Is every sufficiently general monic quartic rational square infinitely often?

Question

Let $f(x)=x^4+b_3 x^3+ b_2 x^2+b_1 x + b_0$.

Let $g(x)=x^4 f(1/x)$. Let $C : g(x)=y^2$.

$C$ is birationally equivalent to $f(x)=y^2$.

The constant coefficient of $g(x)$ is $1$ since $f$ is monic and $(0,1)$ is on $C$.

On the internet we found reduction quartic with square coefficient to Weierstrass model of the elliptic curve with point of infinite order.

One family of failures are $f(x)=x^4+b_2 x^2 + b_0$.

Q1 Is every sufficiently general monic quartic rational square infinitely often?

Q2 Is this known?

sagemath session with the reduction:

sage: K.<x>=QQ[]
sage: f=x^4+5*x^3+x^2+13*x+2;g=numerator(x^4*f(1/x))
sage: E,phi,psi,poi=quarToEll(g)
sage: P2=3*E(poi);P2.order()
+Infinity
sage: Pq=psi[0](P2.xy()),psi[1](P2.xy())
sage: factor(f(1/Pq[0]))
2^-16 * 13^-4 * 137^2 * 5940889^2


def quarToEll(pol,v0=None,q0=None):
# if v^2=a*u^4+b*u^3+c*u^2+d*u+q^2 (i.e. with rat point (0,q))
# this function gives Weierstrass equation and maps
# result is in the format [ell,[x,y],[u,v]] where ell
# is the Weierstrass elliptic curve
# [x,y] are given in terms of u,v and [u,v] given in terms
# of x,y
    qq,d,c,b,a=pol.coefficients(sparse=False) #pol.coeffs()
    #if v0 != None:
    #   q=v0
    #else:  
    try:
        if not qq.is_square():
            print ('non square')
            return []
        #q=ZZ(qq).isqrt()
    except:  pass   
    pr2=PolynomialRing(pol.base_ring(),'z1')
    z1=pr2.gen()
    if q0 is None:
        polr=z1**2-qq
        rots=polr.roots(multiplicities=False)
        q=rots[0]
    else:  q=q0 
    
    a1=d/q
    a2=c-d**2/(4*q**2)
    a3=2*q*b
    a4=-4*q**2*a;
    a6=a2*a4
    eli=[a1,a2,a3,a4,a6]
    ell=EllipticCurve(pol.parent().base_ring(),eli)
    prxy=PolynomialRing(pol.parent().base_ring(),'x,y')
    x,y=prxy.gens()
    pruv=PolynomialRing(pol.parent().base_ring(),'u,v')
    u,v=pruv.gens()

    P1=[(2*q*(v+q)+d*u)/u**2,(4*q**2*(v+q)+2*q*(d*u+c*u**2)-d**2*u**2/(2*q))/u**3]
    f=(2*q*(x+c)-d**2/(2*q))/y
    g=-q+f*(f*x-d)/(2*q)
    P2=[f,g]

    point1= [-a2,a1*a2-a3]

    return [ell,P1,P2,point1]

Answer 1

The genus one curve given by $$y^2=x^4 + b_3 x^3 + b^2 x^2+b_1 x+ b_0$$ in one affine chart and $$z^2 = b_0 w^4 + b_1 w^3 + b_2 w^2 + b_3 w +1$$ in another affine chart (glued by $w = 1/x, z = y/x^2$), has two rational points $w=0, z = \pm 1$. Choosing one, say $(0,1)$ as the origin for the group law, this is an elliptic curve.

Hence the elliptic curve has infinitely many rational points as long as the remaining one, $(0,-1)$, is not a torsion point. If it is a torsion point, then by Mazur's theorem (because everything is defined over $\mathbb Q$) it must have order $1,2,3,4,5,6,7,8,9,10,$ or $12$.

For each of those numbers, the set where $(0,-1)$ is torsion of that order is a closed subset of the space with coordinates $b_0,b_1,b_2,b_3$. To check that it is a proper closed subset, it suffices to find one elliptic curve where that point has infinite order, which is easy to do. (This can also be done by a geometric argument).

So for any $b_0,b_1,b_2,b_3$ outside a finite union of proper closed subsets (i.e. sufficiently general) there are infinitely many rational points. One can find these closed subsets explicitly using the group law of the elliptic curve (division polynomials).

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Why is the minimalist conjecture, suggested by Wojowu, not relevant here? The reason is that we are dealing with a geometric family of elliptic curves, rather than all elliptic curves.

However, the minimalist conjecture can be generalized to apply to an arbitrary family of elliptic curves, and in much greater generality to geometric families of L-functions, as was essentially done by Sarnak, Shin, and Templier.

In this context, the right conjecture to make is that for any family of elliptic curves parameterized by a reasonable variety (e.g. projective space, or even a Fano variety with no obstruction to rational points) 100% of curves should have rank equal to the minimum number greater than the rank of the family (i.e. the number of independent sections in the Mordell-Weil group of the generic fiber), whose parity matches the sign of that curve's L-function.

In this context, the signs are surely equidistributed, so we can predict 50% have rank 1 and 50% have rank 2.

Some families (e.g. K3 surfaces studied by Noam Elkies) have a quite high rank, so generic members have rank distributions that are strange from the perspective of the family of all elliptic curves!