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

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 $$F$$?

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.

• Thanks. I am learning a new technique from your answer, +1. Recently we did something similar. Constructed an infinite family of D4-quartics containing given real quadratic with the additional condition that these quartics contain exceptional units. See arxiv.org/abs/2306.17556 . Ours was ad-hoc. This would be more useful Commented Nov 11, 2023 at 6:13

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.

• Maybe this is obvious but could you please clarify why the extensions can be picked linearly disjointly over the quadratic subfield? Commented Jul 6, 2023 at 16:23
• Thanks, Joachim, for the same reason I thanked Stanley. Commented Nov 11, 2023 at 6:15