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Let $F(x,y) = ax^4 + bx^3y + cx^2y^2 + dxy^3 + ey^4$ be a binary quartic form with integer coefficients. It is well-known that $F$ has two algebraically independent invariants under the action of $\text{GL}_2(\mathbb{Z})$ acting on $F$ via substitution, namely

$$\displaystyle I(F) = 12ae - 3bd + c^2$$

and

$$\displaystyle J(F) = 72ace + 9bcd - 27ad^2 - 27eb^2 - 2c^3.$$

In this paper, Bhargava and Shankar proved (see Theorem 1.7) that if a pair of integers $(A,B)$ satisfy certain congruence conditions modulo 9 and 27 respectively, then there exists a binary quartic form $F$ with integer coefficients such that $I(F) =A$ and $J(F) = B$. Specifically, they determined when $(A,B)$ can appear as $\mathcal{F}(\mathbb{Z},1)$ invariants of monic binary cubic forms, where $\mathcal{F}(\mathbb{Z}, 1)$ is the subgroup of $\text{GL}_2(\mathbb{Z})$ consisting of upper triangular matrices with 1's on the diagonal, and show that each binary cubic form appears as the cubic resolvent of some binary quartic form (under some discriminant preserving map). Indeed, they showed that the binary cubic form

$$\displaystyle x^3 + rx^2y + sxy^2 + ty^3$$

is the cubic resolvent of

$$\displaystyle x^3 y + rx^2y^2 + sxy^3 + ty^4.$$

The problem with this proof is that the quartic form they constructed is always reducible over $\mathbb{Z}$. Certainly, it is possible that this is the only integral quartic form with invariants $A,B$ respectively, but in general there should be other $\text{GL}_2(\mathbb{Z})$ distinct examples since they proved in the same paper that on average, there are $5 \zeta(2)/2$ distinct $\text{GL}_2(\mathbb{Z})$ classes of binary quartic forms with invariants $(A,B)$.

Can anyone give, at least for an infinite family of pairs of $(A,B)$, examples of an irreducible quartic form $F$ whose $I$ and $J$ invariants are equal to $A$ and $B$ respectively?

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  • $\begingroup$ Are you looking for any family of pairs of $(A,B)$, or one of the four families of congruence classes in the paper? $\endgroup$ – Zack Wolske May 30 '16 at 22:30
  • $\begingroup$ They actually proved that pairs satisfying those congruence conditions are the only ones that can arise as invariants of binary quartic forms $\endgroup$ – Stanley Yao Xiao May 30 '16 at 23:13
  • $\begingroup$ Do you want irreducible forms $F$ with, for example, $(I,J)=(12e,0)$, or are you looking for a family of forms for any $I\equiv 0 (3)$, $J\equiv 0 (27)$? $\endgroup$ – Zack Wolske May 31 '16 at 0:35
  • $\begingroup$ I am fine with any infinite family in any of the congruence classes in their paper $\endgroup$ – Stanley Yao Xiao May 31 '16 at 2:20
  • $\begingroup$ You can take $a=1$, $b=c=d=0$, and $e$ squarefree, or just with some prime factor that isn't repeated. Then by Eisenstein, the quartic $F(x,1)$ is irreducible, so the form $F$ is as well. You can have non-zero $J$ in the same congruence class as well, by taking $d$ to be any multiple of the specified prime. $\endgroup$ – Zack Wolske May 31 '16 at 3:20
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Here are constructions using Eisenstein's criterion to get an irreducible $F$ with given invariant $I$ (some restrictions apply).

The congruence conditions given in Bhargava & Shankar are:

  1. $I \equiv 0 \pmod{3}$, $J \equiv 0 \pmod{27}$
  2. $I \equiv 1 \pmod{9}$, $J \equiv \pm2 \pmod{27}$
  3. $I \equiv 4 \pmod{9}$, $J \equiv \pm16 \pmod{27}$
  4. $I \equiv 7 \pmod{9}$, $J \equiv \pm7 \pmod{27}$

Condition 1 has $3|c$, the rest have $c\equiv \mp1 \bmod{3}$, and the sign determines the sign of $J$.

Let $n$ be a non-powerful number, i.e., $n$ has a prime factor $p$ with $p^2\nmid n$. Write $n=mp$.

For the first congruence condition, we'll get $I=12ae-3bd+c^2=3n$, and a $J$ that depends on our choices. To apply Eisenstein, we want $b$,$c$,$d$, and $e$ to have a common factor, and the obvious choice is $p$. The non-powerful number $n$ ensures that $p$ won't divide $a$, and won't divide $e$ more than once, so we'll get an irreducible $F$. Let $d=e=p$, $c=3p$, and $b=kp$ for some integer $k$. Then we want to find $a$ with $$ 4a+(3-k)p=m $$ If $p \neq 2$, we can always choose $k$ such that $(3-k)p \equiv m \pmod{4}$, which gives an integer $a$, and an irreducible form $F$ with $I=3n$, $J=189ap^2-27p^3(k^2-k+2)$.

For the second condition, we can take $d=e=p$, $c=3p\mp1$, $b=kp$, and then choose $k$ such that $$ 4a \mp 2 +(3-k)p =3m $$ has an integral solution for $a$. The other two have similar results.

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