Is $\Bbb Q[a^2b+b^2c+c^2d+d^2a]$ the fixed field of the subgroup $\langle(1234)\rangle$ of Galois group $S_4$?

Question

$f$ is an irreducible quartic polynomial in $\Bbb Q[x]$ with Galois group $S_4$.

$a,b,c,d$ are four distinct roots of $f$.

Let $K$ be the splitting field of $f$. Let $\text{Gal}(K/\Bbb Q)=S_4$.

Then $a^2b+b^2c+c^2d+d^2a$ is fixed by $\langle(1234)\rangle$.

So $\Bbb Q[a^2b+b^2c+c^2d+d^2a]\subseteq K^{\langle(1234)\rangle}$

Is it always true that $\Bbb Q[a^2b+b^2c+c^2d+d^2a]=K^{\langle(1234)\rangle}$?

Equivalently, the minimal polynomial of $a^2b+b^2c+c^2d+d^2a$ has degree $6$.

I think it is not always true, but cannot come up with a counterexample.

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I verified it for $f=x^4-x+1$. It has Galois group $S_4$.

f = QQbar['x'](x^4-x+1)
roots = [f.roots()[n][0] for n in range(4)]
expr = sum([roots[i]**2*roots[(i+1)%4] for i in range(4)])
expr.minpoly()

$x^{6} + 6 x^{5} + 24 x^{4} + 56 x^{3} + 32 x^{2} - 32 x - 256$ indeed has degree $6$.

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I verified it for $f=x^4+2x+2$. It has Galois group $S_4$.

f = QQbar['x'](x^4+2*x+2)
roots = [f.roots()[n][0] for n in range(4)]
expr = sum([roots[i]**2*roots[(i+1)%4] for i in range(4)])
expr.minpoly()

$x^{6} - 12 x^{5} + 96 x^{4} - 448 x^{3} + 1024 x^{2} - 1024 x - 6144$ indeed has degree $6$.

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I verified it for $f=x^4 - 6x^2 +2x+2$.

It has four real roots and Galois group $S_4$ (from Finding a quartic polynomial in $\mathbb{Q}[X]$ with four real roots such that Galois group is ${S_4}$.)

f = QQbar['x'](x^4 - 6*x**2 +2*x+2)
roots = [f.roots()[n-1][0] for n in range(4)]
expr = sum([roots[i]**2*roots[(i+1)%4] for i in range(4)])
expr.minpoly()

$x^{6} - 12 x^{5} - 240 x^{4} + 2240 x^{3} + 10240 x^{2} - 59392 x - 20736$ indeed has degree $6$.

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I verified it for $x^4−4x^2+x+1$.

It has four real roots and Galois group $S_4$.

f = QQbar['x'](x^4 - 4*x**2 + x + 1)
roots = [f.roots()[n-1][0] for n in range(4)]
expr = sum([roots[i]**2*roots[(i+1)%4] for i in range(4)])
expr.minpoly()

$x^{6} - 6 x^{5} - 72 x^{4} + 328 x^{3} + 896 x^{2} - 2464 x - 640$ indeed has degree $6$.

Answer 1

Yes it is always true.

The following argument works for any base field assuming that it doesn't have characteristic $2$, so certainly for the field $\mathbf{Q}$.

Determining the field is the same as determining which subgroup of $S_4$ fixes this element. Suppose it was fixed by a larger group. I claim the only subgroups of $S_4$ which strictly contain $(1,2,3,4)$ are either $S_4$ or

$$\Gamma = \langle (1,2,3,4), (1,2)(3,4) \rangle \simeq D_8$$

To see this, the order of such group is divisible by $4$ and divides $24$, so its order is either $24$, in which case it is $S_4$, or order $12$, in which case it is $A_4$ but that doesn't contain a $4$-cycle, or order $8$, in which case it is a $2$-Sylow. But there is a unique $2$-Sylow of $S_4$ containing this $4$-cycle, namely $\Gamma$. In all cases, any such group contains $\Gamma$.

So we only need to show that $a^2 b + b^2 c + c^2 d + d^2 a $ is not invariant under $\Gamma$, or that it is not fixed by $(12)(34)$. If it is fixed by $(1,2)(3,4)$ then

$$a^2 b + b^2 c + c^2 d + d^2 a = a b^2 + a^2 d + d^2 c + c^2 b,$$

and taking the difference of these two elements it follows after factorizing that

$$(a-c)(b-d)(a+c-b-d) = 0.$$

The roots are all distinct so if this holds it must be the case that $a+c = b + d$. But since the Galois group is $S_4$ (the only point where this is used), there exists an automorphism which sends $a$ to $b$ and fixes $c$ and $d$. Applying this to $a+c=b+d$ gives $b+c=a+d$, and then subtracting one equality from the other, it follows that $a-b=b-a$ or $2a=2b$ so (when the characteristic is not $2$) $a=b$, a contradiction.