Expressing a root of an irreducible polynomials using the other roots

Question

Here is my question. Below I give some context and my thoughts.

Let $P \in \Bbb Q[X]$ be an irreducible polynomial, of degree $n$. We denote by $r_1,\dots,r_n$ its roots. Is there a subfield $K \subset K_P := \Bbb Q(r_1,\dots,r_n)$ such that the Galois extension $ K_P/K$ has a solvable Galois group, and such that $\Bbb Q(r_1) \cap K = \Bbb Q$ ?

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Motivation:

If $K_P/K$ has a solvable Galois group, then it would mean that we could find a chain of subfields $K \subset K_1 \subset \cdots \subset K_n \subset K_P$ such that for any $1≤j≤n-1$, $K_{j+1}=K_j(a_j)$ where $a_j^{m_j} \in K_j$ for some $m_j ≥ 1$. Since $\Bbb Q(r_1) \cap K = \Bbb Q$, it means that we can express $r = r_1 \in K_P$ using radicals and other roots (this condition about the intersection ensures that $K$ has "nothing in common" with $r_1$, which is actually not completely true in some sense...).

Context:

I was thinking about the relations between the roots of irreducible polynomials (you can choose the roots of reducible polynomials as you want, e.g. $(x-a_1)\cdots(x-a_n)$ so that's not interesting). The Viète's formulas are well-known, but I wanted to express one particular root $r=r_1$ using the other roots. [More generally, I was wondering about the possible relations between the conjugates of some algebraic number $z_0$.]

It is not true that $r_1$ is a polynomial function of some other root. For instance, $\sqrt{2-\sqrt 5} \neq P(\sqrt{2+\sqrt 5})$ for any $P \in \Bbb Q[X]$. Indeed, $\sqrt{2-\sqrt 5}$ has degree $4$ over the rationals, while $\Bbb Q(\sqrt{2+\sqrt 5})$ is not Galois (its Galois closure has Galois group $D_8$). [Probably similar situations happen for $\sqrt{\sqrt 2 + \sqrt 3}$ which doesn't seem to be equal to some $P(\sqrt{\sqrt 2 - \sqrt 3})$, where $\sqrt{\sqrt 2 - \sqrt 3}$ is a conjugate of $\sqrt{\sqrt 2 + \sqrt 3}$ (by the way, recall that $\Bbb Q(\sqrt 2+ \sqrt 3)=\Bbb Q(\sqrt 2, \sqrt 3)=\Bbb Q(\sqrt 2- \sqrt 3)$, or happen for roots of $X^5-X-1$, I think.]

We also see that $\zeta_3 \sqrt[3]{2}$ is a conjugate of $\sqrt[3]{2}$ but it can't be written as $P(\sqrt[3]{2})$ with $P \in \Bbb Q[X]$. So my idea was not to focus on polynomials/rational functions of roots, but rather on radicals.

My thoughts :

Let $A$ be the collection of subfields $K \subset K_P$ such that $\text{Gal}(K_P/K)$ is solvable, ordered with $K≤L \iff K \supset L$. As $K_P/K$ is a finite separable extension of $\Bbb Q$, it has finitely many subextensions, so that $A$ is finite (and non-empty since $K_P \in A$). Thus it is clearly an inductive set. By Zorn's lemma (!) I get a $≤$-maximal subextension $K$, i.e. $\subset$-minimal. I only have to show that it has trivial intersection with $\Bbb Q(r_1)$, which seems difficult to do.

Possibly related: (1).

Thank you for your help!

Answer 1

The following is too long for a comment, but I felt it adds some clarity to the question, even though it does not provide an answer:

Let $G = \text{Gal}_{\mathbb{Q}}(K_P)$, let $F = \mathbb{Q}(r_1)$ and $H = \text{Gal}_F(K_P)$. Then, by this answer, your question is equivalent to asking

Does there exist a solvable subgroup $L < G$ such that $\langle LH\rangle = G$.

What we know:

1. Consider the injective homomorphism $$ G\hookrightarrow S_n $$ via the (faithful, transitive) action of $G$ on the roots $X:= \{r_1,r_2,\ldots, r_n\}$, then it follows that $$ H = G\cap S_{n-1} $$
2. $[G:H] = [\mathbb{Q}(r_1):\mathbb{Q}] = \deg(P(x)) = n$, since $P$ is separable.
3. If $G$ is solvable, there is nothing to prove, so we may assume $n\geq 5$.
4. If $G = S_n$, then $L = \langle (12)\rangle$ works.
5. If $G = A_n$, then (I suspect), $L = \langle (12)(34)\rangle$ works.
6. If $n=5$, the only possibilities for $G$ are $S_5, A_5$ and 3 solvable groups ($\mathbb{Z}_5, D_5$, and $F_{20}$), so the result you want is true in this case.
7. These notes of Keith Conrad might be useful, particularly Theorem 2.2

Answer 2

No.

Let $L/\mathbf{Q}$ be any Galois extension with non-abelian Galois group $G$. (Such fields exist, by Galois.) By the primitive element theorem, $L = \mathbf{Q}[x]/P(x)$ for a polynomial $P$ of degree $|G|$. But now $L = \mathbf{Q}(r_1)$ by construction, hence $\mathbf{Q}(r_1) \cap K = \mathbf{Q}$ for $K \subset L$ implies that $K = \mathbf{Q}$. But this implies that $\mathrm{Gal}(L/K) = \mathrm{Gal}(L/\mathbf{Q}) = G$ cannot be solvable.