Algebraic numbers which prescribed degree which does not belong to some fields

Question

In my research it would be great if the following result is valid. In what follows, $\overline{\mathbb{Q}}$, $\overline{\mathbb{Q}}_n$ and $\overline{\mathbb{Q}}_{<n}$ denotes the set of algebraic numbers, $n$-degree algebraic numbers and algebraic numbers with degree at most $n-1$, respectively. Also $\mathcal{R}_{\mathcal{F}}$ denotes the set of zeroes of a family of polynomials $\mathcal{F}$.

Thus, my question is if the following two propositions are true (clearly, Prop. 2 implies in Prop. 1):

Prop 1. If $K_n:=\mathbb{Q}(\{\alpha: \alpha\in \overline{\mathbb{Q}}_{<n}\})$, then there exists an $n$-degree algebraic number $\gamma$ which does not belong to $K_n$.

Prop 2. Let $t<n$ be a positive integer and set $K_n:=\mathbb{Q}(\mathcal{R}_{\mathcal{F}})$, where $\mathcal{F}$ is the set of all $t$-degree polynomials in $\overline{\mathbb{Q}}_{<n}[x]$. Then there exists an $n$-degree algebraic number $\gamma$ which does not belong to $K_n$.

By using some results of the paper https://arxiv.org/pdf/0901.1335.pdf it is possible to prove the existence of the algebraic $\gamma\not\in K_n$ with arbitrarily large degree. However, I was not able to find any quantitative version of their result for degree exactly equals to $n$.

Any suggestions?

Answer 1

Proposition 1 is true.

Let $x$ be an element of degree $n$ whose Galois group is isomorphic to $S_n$. Let $F$ be a finite subset of algebraic elements of degree $<n$. I claim that $x$ is not in the subfield generated by $F$.

Let $K$ be a finite Galois extension containing $F\cup\{x\}$. Let $G$ be its Galois group. Galois corresponds yields an order-reversing bijection between the set of subfields of $K$ and the set of subgroups of $G$. Under this bijection, for $z\in K$, the subfield $\mathbf{Q}(z)$ corresponds to a subgroup $H_z$ of $G$, of index $\deg(z)$. In particular, for $y\in F$, $H_y$ has index $<n$. The condition $\mathbf{Q}(x)\subset\mathbf{Q}(F)$ (which we assume by contradiction) means, since $\mathbf{Q}(F)$ is the smallest upper bound to $\{\mathbf{Q}(y):y\in F\}$. Hence it means that $H_x$ contains the intersection $\bigcap_{y\in F} H_y$. Let $F'$, resp. $X$ be the set of all conjugates of elements of $F$, resp of $x$ and $N=\bigcap_{y\in F'}H_y$, and $M=\bigcap_{x'\in X}H_{x'}$. So $N$ is a normal subgroup and $N\subset H_x$. Hence $N\subset M$. Since $N$ is an intersection of subgroups of index $<n$, $G/N$ embeds into $S_{n-1}^k$ for some $k$. Hence its quotient $G/M\simeq S_n$ is a subquotient of $S_{n-1}^k$. This is absurd.

[If $n\ge 5$ the simple group $A_n$ is not a subquotient of $S_{n-1}$, hence not of $S_{n-1}^k$. Also $S_n$ is not a subquotient of $S_{n-1}^k$ for $n=2,3,4$: for $n=2$ this is clear since $S_2\neq 1=S_1$; for $n=3$ $S_3$ has elements of order 3 but not $S_2^k$; for $n=4$ $S_4$ is not metabelian unlike $S_3^k$.]

Answer 2

Proposition 2 is false, although perhaps only for $n = 4$ (and $t = 2$ or $t = 3$). If $t = 2$, every algebraic number $\gamma$ of degree $4$ is contained in $K_{4}$.

Let $K$ be the Galois closure of $\mathbb{Q}(\gamma)$, and $G = {\rm Gal}(K/\mathbb{Q})$. If $G \simeq (\mathbb{Z}/2\mathbb{Z}) \times (\mathbb{Z}/2\mathbb{Z})$, $\mathbb{Z}/4\mathbb{Z}$, or the dihedral group of order $8$, then there is a quadratic field $L$ with $\mathbb{Q} \subseteq L \subseteq \mathbb{Q}(\gamma)$, and hence $\gamma$ is a root of a degree $2$ polynomial with coefficients in $L$ and $L \subseteq \overline{\mathbb{Q}}_{<4}$. This implies that $\gamma \in \mathcal{R}_{\mathcal{F}}$ and so $\gamma \in K_{4}$.

If $G \simeq A_{4}$, then is a degree $3$ extension $M/\mathbb{Q}$ that is Galois and contained in $K$. There are three quadratic extensions of $M$, say $M_{1}$, $M_{2}$ and $M_{3}$, that are contained in $K$. Each $M_{i}$ is a quadratic extension of $M$, and so if $M_{i} = \mathbb{Q}(\beta_{i})$, then $\beta_{i}$ is a root of a degree $2$ polynomial with coefficients in $M$, and hence a root of a degree $2$ polynomial with coefficients in $\overline{\mathbb{Q}}_{<4}$. This implies that $M_{i} \subseteq K_{4}$ and since $K = \langle M_{1}, M_{2}, M_{3} \rangle$, we have $K \subseteq K_{4}$.

If $G \simeq S_{4}$, the fact that $K/\mathbb{Q}$ and $K_{4}/\mathbb{Q}$ are both normal implies that $K \cap K_{4}$ is a normal extension of $\mathbb{Q}$. For an $S_{4}$ extension, there are only two normal subextensions: the discriminant field $\mathbb{Q}(\sqrt{D})$, and the splitting field of the resolvent cubic. However, there is a non-normal cubic extension $M$ (corresponding to a Sylow $2$-subgroup of $S_{4}$), and quadratic extensions $M_{1}$, $M_{2}$ and $M_{3}$ of $M$. As in the $A_{4}$ case, $M_{i} \subseteq K_{4}$. However, precisely one of the $M_{i}$ is the splitting field of the resolvent cubic, and so $\langle M_{1}, M_{2}, M_{3} \rangle \subseteq K_{4}$ has degree over $\mathbb{Q}$ greater than $6$. Since $K \cap K_{4}$ is normal and $K \cap K_{4} \supseteq \langle M_{1}, M_{2}, M_{3} \rangle$, we have $K \cap K_{4} = K$ and thus $K \subseteq K_{4}$.