Is there a quadratically closed field strictly between the quadratic closures of $\mathbb{Q}$ and $\mathbb{Q}(\sqrt[3]{2})$?

Solution 1:

The question has been answered positively on Mathoverflow in two different ways. I post here a copy of my solution: Let $K$ be the quadratic closure of $\mathbb{Q}$, and $L$ be the quadratic closure of $\mathbb{Q}(\sqrt[3]{2})$. The roots of the polynomial $P(x) := x^4 + 36x + 54$ are not in $K$, but they are in $L$, and $P$ is irreducible because it is irreducible modulo $5$, hence any root $\alpha$ of it is of degree $4$, and the quadratic closure of $\mathbb{Q}(\alpha)$ is strictly between $K$ and $L$. (I used https://www.alpertron.com.ar/POLFACT.HTM for the calculations modulo 5.)

First, why are the roots of $P$ not in $K$? It is because mod $7$, it decomposes in $(x+6)(x^3+x^2+x+2)$, which are irreducible, and the discriminant of this polynomial is $-5038848$, which is not divisible by $7$, therefore by Dedekind's theorem there is a cycle of order $3$ in the Galois group of $P$ (Again, calculation modulo 7 with https://www.alpertron.com.ar/POLFACT.HTM, and calculation of the discriminant with https://planetcalc.com/8188/.)

Finally, it is in the quadratic closure of $\mathbb{Q}(\sqrt[3]{2})$, because we can look at the explicit formula for the roots of a quartic:

In our case, $a = b = 0$, therefore $e = \sqrt{(27c^2)^2-4(12d)^3} = 11664$, and $f = \sqrt[3]{27c^2 + e} = \sqrt[3]{2^6 \cdot 3^6} = 2^2 \cdot 3^2$. Except that, only square roots and $\sqrt[3]{2}$ appear in the formula, thus the roots are in $L$.