Recently I was assigned a homework problem to determine the possible Galois groups of an irreducible quartic polynomial over $\mathbb{Q}$ that has exactly two real roots (equivalently has negative discriminant) (Artin Exercise 16.9.10). By Galois group of a polynomial, we mean if $K$ is the splitting field for $f(x)$ over $\mathbb{Q}$, then the Galois group of $f(x)$ is the Galois group of the extension $K/\mathbb{Q}$. We determined that the possible Galois groups are $D_4$ or $S_4$. Artin gave an example of a polynomial of this form whose Galois group is $D_4$, but he did not give one for $S_4$. We were eventually able to show that the Galois groups for $x^4 – 3x + 1$ and for $x^4 – x – 1$ are isomorphic to $S_4$, but we did not find a systematic method to construct these polynomials, we guessed and checked until we found some that worked.
This leads to my question. We knew that if we had an irreducible cubic polynomial $g(x)$ with negative discriminant and if we had an irreducible quartic polynomial $f(x)$ with negative discriminant whose resolvent cubic is $g(x)$, then the Galois group for $f(x)$ over $\mathbb{Q}$ will be isomorphic to $S_4$. Thus in an attempt to find an algorithm to construct these polynomials, we wanted to be able to start with the cubic polynomial $g(x)$ and then construct a quartic polynomial $f(x)$ in $\mathbb{Q}[x]$ whose resolvent cubic is $g(x)$. Is there a way to do this? Are there cubic polynomials that are not the resolvent cubic of any quartic polynomial? Even special cases of when this is possible would be helpful.
Best Answer
One way to construct quartic polynomials with a given cubic resolvent is to consider quartic polynomials of the form $f(x)=x^4+bx^2+cx+d$. The cubic resolvent is $g(x)=x^3-2bx^2+(b^2-4d)x+c^2$. If you write $g(x)=x^3+rx^2+sx+t$ and solve for $b,c,d$ in terms of $r,s,t$, this gives $b = -r/2$, $c=\pm\sqrt{t}$, and $d=(r^2-4s)/16$.
For example, if we choose $r=6$, $s=5$, and $t=4$, this produces $f(x)=x^4-3x^2 \pm 2x+1$, which has cubic resolvent $g(x)=x^3+6x^2+5x+4$.