Question: If $x$ is an algebraic number, then it occurs as a root of some polynomial with integer (or rational) coefficients. Is there also a polynomial with integer (rational) coefficients such that $x$ is the smallest root of this polynomial (w.r.t. the usual ordering of the reals)?

The question is inspired by a "solution" handed in by a student. The problem was to show that the set of algebraic numbers is countable. The solution started by the observation that $\mathbb Q[x]$ is countable. And then it continued more-or-less like this: "We take an enumeration of polynomials in $\mathbb Q[x]$. Then we omit the polynomials without real roots. We assign to each of the remaining polynomials the smallest root. If we omit the duplicates from this new list, we get an enumeration of the set of all algebraic numbers." (I will add that it is not difficult to fix this proof so that it works even without relying on using smallest root. We have countably many polynomials, each of them has only finitely many roots. Union of countably many finite sets if countable. But this is a digression to topic of other posts on this site, such as Prove that the set of all algebraic numbers is countable or How do we prove the existence of uncountably many transcendental numbers?)

Unless I have missed something, I can shown the answer to the question is No. (Simply by taking some irreducible polynomial with more than two real roots. Unless somebody else does this, I can expand this into an answer and post it.)
Still I consider the problem interesting enough to post it here. (And it is often the case that even when I ask a question to which I know a solution, I am pleasantly surprised by interesting alternative solutions I get as answers.)

  • $\begingroup$ Your student could have fixed his argument by diagonalizing the roots. Let $r(i,j)$ be the $i$th root of the $j$th polynomial (and $0$ if the $j$th polynomial does not have at least $i$ roots) enumerating $r(1, 1), r(1, 2), r(2, 1), r(1, 3), r(2, 2), r(3, 1), \dots$ would cover all algebraic numbers, and after removing duplicates/unwanted roots would enumerate them. Might be useful feedback for him/her :) $\endgroup$
    – orlp
    May 21 '18 at 12:32
  • $\begingroup$ Your argument looks solid and I'd have said it was optimal. Every polynomial with $p(\alpha)=0$ is divisible by the minimal polynomial for $\alpha$. $\endgroup$
    – lulu
    May 21 '18 at 12:40
  • $\begingroup$ @orlp Yes, the argument can be fixed by various (more-or-less standard) ways. I have edited the post to mention this fact. (I have included the attempted solution mainly to show what was the motivation that lead me to thinking about this problem.) $\endgroup$ May 21 '18 at 12:40
  • $\begingroup$ Isn't $\sqrt 2$ an immediate counterexample? $\endgroup$
    – lhf
    May 21 '18 at 14:49
  • $\begingroup$ @lhf Yes it is. (Since whenever it is a root of a polynomial, then the polynomial is divisible by $x^2-2$ and $-\sqrt2$ is a root.) Feel free to post it as an answer - it turns out that this was much easier than I though. Thanks for spotting this! $\endgroup$ May 21 '18 at 14:57

The claim is not true.

Every integer polynomial that has $\sqrt 2$ as a root also has $-\sqrt{2}$ as a root and so $\sqrt 2$ is never the smallest root.


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