# 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

## 2 Answers


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.