For any integer $r \geq 2$, et $V_r$ be the set of polynomials $Q \in {\mathbb Q}[X]$ of degree $r1$ such that there is an algebraic number $\alpha$ of degree $r$ , such that $Q(\alpha)$ is a conjugate of $\alpha$. It is not hard to see that $V_2$ consists exactly of $X$ and all the polynomials $a_0X$, for $a_0\in {\mathbb Q}$. Have the $V_r(r \geq 3)$ been studied? Is anything known about $V_3$ ?

4$\begingroup$ Maybe I am missing something simple, but why is $x+1\in V_2$? $\endgroup$– Tim DokchitserJan 28, 2011 at 14:10

$\begingroup$ In order for $ax + b \in V_2$ it is necessary that $(a+1)\alpha + b \in \mathbb{Q}$, where $\alpha$ is the algebraic number in question. Thus, it is necessary that $a = 1$. This shows that $\alpha$ is a root of $x^2  b + c$, for some $c \in \mathbb{Q}$. In order that $\deg \alpha = 2$ it is necessary and sufficient that $(b^2  4c)$ not be a square. Thus $V_2$ consists of the polynomials $bx$ for all $b \in mathbb{Q}$. $\endgroup$– Victor MillerJan 28, 2011 at 16:15

$\begingroup$ if g(x)=x+1 for an element g of the galois group, then g^n(x)=x+n... Cannot be. $\endgroup$– Yaakov BaruchJan 28, 2011 at 16:21

