1
$\begingroup$

Let $f(x) \in \mathbb{Q}[x]$ be an irreducible polynomial. Denote by $K$ the splitting field for $f(x)$. Let $G$ be Galois group of $K$ over $\mathbb{Q}$. Let $\alpha_1, \dots, \alpha_n \in K$ be all the roots of $f(x) = \Pi (x - \alpha_i)$.

Question: Does there exist such an $f(x)$ that for each $g \in G$ there exist $\alpha_i$ such that $g( \alpha_i ) = \alpha_i$?

Edited: I am interested in the case $\deg f \geq 2$.

$\endgroup$
  • 1
    $\begingroup$ For a trivial extension – yes $\endgroup$ – johnnycrab Dec 21 '16 at 10:14
1
$\begingroup$

$\newcommand{\Set}[1]{\left\{ #1 \right\}}$$\newcommand{\Size}[1]{\left\lvert #1 \right\rvert}$No when $G \ne \{ 1 \}$, that is, when the degree of $f$ is greater than $1$, a requirement just added by OP.

The reason is that the Galois group acts transitively on the set $\Omega$ of the roots, as $f$ is irreducible.

Thus all stabilizers are conjugate. And it easy to prove that a finite group cannot be the set-theoretic union of the conjugates of a proper subgroup - I have appended the standard proof below.


In fact, suppose $G$ is a finite group, $H < G$, and consider the set $$ \bigcup_{g \in G} H^{g}, $$ where $H^{g} = g^{-1} H g$. We have $$ \Size{\bigcup_{g \in G} H^{g}} = 1 + \Size{\bigcup_{g \in G} (H^{\#})^{g}} \le 1 + \Size{H^{\#}} \cdot \Size{G : N_{G}(H)} \le\\\le 1 + (\Size{H} - 1) \cdot \Size{G : H} = \Size{G} - \Size{G : H} + 1 < \Size{G}, $$ where I have used the fact that $H < G$, so that $\Size{G : H} > 1$. Here $H^{\#} = H \setminus \Set{1}$.

$\endgroup$
  • $\begingroup$ And read also the comment below the question: if $\;f(x)=x-q\;,\;\;q\in\Bbb Q\;$ , then $\;G=1\;$ , and thus the answer seems to be yes ... $\endgroup$ – DonAntonio Dec 21 '16 at 10:20
  • $\begingroup$ I have noted that, and added that the answer is no except for (what I would consider the trivial case) $G = 1$. $\endgroup$ – Andreas Caranti Dec 21 '16 at 10:30
  • $\begingroup$ OP has just excluded the trivial case. $\endgroup$ – Andreas Caranti Dec 21 '16 at 10:32
  • $\begingroup$ @An Indeed so...25 minutes after he posted it. $\endgroup$ – DonAntonio Dec 21 '16 at 10:34
1
$\begingroup$

No, unless $G$ is trivial.

The action of the Galois group of an irreducible polynomial is transitive and every transitive permutation group on a finite set has an element that fixes no point.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.