Find all possible rational values of a parametric quartic such that it is reducible

Question

Description: Given the following parametric quartic polynomial

$y^4 - 28 z y^3 - 14 (656 - 328 z + 83 z^2) y^2 + 
4 z (-20464 + 10232 z + 3409 z^2) y + 
91 (62208 - 62208 z + 41504 z^2 - 12976 z^3 + 675 z^4)$

how do I find all possible rational values of $z$ such that this polynomial is reducible.

I am performing Galois group classification on a family of polynomials of which this example is one case. Rational searching has identified the following values of $z$ that results in factorization: $z \in \left\{{-44,-2,1,\frac{44}{23},\frac{67}{19}, 4, \frac{134}{29}}\right\}$.

I do not know if there is an infinite or finite number of such points (I suspect that there is a finite number). Reposing the question: what is the procedure by which I can generate all know points $z$ in a systematic manner.

Answer 1

I essentially follow Igor Rivin's approach here. There are two possibilities for how the polynomial can factor: (degree 1)$\times$(degree 3) or (degree 2)$\times$(degree 2). Writing $$ (y + a)(y^3 + by^2 + cy + d) = f $$ or $$ (y^2 + ay + b)(y^2 + cy + d) = f, $$ where $f$ is your polynomial, and equating coefficents, we obtain two curves $C_1$ and $C_2$ in the 5-dimensional affine space with coordinates $a,b,c,d,z$, and what you want is the $z$-coordinates of their rational points.

EDIT: Actually, there is a somewhat simpler approach. $C_1$ classifies linear factors of $f$, which is equivalent to roots of $f$, so $C_1$ is simply the affine plane curve $f(y,z) = 0$. Its projective closure is a smooth quartic curve, so the genus of $C_1$ is 3. Alternatively, since the discriminant of $f$ as a polynomial in $y$ is a degree 12 polynomial in $z$ without multiple roots, it follows from Riemann-Hurwitz that the genus of $C_1$ is 3.

One finds 16 rational points on it; their $z$-coordinates are those you also found (three of them ($z = -2, 1, 4$) give a factorization into four linear factors, so they each give rise to four points on $C_1$, the other four give only one linear factor).

EDIT: $C_2$ classifies unordered pairs of (distinct) roots, so it can be obtained as a degree 6 cover of the affine $z$-line by expressing the coefficients of the polynomial whose roots are the products (or sums, or any other symmetric expression) of pairs of roots in terms of those of the original polynomial. This ramifies in 24 points (two above each branch point of $C_1 \to \mathbb P^1$), so by Riemann-Hurwitz its genus is 7.

One finds 18 rational points on it; they all come from $z = -2, 1, 4$ by writing the product of four linear factors as a product of two quadratics (which can be done in six different ways).

Since both curves have genus $\ge 2$, they have only finitely many rational points by Faltings's Theorem. So there are only finitely many $z \in \mathbb Q$ such that your polynomial factors. It is also very likely that there are no further rational points on the curves than those mentioned above (points on such curves are usually fairly small, so one is likely to find them all by search), but this is also very likely very hard to prove. But for all practical purposes, you can safely assume that the values you found constitute the complete list.

Answer 2

Factoring a polynomial $q(x)$ over $\mathbb{Q}$ means solving the equation $$ r(x) s(x) = q(x),$$ which gives you a quadratic system of equations over the rationals in the coefficients of $r, s.$ You can eliminate all but one variable (by taking resultants, or some more efficient method of your choosing) to get a single equation in one of the unknowns, of which you want to find rational roots. This is not so difficult, and completely systematic. In this case you have two different systems: one for $r$ of degree $1,$ $s$ of degree $3,$ and the other where they each have degree $2.$ Resultants will reduce both of these to a single equation of degree $8,$ which is not very high.

EDIT In the above, I did not consider $z,$ if I did, you would have one equation with two unknowns, but there is a trick: notice that for the constant term, the equation is $r_0 s_0 = q_0.$ If you are looking for a solution in integers, then you first factor $q_0$ and for each pair of complimentary factors, you have one less equation, and TWO fewer unknowns, so what I say above applies. For rational solution, the same idea would work if you had an a priori height estimate.