# Characteristic Polynomial of Galois automorphism

Let $K/F$ be a finite Galois extension.

Let $g$ be an element of $Gal(K/F)$

How do I compute the characteristic polynomial of $g$, where $g$ is considered as a $F$-linear map $K \rightarrow K$?

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What's wrong about the way you would compute the characteristic polynomial of any linear mapping from a finite-dimensional vector space to itself? Pick a basis (choice of basis is immaterial). Write the matrix of the said linear mapping w.r.t. that basis. Compute the characteristic polynomial of thas matrix. – Jyrki Lahtonen May 7 '14 at 6:10
This is a practice prelim problem; the answer should be something based on the cycle decomposition of g and in particular it should be the same for every cyclic extension. But knowing that the galois group is cyclic (for example) doesn't tell me how to find a nice choice of basis. – Dean Menezes May 7 '14 at 6:18
Ok. Sorry about being an ass. – Jyrki Lahtonen May 7 '14 at 7:52

## 2 Answers

A Galois extension $K/F$ with Galois group $g$ has a normal basis: that means, there is an $a\in K$ such that $B=\{g(a):g\in G\}$ is an $F$-basis of $K$. If $h$ is in $G$ then the matrix of $h$ with respect to the basis $B$ is easy to describe. Can you do that?

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We have $K\cong F[G]$ as an $F[G]$-module; this is the content of the normal basis theorem. So it suffices to consider the action of $g$ on the space $F[G]$. Observe that, with the basis $G$, $g$ essentially acts as a permutation matrix, associated to the permutation $g$ acts as on $G$ itself.

The characteristic polynomial of a permutation matrix depends only on the cycle structure of the permutation (which makes sense, since cycle types enumerate conjugacy classes and characteristic polynomials are conjugacy-invariant).

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