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Fall 2016 - Problem 5

Galois theory

Let $f \in F\br{X}$ be an irreducible separable polynomial of prime degree over a field $F$ and let $K/F$ be a splitting field of $f$ . Prove that there is an element in the Galois group of $K/F$ permuting cyclically all roots of $f$ in $K$ .

Solution.

Let $p = \deg{f}$ .

Since $f$ is separable, it follows that $K$ is Galois. Moreover, because $f$ is irreducible, we have the tower $F \subseteq F/\gen{f} \subseteq K$ . In particular, $p = \br{F/\gen{f} : F}$ divides $\br{K : F} = \abs{\Gal\p{K/F}}$ , so by Cauchy's theorem, $\Gal\p{K/F}$ admits an order $p$ element, say $\sigma$ .

Let $\alpha$ be a root of $f$ not fixed by $\sigma$ . Then the orbit of $\alpha$ under $\sigma$ gives all the roots of $f$ : if this were not the case, then there exists $0 < k < p$ minimal such that $\sigma^k\p{\alpha} = \alpha$ . Since $p$ is prime, we can write $p = qk + r$ with $0 < r < k$ so that

\alpha = \sigma^p\p{\alpha} = \sigma^r\p{\sigma^{qk}\p{\alpha}} = \sigma^r\p{\alpha},

which contradicts minimality of $k$ . Thus, $\sigma$ permutes the roots of $f$ cyclically, which was what we wanted to show.