Is it possible that each element of Galois group preserve a root?

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 $\deg f \geq 2$.

• For a trivial extension – yes – johnnycrab Dec 21 '16 at 10:14

$\newcommand{\Set}{\left\{ #1 \right\}}$$\newcommand{\Size}{\left\lvert #1 \right\rvert}No when G \ne \{ 1 \}, that is, when the degree of f is greater than 1, a requirement just added by OP. The reason is that the Galois group acts transitively on the set \Omega of the roots, as f is irreducible. Thus all stabilizers are conjugate. And it easy to prove that a finite group cannot be the set-theoretic union of the conjugates of a proper subgroup - I have appended the standard proof below. In fact, suppose G is a finite group, H < G, and consider the set$$ \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$H^{\#} = H \setminus \Set{1}$. • 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 ... – DonAntonio Dec 21 '16 at 10:20 • I have noted that, and added that the answer is no except for (what I would consider the trivial case)$G = 1$. – Andreas Caranti Dec 21 '16 at 10:30 • OP has just excluded the trivial case. – Andreas Caranti Dec 21 '16 at 10:32 • @An Indeed so...25 minutes after he posted it. – DonAntonio Dec 21 '16 at 10:34 No, unless$G\$ is trivial.