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

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.

• If you are identical with "Lorenz H Menke" who asked this question: mathoverflow.net/questions/188579/…, then you should consider merging the accounts... – Michael Stoll Apr 10 '15 at 7:37
• Thank you for the suggestion. My two accounts have been merged. Also thank you for all the help. – Lorenz H Menke Apr 14 '15 at 0:06

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.

• I am curious: how did you find the rational points? – Igor Rivin Apr 8 '15 at 21:36
• @IgorRivin I did a "PointSearch" in Magma on a (somewhat minimized and reduced) plane model of each of the curves, up to reasonably large height. – Michael Stoll Apr 9 '15 at 7:41
• By the way, the genus 3 curve $C_{1}$ has an automorphism of order 2, and the quotient is an elliptic curve. I would guess this elliptic curve has positive rank, but I haven't actually computed it. – Jeremy Rouse Apr 9 '15 at 12:02
• @JeremyRouse It is the curve $$y^2 + y = x^3 - 69975 x + 7291406$$ and has rank 3. Since there are so many rational points, it was very likely that it would have fairly large rank... – Michael Stoll Apr 9 '15 at 14:15

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.

• My usual approach has been to for the discriminant. In the case of the main variable y, the discriminant leads to an hyper elliptic curve. Using the inline MAGMA for example, it will identify the points $z = -2, 1, 4$ which result is the complete factoring. The other part is to use the substitution $y = z\, w$ and from there form the discriminant. In this case there is a quartic degree which is then converted into an elliptic curve. The problem is that this elliptic curve does not provide the solution for the few rational points that have already been found. – Lorenz H Menke Apr 6 '15 at 21:43
• @LorenzMenke As I say in my answer, the method I suggest is completely algorithmic (and simple to implement, if that's what you seek). Of course, if the equation of high degree $n$, the resultant polynomial will be of degree $2^n,$ but for $n<10$ this should work very quickly. – Igor Rivin Apr 6 '15 at 21:49
• Do you really get one equation in one unknown? E.g. for the case of degrees 1 and 3, taking the leading coefficients as $1$, you have $5$ unknowns ($z$, one coefficient in the degree-1 factor, three in the degree-3 factor), and four equations, and, as might be expected, I end up with an equation in two unknowns, of which it doesn't seem so easy to find rational solutions. – Robert Israel Apr 6 '15 at 22:33
• In this example, eliminating all but $r_0$ (the constant coefficient in $r$) and $z$, I get $$61425\,{z}^{4}-13636\,{z}^{3}r_{{0}}-1162\,{z}^{2}{r_{{0}}}^{2}+28\,z{ r_{{0}}}^{3}+{r_{{0}}}^{4}-1180816\,{z}^{3}-40928\,{z}^{2}r_{{0}}+4592 \,z{r_{{0}}}^{2}$$ $$+3776864\,{z}^{2}+81856\,zr_{{0}}-9184\,{r_{{0}}}^{2}- 5660928\,z+5660928$$ – Robert Israel Apr 6 '15 at 22:34
• On some of my other cases the parameter appears as a quadratic. I can then solve the quadratic and the discriminant is a polynomial in the main variable. If the degree is 3 or 4 then I have an elliptic curve or hyper elliptic curve. In either case I have a perfect rational square requirement for rational solutions then the problem is solved. In this case I have not been able to find a perfect square condition. I have tried setting the discriminant of the above answer to a perfect square. The resultant elliptic curve is of rank zero and does not generate all of the known solutions. – Lorenz H Menke Apr 7 '15 at 0:31