Let $s,t$ be two independent real parameters, and let $a_2(s,t), a_1(s,t), a_0(s,t)$ be linear forms in $s, t$ with real coefficients. Put $a_4 = s, a_3 = t$ and consider the quadratic form

$$\displaystyle Q(s,t) = 12a_4 a_0 - 3a_3 a_1 + a_2^2$$ and the cubic form

$$\displaystyle R(s,t) = 72a_4 a_2 a_0 + 9 a_3 a_2 a_1 - 27 a_4 a_1^2 - 27 a_0 a_3^2 - 2a_2^3.$$

Can anyone suggest a method to find coefficients of $a_2, a_1, a_0$ such that

$$Q(s,t) = s^2 - st + t^2$$ and

$$R(s,t) = s^3 + 3s^2 t - 6 s t^2 + t^3?$$

The astute reader may notice that $Q,R$ are respectively the $I$ and $J$ invariants of a binary quartic form $F = a_4 x^4 + a_3 x^3 y + a_2 x^2 y^2 + a_1 xy^3 + a_0 y^4$. Parametrizing them in this way reveals a way to parametrize those binary quartic forms whose Galois group is isomorphic to $A_4$.


1 Answer 1


Is the obvious method unsatisfactory?. I indicate how to get all solutions for $Q$:

Writing $a_j(s,t)= p_js+q_jt,\ j=0,1,2$ and substituting them in your definition of $Q(s,t)$ and equating them to the $s^2-st+t^2$ gives rise to 3 equations of degree 2 in the real variables $p_j,q_j,\ j=0,1,2$

This leads to $$ p_2=\sqrt{1-12p_0},\quad q_2=\sqrt{1+3q_1}, \quad p_1 =1+12q_0+ \sqrt{2(1-12p_0)(1+3q_1)}$$ Choose $p_0,q_1$ such that the quantities under square roots are nonnegative, and $q_0$ arbitrarily and the other values as per the formula above, we see that all of them arise this way. (this is an affine variety of dimension 3 in a 6-dimensional affine space: rather real points of such a variety).


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