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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 $\Bbb Q(r_1) \cap K = \Bbb Q$ ?


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!

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  • $\begingroup$ The part "we could find a chain of subfields…" is proved in theorem 7.2., Algebra, S. Lang. $\endgroup$ – Watson Oct 9 '16 at 15:33
  • $\begingroup$ Related: math.stackexchange.com/questions/2100828 $\endgroup$ – Watson Jan 18 '17 at 10:25
  • $\begingroup$ When I wrote "this condition about the intersection ensures that K has "nothing in common" with $r_1$", this is actually not completely true, since we may have $\sqrt{r_1} \in K$, for instance... $\endgroup$ – Watson Oct 28 '18 at 8:54
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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.

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  • $\begingroup$ Thank you. You might be also interested in this question. $\endgroup$ – Watson Oct 28 '18 at 10:20
  • $\begingroup$ Read it, answered it to my satisfaction, and then noticed you don't even vote up correct answers to your own question, and tossed the answer in the trash, never to return. $\endgroup$ – user608470 Oct 28 '18 at 23:57
  • $\begingroup$ Sorry for upvoting a bit late your post (I did not forget to do so!). It would have been a pleasure to probably accept and upvote another answer of yours… $\endgroup$ – Watson Oct 29 '18 at 8:21
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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
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