For which non-constant rational functions $f(x)$ in $\mathbb{Q}(x)$ is there $\alpha$, algebraic over $\mathbb{Q}$, such that $\alpha$ and $f(\alpha) \neq \alpha$ are algebraic conjugates? More generally, can one describe the set of such $\alpha$ (empty/non-empty, finite/infinite etc.) if one is given $f$?


  • $f(x)=-x$: These are precisely the square roots of algebraic numbers $\beta$ such that there is no square root of $\beta$ in $\mathbb{Q}(\beta)$. There are infinitely many $\alpha$ and even infinitely many of degree $2$.
  • $f(x)=x^2$: These are precisely the roots of unity of odd order, since $H(\alpha)=H(\alpha^2)=H(\alpha)^2$ implies $H(\alpha)=1$, so $\alpha$ a root of unity. Here $H(\alpha)$ is the absolute multiplicative Weil height of $\alpha$. There are infinitely many $\alpha$, but only finitely many of degree $\leq D$ for any $D$.
  • $f(x)=x+1$: There is no $\alpha$. If there was and $P(x)$ was its minimal polynomial, then $P(x+1)$ would be another irreducible polynomial, vanishing at $\alpha$, with the same leading coefficient and hence $P(x+1)=P(x)$. Looking at the coefficients of the second highest power of $x$ now leads to a contradiction. Analogously for $f(x)=x+a$ if $a$ is any non-zero rational number.

So the existence of such an $\alpha$ and the set of all possible $\alpha$ seem to depend rather intricately on $f(x)$, which seems interesting to me. As I found no discussion of this question in the literature, I post it here.

UPDATE: Firstly thanks to all who have contributed so far! As Eric Wofsey pointed out, any solution $\alpha$ will satisfy $f^n(\alpha)=\alpha$ for some $n>1$, where $f^n$ is the $n$-th iterate of $f$. So one should consider solutions of the equation $f^n(x)-x=0$ or $f^p(x)-x=0$ for $p$ prime.

If the degree of $f$ is at least 2, one can always find irrational such solutions $\alpha$ with $f(\alpha) \neq \alpha$ by the answer of Joe Silverman. However, for his proof to work, we'd need to know that $f^k(\alpha)$ and $\alpha$ are conjugate for some $k$ with $0 < k <p$. I'm not enough of an expert to follow through with his hints for proving this, but if someone does, I'd be very happy about an answer!

If the degree of $f$ is 1, then $f$ is a Möbius transformation and all $f^n$ will have the same fixed points as $f$ (so there's no solution) unless $f$ is of finite order. In that case, if $f(x) \neq x$, the order is 2, 3, 4 or 6 (see http://home.wlu.edu/~dresdeng/papers/nine.pdf). By the same reference, in the latter three cases, $f$ is conjugate to $\frac{-1}{x+1}$, $\frac{x-1}{x+1}$ or $\frac{2x-1}{x+1}$, so it suffices to consider these $f$, which give rise to the minimal polynomials $x^3-nx^2-(n+3)x-1$ (closely related to the polynomial in GNiklasch's answer), $x^4+nx^3-6x^2-nx+1$ and $x^6-2nx^5+5(n-3)x^4+20x^3-5nx^2+2(n-3)x+1$, if my calculations are correct. If the order is 2, the map is of the form $\frac{-x+a}{bx+1}$ or $\frac{a}{x}$, which leads to $x^2+(bn-a)x+n$ or $x^2+nx+a$ respectively. So this case is somewhat degenerate, which explains the unusual behavior of $f(x)=-x$ and $f(x)=x+1$ above.

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    $\begingroup$ Looking at the "Related" questions: mathoverflow.net/questions/53610/… Note that it requires $\deg(\alpha)=\deg(f)+1$, and so in particular $f(x)=x^2$ is not applicable in that setting. $\endgroup$
    – kantelope
    Commented Aug 25, 2015 at 12:34
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    $\begingroup$ Clearly the orbit of $\alpha$ when you iterate $f$ must be finite, which gives a simpler argument for your last two examples. $\endgroup$ Commented Aug 25, 2015 at 12:41
  • $\begingroup$ Have a look at mif.vu.lt/~jonajank/files/linear.pdf (not same, but related) $\endgroup$
    – Boris Bukh
    Commented Aug 25, 2015 at 12:47

2 Answers 2


A first observation is that the field $\mathbb{Q}(\alpha)$ must admit a non-identity automorphism $\sigma$, since applying a rational function to $\alpha$ will not take us outside this field. Given such $\alpha$ and $\sigma$, you can immediately read off an $f$ by expressing $\sigma(\alpha)$ as a $\mathbb{Q}$-linear combination of $1, \alpha, \alpha^2,\ldots$, and there'll be additional non-polynomial (rational) expressions in terms of $\alpha$. This leads to plenty of examples, but now we want to turn this around and start from a given $f$.

Next, the application of $f$ can be iterated. If $g$ is the minimal polynomial of $\alpha$ over the rationals, $g \circ f$ will be a rational function whose numerator vanishes at $\alpha$, hence is divisible in $\mathbb{Q}[x]$ by $g$, hence vanishes at $f(\alpha)$ as well, so $g(f(f(\alpha))) = (g\circ f)(f(\alpha))=0$, so $f(f(\alpha))$ is another conjugate of $\alpha$, and so forth.

