I was reading this expository paper by Yves André in which he states a nice result: every transcendental number is the root of a power series over $\mathbb Q$. He accredits this theorem to Hurwitz in paragraph 2.3, but doesn't give a reference for it. I haven't been able to find the relevant paper (or another exposition of te proof) myself, so I'm hoping someone here knows where to look!

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    $\begingroup$ @David C. Ullrich has already answered, so I'll do the usual (for me) and give some references. I posted some references in this 18 December 2006 sci.math post archived at Math Forum. The 1890/1891 Hurwitz paper was probably not online at the time (maybe it was, but I suspect I would have given a link if it had been), but you can now find it here. $\endgroup$ – Dave L. Renfro Dec 26 '18 at 21:37

Edit: There's a straightforward proof below for the result exactly as mentioned in the OP. However, the paper linked to cites a stronger result, specifying that the power series should define an entire function of exponential growth. I don't see how the argument below gives that; see Comments at the bottom.

Say $\alpha\in\Bbb C$, $\alpha\ne0$.

First, if $\alpha$ is real it's trivial that it's a zero of some power series with rational coefficients: If $r_0,\dots,r_n\in\Bbb Q$ have been chosen, there exists $r_{n+1}\in\Bbb Q$ with $$|r_0+\dots+r_{n+1}\alpha^{n+1}|<1/n;$$hence $$r_0+r_1\alpha+\dots=0.$$

Now say $\alpha=\rho e^{it}$, $\rho>0$, $t\in\Bbb R$. If $t/\pi$ is rational then there exists a positive integer $N$ so that $\beta=\alpha^N\in\Bbb R$; so $\beta$ is a root of some rational power series, hence so is $\alpha$.

Finally, suppose $t/\pi$ is irrational. Then $\{e^{ikt}:k=1,2\dots\}$ is dense in the unit circle. Hence for every $n$ the set $\{r\alpha^k:r\in\Bbb Q, k=n+1,n+2,\dots\}$ is dense in $\Bbb C$ (to approximate $z$ by $r\alpha^k$, first choose $k$ so as to get the argument approximately right, then choose $r$ to fix up the modulus). So as above we can recursively construct a sequence $r_j$ of rationals and a strictly increasing sequence $n_j$ of positive integers so that $$\sum r_j\alpha^{n_j}=0.$$

Comments. Now what about getting an entire function of exponential growth?

If $\alpha$ is real this is no problem: Say wlog $\alpha>1$ to keep the inequalities clean and replace the main inequality above by $$|r_0+\dots+r_{n+1}\alpha^{n+1}|<1/(n+2)!;$$it follows that $r_n=O(1/n!)$, hence the power series is an entire function of exponential growth.

And so we're done if $\alpha^N$ is real. But the case $t/\pi$ irrational is not so simple, as far as I can see. We can make $\left|\sum_{j=0}^k r_j\alpha^{n_j}\right|$ as small as we want as a function of $k$, but the doesn't help; saying for example $$\left|\sum_{j=0}^k r_j\alpha^{n_j}\right|\le1/k!!!$$ says nothing about the radius of convergence. The problem is that in order to make $\left|\sum_{j=0}^k r_j\alpha^{n_j}\right|$ small, by the trivial argument above, we may be forced to take $n_k$ large, so we don't get anything analogous to$$\left|\sum_{j=0}^k r_j\alpha^{n_j}\right|\le1/(n_k)!,$$which is what we need.

Well, the paper cited in the OP calls this an elementary result, so it can't be that hard. Probably there's a simple proof that's nothing like what above; the argument above is after all a simple brute-force sort of thing, surely one can do something more subtle?

Perhaps one can fix the argument above, showing that you don't need to take $n_k$ too large to make $\left|\sum_{j=0}^k r_j\alpha^{n_j}\right|$ small.

Or something I just thought of. The proof that the sum of two algebraic numbers is algebraic is fairly simple if you look at it right, but it's not a priori obvious. Maybe some extension of that argument shows that if $\alpha=x+iy$ then the existence of suitable power series for $x$ and for $y$, proved above, implies the same for $\alpha$?

  • $\begingroup$ Very nice! I was hoping there was some way to distill some kind of "representative" power series for a transcendental number out of the proof, similar to how the minimal polynomial of an algebraic number $a$ is a "representative" for all polynomials vanishing at $a$, but this seems a rather hopeless task. Thank you anyway for the insight! $\endgroup$ – user Dec 20 '18 at 22:35
  • $\begingroup$ @user See edit - it may that I've given you what you asked for, but not what you meant to ask for. $\endgroup$ – David C. Ullrich Dec 21 '18 at 14:39

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