x_n|<\infty$.

Proof: Since $x_n \ne 0$ given any real $a_n$ and any $b_n > 0$ we can pick $c_n \ne 0$ rational st $c_nx_n=a_n+\epsilon_n, |\epsilon_n|<b_n$ by rationally approximating $a_n/x_n$ with error at most $|b_n/x_n|$

So we can pick rational $c_1$ st $c_1x_1=1+\epsilon_1, |\epsilon_1|<1/16$ and then pick $c_2$ rational st $c_2x_2=-1/2-\epsilon_1+\epsilon_2, |\epsilon_2|<1/32$ so $$c_1x_1+c_2x_2 =1/2+\epsilon_2, \hspace{2mm} |c_1x_1|+|c_2x_2|< 1+1/2+2|\epsilon_1|+|\epsilon_2|$$

Now let's pick $c_3$ rational st $c_3x_3=-1/4-\epsilon_2+\epsilon_3, |\epsilon_3|<1/64$ so $$c_1x_1+c_2x_2+c_3x_3=1/4+\epsilon_3$$ $$|c_1x_1|+|c_2x_2|+|c_3x_3|\le (1+1/2+1/4)+2(|\epsilon_1|+|\epsilon_2|)+|\epsilon_3|$$

Now we choose $c_4$ rational st $c_4x_4=-1/8-\epsilon_3+\epsilon_4, |\epsilon_4|<1/2^{4+3}$ so $$c_1x_1+c_2x_2+c_3x_3+c_4x_4=1/8+\epsilon_4$$ and clearly $|c_4x_4|\le 1/8+|\epsilon_3|+|\epsilon_4|$ so the construction iterates an we get inductively $c_n$ rational st $$\sum_{k=1}^nc_kx_k=1/2^{n-1}+\epsilon_n, |\epsilon_n|<1/2^{n+3}$$ while $$\sum_{k=1}^n|c_kx_k|< \sum 1/2^k+2\sum 1/2^{k+3}<\infty$$

Note that $0$ was chosen for convenience as we can start with $c_1x_1=q+1+\epsilon_1$ and repeat same construction from step $2$ on and get $\sum c_kx_k=q$ with absolute convergence.

Conrad
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  • Thanks for the construction!
    Just to clarify: this shows that absolute convergence alone imposes no restriction.
    However, my question was specifically about whether algebraic independence of the family imposes an obstruction: if the $x_i$ are algebraically independent, can an absolutely convergent infinite linear combination still be algebraic (unless only finitely many coefficients are non-zero)?

    Your construction doesn’t seem to use algebraic independence, so I’m wondering whether there is a result addressing the algebraic-independence aspect.

     – KELVIN LAM Nov 18 '25 at 05:39
  • Kelvin, this answers completely the question in the body of your post. Please edit your title, which is misleading as it asks something different. – Anne Bauval Nov 18 '25 at 06:14
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    In the construction above $x_k$ is any sequence of non zero real numbers, in particular they can be algebraically independent so no, algebraic independence imposes no constraints – Conrad Nov 18 '25 at 13:17