Since there are only finitely many conjugates to choose from, some iterate $f^{\circ n}(\alpha)$ must return to $\alpha$: in other words, $\alpha$ will be a root of (the numerator of) $f^{\circ n}(x)-x$ for some integer $n > 1$. (This leads to another argument ruling out $f(x)=x+1$ and friends, as has been noted in the comments.)

So in many cases you'll be able, given $f$, to find the $\alpha$ by picking an $n$ and factoring the numerator of $f^{\circ n}(x)-x$ into irreducible polynomials.

However, the $n$-th iterate of $f$ might be the identity function $x\in \mathbb{Q}(x)$ already, and the numerator to be factored might thus be zero! This happens in your first example $f(x)=-x$ with $n=2$, or with $$f(x)=1/x$$ again with $n=2$ which leads to reciprocal minimal polynomials $g(x)=x^{\mathrm{deg}(g)}g(1/x)$, or with (e.g.) $$f(x)=1-1/x$$ with $n=3$ which gives rise (among other things) to what D. Shanks in 1974 called The Simplest Cubic Fields (Math.Comp. 28, 1137-1152), with $g(x)=x^3+ax^2-(a+3)x+1$. Here, you can proceed by prescribing a degree for $g$, spelling out the condition that $g$ should divide the numerator of $g\circ f$, and comparing coefficients.


The answer to the question you ask, namely "for which $f(x)$...?", is that there exists such an $\alpha$ for every $f(x)$ of degree at least $2$. Proof: If $p$ is a sufficiently large prime, then there is a $p$-periodic point $\alpha$ of $f(x)$ that is not in $\mathbb Q$. (This follows, e.g., by computing heights.) So $f^p(\alpha)=\alpha$ and $f(\alpha)\ne\alpha$. The orbit $\{f^i(\alpha):0\le i<p\}$ is Galois invariant [not necessarily true, see below] and contains an element $\alpha$ not in $\mathbb Q$, so there is some $\sigma\in\text{Gal}(\overline{\mathbb Q}/\mathbb Q)$ such that $\sigma(\alpha)=f^k(\alpha)$ for some $1\le k<p$. Now let $j$ satisfy $jk\equiv1\pmod{p}$. Then $$ f(\alpha) = f^{jk}(\alpha) = \underbrace{f^k\circ\cdots\circ f^k }_{\text{$j$ iterates}}(\alpha) = \sigma^j(\alpha). $$ This shows that $f(\alpha)$ is a Galois conjugate of $\alpha$.

For your second question, as has been noted, if $f(\alpha)=\sigma(\alpha)$, then since $\sigma$ has finite order, one must have that $\alpha$ is a periodic point for iteration of $f$. So your question seems to come down to more-or-less classifying all irrational numbers that are periodic (but not fixed) for a given $f(x)\in\mathbb{Q}(x)$. There's been a lot of study of the Galois properties of fields generated by periodic points of rational functions. See Section 3.9 of The Arithmetic of Dynamical Systems (Springer GTM 241) for an introduction and some further references.

--- ADDITION ---

As Gabriel Dill pointed out, it is the set of points of period $p$ that is invariant under Galois, not the individual orbits. In general, the Galois group $G_p$ of the points of period $p$ is a subgroup of a wreath product $W_p$ of a cyclic group $C_p$ and a symmetric group $S_N$, where $N_p=(d^p-d)/p$. [This is assuming that the $p$-periodic points don't appear with multiplicities.] It is conjectured (I believe) that $G_p=W_p$ for all but finitely many $p$ (except maybe for $x^d$, Chebyshev, and Lattes maps?). This would certainly be enough. Alternatively, if one could find primes with $p\nmid N_p$, then it would suffice to know that $p\mid\#G_p$. (Of course, even for $d=2$ we don't know how to find infinitely many such primes.) Clearly one can get away with making various weaker assumptions about $G_p$, and possibly they could be proven.

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    $\begingroup$ I'm not sure I completely understand your argument. Take e.g. $f(x)=x^2$ and $p=3$. Then $\alpha=\exp(2\pi i/7)$ is irrational with $f^p(\alpha)=\alpha$ and $f(\alpha) \neq \alpha$. But the orbit of $\alpha$ isn't Galois invariant (it contains only 3 elements). It still contains at least one conjugate of $\alpha$, different from $\alpha$, which is the important thing, but why does this hold in general? $\endgroup$ Commented Aug 25, 2015 at 17:39
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    $\begingroup$ @GabrielDill Sorry, you're right. The complete set of $p$-periodic points is a Galois invariant set (this is obvious), it breaks up into disjoint orbits of size $p$, so one needs to find an element of Galois that maps one of the orbits to itself, but doesn't fix the orbit. So that makes things a little trickier. The Galois group is a subgroup of a wreath product of a cyclic group $C_p$ and a permutation groups $S_N$ for an appropriate $N$. I withdraw my answer as a complete proof, but will edit it to indicate how one might proceed. $\endgroup$ Commented Aug 25, 2015 at 19:44

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