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Let $L$ be a dihedral quartic field. That is, we have $[L : \mathbb{Q}] = 4$ and the Galois closure $M$ of $L$ is a degree 8 extension of $\mathbb{Q}$ with Galois group isomorphic to the dihedral group $D_4$ of order 8. It is well-known (see the diagram on page 6 of this paper: https://arxiv.org/abs/1704.01729) that $M$ contains five subfields which are degree 4 extensions of $\mathbb{Q}$; say $L = L_1, L_2, L_0, L_3, L_4$. In particular, $L_1, L_2$ are Galois conjugates and contain a common quadratic field $K_1$, $L_0$ is a Galois extension of $\mathbb{Q}$ and has Galois group isomorphic to $C_2 \times C_2$, while $(L_3, L_4)$ is related to $(L_1, L_2)$ via an outer automorphism which comes from the group theory of $D_4$, which we denote by $\phi$. In this language, we identify $L_3 = \phi(L_1), L_4 = \phi(L_2)$ and $L_3, L_4$ are Galois conjugates.

Given $L$, it is of course possible (say via the primitive element theorem) to find a quartic polynomial $f$ such that for some root $\alpha$ of $f$ we have $L = \mathbb{Q}(\alpha)$.

My question is, given a quartic polynomial $f$ such that for some root $\alpha$ of $f$ we have $L = \mathbb{Q}(\alpha)$ is a dihedral quartic field, how to find a polynomial $g$ whose coefficients are algebraic functions of the coefficients of $f$ such that for some root $\beta$ of $g$ we have $\phi(L) = \mathbb{Q}(\beta)$? Moreover, can one take advantage of the invariant theory of the homogenized binary quartic form $F(x,y) = y^4 f(x/y)$? That is, is $g$ naturally given by some quartic covariant of $F$?

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Let $\alpha=\alpha_1, \alpha_2, \alpha_3, \alpha_4$ be the conjugates of $\alpha$ in $M$, numbered so that one of the 4-cycles in the Galois group permutes them in the order $\alpha_1 \mapsto \alpha_2 \mapsto \alpha_3 \mapsto \alpha_4 \mapsto \alpha_1$.

Thus $L_1=\mathbb{Q}(\alpha_1) = \mathbb{Q}(\alpha_3)$ is fixed by that automorphism which interchanges $\alpha_2$ and $\alpha_4$. (Think of a square with corners labeled $1,2,3,4$ being reflected through the diagonal passing through corner number $1$. And $L_0$ corresponds to the central inversion symmetry of the square.)

The outer automorphism of $D_4$ amounts to looking at how rigid motions act on the set of unoriented edges of the square instead of on the corners: what fixes an edge is reflecting the square across the bisector. We're thus looking for an element fixed e.g. under simultaneously exchanging $\alpha_1$ with $\alpha_4$ and $\alpha_2$ with $\alpha_3$, and not fixed under the other non-identity permutations coming from the Galois group. In particular, it must not remain fixed when we reflect the chosen edge onto its opposite edge.

My first suggestion had been to set $\beta_1=(\alpha_4-\alpha_1)^2 - (\alpha_3-\alpha_2)^2$, but that turns out not to work at all - it always ends up in one of the quadratic subfields (and as has been pointed out in the comments, it sometimes vanishes outright!).

Revised approach: By adding a rational number if necessary, we can assume from the start that $\alpha_1 = -\alpha_3$ is traceless over the quadratic subfield of $L_1$, and thus of the form $\pm\sqrt{m\pm\sqrt{n}}$ with rational $m$ and $n\ne 0$, and its conjugates correspond to all the sign combinations.

As a warm-up exercise, $\gamma_1=\alpha_1/\alpha_4$ is then conjugate to its inverse (under $1\leftrightarrow 4, 2\leftrightarrow 3$) and also to its negative inverse $\alpha_2/\alpha_1$ (under a 4-cycle), and not fixed by either operation; it is fixed however by the central involution $1\leftrightarrow 3, 2\leftrightarrow 4$, and thus it generates the abelian quartic field $L_0$.

Shifting things away from the traceless subspace destroys these relationships: $(1+\alpha_1)/(1+\alpha_4)$ generically generates the normal closure $M$. (When it doesn't, replace $1$ with another rational number.) And adding its inverse to it produces an element invariant under the desired involution: $$\beta_1=\frac{1+\alpha_1}{1+\alpha_4} + \frac{1+\alpha_4}{1+\alpha_1}$$

It remains to compute the symmetric functions of $\beta_1$ and its conjugates and to express them in terms of those of the $\alpha_i$.

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  • $\begingroup$ I am concerned that when $f$ is bi-quadratic, that’s is $f(x) = ax^4 + bx^2 + c$ that the terms $(\alpha_1 - \alpha_4)^2 - (\alpha_3 - \alpha_4)^2$ collapse (they are all equal to zero in this case)... $\endgroup$ – Stanley Yao Xiao Oct 15 '18 at 22:14

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