$\begingroup$ @Tim, Vicro and Yaakov : I corrected the equality for $V_2$ in the OP. $\endgroup$– Ewan DelanoyJan 28, 2011 at 16:46
2 Answers
$\newcommand\Z{\mathbf{Z}}$ $\newcommand\Q{\mathbf{Q}}$
If we replace $\alpha$ by $\alpha + \lambda$ for $\lambda \in \Q$ (translation), we may replace $Q(X)$ by $Q(X  \lambda) + \lambda$. Similarly, if we replace $\alpha$ by $\mu \cdot \alpha$ (dilation), we may replace $Q(X)$ by $\mu \cdot Q(X/\mu)$.
Let's discuss the case $[\Q(\alpha):\Q] = 3$. By translation, we may assume that the coefficient of $X$ is trivial. By dilation, we may assume that the leading coefficient is $1$ (this specifically uses the fact that we are in degree $3$). Hence $Q(X) = X^2 + c$ for some $c \in \Q$.
If a degree $3$ field $K = \Q(\alpha)$ contains at least two conjugates of $\alpha$ then it contains all the conjugates, and is therefore Galois with cyclic Galois group. Thus $Q$ induces an isomorphism of $K$ of order three, and $\alpha$ must be a root of $Q(Q(Q(X)))  X$ but not a root of $Q(X) X$. Hence it is a root of $F(X) = (Q(Q(Q(X)))  X)/(Q(X)  X)$, which is an explicit degree $6$ polynomial.
Suppose that $\beta = u + v \sqrt{d}$ is a root of $F(X)$, where $u$ and $v$ are rational and $d$ is not a square. We may write $F(u + v \sqrt{d}) = R(u,v) + S(u,v) \sqrt{d}$, where $R$ and $S$ are polynomials with coefficients in $\Q[a,b]$. Since $F(\beta) = 0$, then $R(u,v) = S(u,v) = 0$. Computing the resultant of these polynomials with respect to $c$, we obtain the equation: $$(1 + 2 u^2 + 4 u^2)(1  18 u + 4u^2 + 8 u^3) v = 0.$$ Since $u$ is rational, it follows that $v = 0$. Thus $F(X)$ does not have any genuine quadratic solutions, and hence $F(X)$ has no factors that are quadratic. (If we assume that $F(X)$ has a cubic factor, we can prove this in a slightly cleaner way. Since $F(X)$ has a cubic factor, it has at most one quadratic factor. Yet $Q(\beta)$ and $Q(Q(\beta))$ are also quadratic roots of $F(X)$, and hence either $Q(\beta) = \beta$ or $Q(\beta) = \sigma \beta$. Both possibilities are impossible.)
We are assuming that $F(X)$ has a cubic factor corresponding to $\alpha$. Then either $F(X)$ factors as a product of two cubics, or as a cubic times three linear factors. In either case, the cubics correspond to fields which are Galois and hence cyclic of degree three. Hence the discriminant of $F(X)$ is a square (easy exercise). We compute explicitly that $$\Delta_F =  (7 + 4 c)^3 (7 + 4 c + 16 c^2)^2,$$ and hence we deduce the necessary relation: $$7 + 4 c =  t^2$$ for some $t \in \Q$. Making the substitution $c = (7  t^2)/4$, the polynomial $2^{6} \cdot F(X)$ factors as $A(X,t) A(X,t)$, where $$A(X,t) = 1 + 7 t  t^2 + t^3 + 18 x  4 t x + 2 t^2 x  4 x^2  4 t x^2  8 x^3.$$ As long as either this polynomial or its cousin $A(X,t)$ are irreducible, we obtain a cubic with a root $\alpha$ such that $Q(\alpha)$ is a conjugate of $\alpha$. The polynomial $A(X,t)$ is reducible in $t$ if and only if there exists a rational point $A(X,t) = 0$. This turns out to be a rational curve, and we deduce that it is reducible if and only if there exists a $u \in \Q$ such that $$t = \frac{1 + 2 u  u^2  u^3}{u(u+1)}.$$ On the other hand, $A(X,t)$ is reducible if and only if there exists a $v \in \Q$ such that $$t = \frac{1 + 2 v  v^2  v^3}{v(v+1)}.$$ Hence any forbidden $t$ corresponds to a solution to the equation $$\frac{1 + 2 u  u^2  u^3}{u(u+1)} =  \frac{1 + 2 v  v^2  v^3}{v(v+1)},$$ which correspond to points on the curve $$C:u + u^2 + v + 4 u v + u^2 v  u^3 v + v^2 + u v^2  2 u^2 v^2  u^3 v^2  u v^3  u^2 v^3 = 0.$$ This curve is smooth in the affine locus. The corresponding projective curve has three points at $\infty$, and two of the corresponding points are singular.
The singularities at these points are nodes (I think), and thus using Plücker's formula, we deduce that $C$ has genus: $$ g = (d1)(d2)/2  n = (4 \cdot 3)/2  2 =4.$$ Thus $C$ has finitely many rational points (Faltings). (Note: this calculation may have been wrong, but we won't actually use it.) The curve $C$ has some obvious points at $\infty$ and when $u(u+1)v(v+1) = 0$. Make the substitution $u = x+y$ and $v = xy$. Then the equation becomes: $$x  3 x^2  x^3 + 2 x^4 + x^5 + y^2 + x y^2  2 x^2 y^2  2 x^3 y^2 + x y^4 = 0.$$ This is a quadratic in $y^2$. Hence we obtain a degree two covering $C \rightarrow E$, where $E$ is the curve $$E:x  3 x^2  x^3 + 2 x^4 + x^5 + z + x z  2 x^2 z  2 x^3 z + x z^2 = 0.$$ Given a rational point on this curve, the discriminant $\Delta$ is rational, and hence $E$ is isomorphic to: $$\Delta^2 = 4(4 x^4 + 3 x^3 + x^2 + 2 x + 1) = 4(1+x)(1+2x)(1x+2x^2)$$ This is birational to an elliptic curve which turns out to have conductor $112$. Cremona's program mwrank tells me has MordellWeil group $\Z/2\Z$. Hence the only rational points correspond to $x = 1/2$ and $x = 1$, which pull back to the "obvious" rational points on $C$. Hence we have determined $C(\Q)$ completely, and we see that one of $A(X,t)$ or $A(X,t)$ is always irreducible. Hence we deduce:
If $\alpha$ is a cubic irrationality with $Q(\alpha) = \sigma \alpha \ne \alpha$, then, replacing $\alpha$ by $\lambda \alpha + \mu$, $Q(X)$ is of the form: $$Q(X) = X^2  \frac{7+t^2}{4}$$ with $t \in \Q$. Conversely, for any such $Q(X)$, there exists a cubic irrationality $\alpha$ such that $\sigma \alpha = Q(\alpha)$.
Example: $t = 1$, and $Q(X) = X^2  2$. Then $\alpha = 2 \cos(2 \pi/7)$.
For higher degree polynomials, things will be even more of a mess, because there will be various possibilities corresponding to what the Galois closure is, &. &. The answer will have the following flavor: It will correspond to the rational points on a bunch of varieties minus rational points on subvarieties corresponding to degeneracies (which in this case turn out to be empty).

$\begingroup$ "This question should probably be tagged nt.numbertheory" Done. $\endgroup$ Jan 29, 2011 at 0:00

$\begingroup$ Nice. I don't see the need for paragraph 4; if $\alpha$ is degree 3, as we are assuming, then it obviously can't be $u+v\sqrt d$, which is of degree 2. $\endgroup$ Jan 29, 2011 at 5:06

If $f(x)$ is in $V_r$, and $g(x)$ is the minimal polynomial of a corresponding element $\alpha$, then $f(\alpha)$ is also a root of $g(x)$, so $g(f(\alpha))=0$; hence $g(x)$ divides $g(f(x))$. Thus, $V_r$ consists of the polynomials $f(x)\in\mathbb{Q}[x]$ of degree $r1$ for which there is an irreducible degree$r$ polynomial $g(x)\in\mathbb{Q}[x]$ satisfying $g(x)\mid g(f(x))$. This perspective explains the result for $r=2$: the map $\beta\mapsto f(\beta)$ must permute the two roots of $g(x)$, and the only linear polynomials which permute two numbers are $x$ and $a_0x$ (this doesn't quite show $a_0x\in V_2$, one has to finish as Victor did).