Splitting fields of degree 4 irreducible polynomials containing a fixed quadratic extension

Question

Let $f_1(x)\in \mathbb{Z}[x]$ be a fixed irreducible degree 4 polynomial such that its splitting field $F_1$ is an $S_4$-Galois extension over $\mathbb{Q}$ and the discriminant of $F_1$ is of the form $-k^2$ for some integer $k$. It is possible to show that $F_1$ contains $\mathbb{Q}(\sqrt{-1})$.

Does there always exist another irreducible degree 4 polynomial $g(x)\in \mathbb{Z}[x]$ whose splitting field $F$ is an $S_4$-Galois extension over $\mathbb{Q}$ and such that $F\cap F_1 = \mathbb{Q}(i)$?

In fact, I suspect that there should be infinitely many such extensions $F/\mathbb{Q}$ which are $S_4$-Galois over $\mathbb{Q}$ satisfying $F\cap F_1 = \mathbb{Q}(i)$? If my guess is actually correct, is there a generic way to construct a family of such polynomials which give rise to these distinct $F$?

Answer 1

Yes, this can be done. Let $K$ be a $S_4$-quartic field, $C$ its cubic resolvent field, and $L$ the Galois closure of $K$. By Galois correspondence, $L$ contains a unique quadratic subfield $Q$ which corresponds to the alternating group $A_4$; by our hypothesis, this quadratic subfield is equal to $\mathbb{Q}(\sqrt{-1})$. Note that the Galois closure $C^\prime$ of $C$ is also contained in $L$, and by Galois correspondence again $C^\prime$ has a unique quadratic subfield, which is then necessarily equal to $Q = \mathbb{Q}(\sqrt{-1})$. Therefore, to construct the required quartic polynomials (fields) we first look at the possible cubic resolvent fields.

It is known that the discriminant of a cubic field $C$ can be expressed uniquely in the form $\Delta(C) = df^2$, where $d$ is the discriminant of the quadratic resolvent field of $C$. By our assumption, we must have $d = -4$. We are thus looking for cubic fields whose discriminants are equal to $-(2k)^2$ for some $k \geq 1$. By the Delone-Faddeev correspondence, this is the same as looking for ($\text{GL}_2(\mathbb{Z})$-equivalence classes of) binary cubic forms with integer coefficients and discriminant $-4k^2$.

Restricting to monic binary cubic forms of the shape $F(x,y) = x^3 - Axy^2 + By^3$ whose discriminant is $4A^3 - 27B^2$, this gives the equation

$$\displaystyle 4A^3 = 27B^2 - 4k^2 \text{ giving } A^3 = 27b^2 - k^2, b = B/2.$$

This equation defines a genus 0 curve and can be explicitly parametrized.

Now to answer the question: given an initial $F_1$ with discriminant $-k^2$, construct infinitely many cubic fields $C_\ell$ with discriminant equal to $-4\ell^2$ and $\gcd(k, \ell)$ equal a power of 2 as above. Then, using the results of this paper (Dirichlet series associated to quartic fields with given cubic resolvent) by Cohen and Thorne, one can give infinitely many quartic fields with $K$ with cubic resolvent field equal to $C_\ell$. The resulting Galois closures of the fields $K$ generated this way will satisfy your constraints.

Answer 2

This works even for any symmetric group and any quadratic number field, and can be done very explicitly.

When $n$ is even, let $f(t,X) = X^{n-1}((n-1)X-n) + t$, and when $n$ is odd, let $f(t,X) = X^n-t(nX-(n-1))$.

These are trinomials with Galois group $S_n$ over $\mathbb{Q}(t)$ (actually, they're basically the same kind of trinomial, just shifted a bit according to parity, in order to reach the same kind of discriminant). From available explicit discriminant formulas, one gets discriminant equivalent (up to squares) to $c\cdot (t-1)$, for some (also explicit, namely basically $n$ or $n-1$ up to sign) integer constant $c$. So now, substitute $t = c\cdot \alpha s^2 + 1$ with a variable $s$ and $\alpha\in \mathbb{Z}$ your favorite non-square, e.g. $\alpha=-1$. The discriminant, up to squares, now equals $\alpha$, so that that this is an $S_n$-extension of $\mathbb{Q}(s)$ whose constant extension equals $\mathbb{Q}(\sqrt{\alpha})/\mathbb{Q}$ (corresponding to the fixed field of $A_n$. From Hilbert's irreducibility theorem, specializing $s$ to integer values "mostly" gives $S_n$-extensions with quadratic subfield $\mathbb{Q}(\sqrt{\alpha})$, and also those extensions can be picked linearly disjoint over that quadratic subfield.