Let $f(x) \in \mathbb Z[x]$ be an irreducible polynomial of degree $d \geq 3$ such that for some distinct roots $\alpha$ and $\beta$ of $f(x)$ it is the case that

$$\beta = \frac{a\alpha + b}{c\alpha + d}$$

for some integers $a, b, c$ and $d$. Is there an easy way to classify polynomials of such kind? I suspect that such a nice relationship between the roots can appear if and only if the binary form $F(x,y)$, which is the result of homogenization of $f(x)$, satisfies

$F_A(x,y) = F(x,y)$

for some matrix $A \in \operatorname{SL}_2(\mathbb Z)$ different from identity. Is this true, or perhaps there is an obvious counter example which I don't see?


The matrix $M = \left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right)$ need not be in $\operatorname{GL}_2(\mathbb{Z})$. For example, take the quartic form

$$\displaystyle F(x,y) = x^4 + 3x^3 y + 5x^2 y^2 - 21xy^3 + 49y^4.$$

One checks that $F$ is fixed by the matrix given by $\frac{1}{\sqrt{7}} \left(\begin{smallmatrix} 0 & 7 \\ -1 & 0 \end{smallmatrix} \right)$ under substitution, and this implies that the roots of $F(x,1)$ are permuted by the Mobius transformation

$$\displaystyle x \mapsto \frac{7}{-x}.$$

If you assume a priori that $M = \left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right) \in \operatorname{SL}_2(\mathbb{Z})$ and $F$ is irreducible, then $M$ must be an automorphism. Indeed, since the Galois group of $F(x,1)$ acts transitively on the roots of $F(x,1)$ since $F$ is assumed to be irreducible, it follows that you can apply Galois action on $\beta, \alpha$ to obtain an equation of the shape

$$\displaystyle \theta_1 = \frac{a \theta_2 + b}{c \theta_2 + d}$$

for any root $\theta_1$ of $F(x,1)$, so that all of the roots of $F(x,1)$ are related to another root by the same Mobius transformation (note that the Galois group of $F$ acts trivially on $a,b,c,d$, since we assumed that these are rational integers). Thus the Mobius transformation given by the matrix $M = \left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right)$ permutes the roots of $F(x,1)$. Since $M \in \operatorname{SL}_2(\mathbb{Z})$, it is discriminant preserving and one can check that it fixes the leading coefficient of $F(x,y)$ as well, and since it permutes the roots and fixes the leading coefficient, it must be an automorphism.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.