Special case of normal basis theorem - Artin 16.M.13

Question

The very last problem in Artin's Algebra, second edition, reads:

Let $K/F$ be a Galois extension with Galois group $G$. If we think of $K$ as an $F$-vector space, we obtain a representation of $G$ on $K$. Let $\chi$ denote the character of this representation. Show that if $F$ contains enough roots of unity, then $\chi$ is the character of the regular representation.

I know this is a corollary of the normal basis theorem, but finding a proof for the normal basis theorem is not that easy, and going this route seems to ignore the "enough roots of unity" condition. How can we use this condition to obtain a simpler proof?

Answer 1

As I commented, the proof of Normal Basis Theorem is not very hard. Please comment me any questions regarding the proof.

As for the exercise, the "enough roots of unity" condition simplifies the proof of this problem. In this proof, we avoid using the Normal Basis Theorem.

Let $n=|G|$ and assume that $F$ contains a primitive $n$-th root of unity (that should be a more precise statement about having enough roots of unity). Regarding $g\in G$ as a $F$-linear endomorphism on $K$. Since $g^n=1$, the minimal polynomial of $g$ divides $X^n-1 \in F[X]$. It is easy to see that if $g=1$ then $\chi(1)=|G|=n$.

So, we assume $g\neq 1$. Let $1<d|n$ be the order of $g$ in $G$. Then $g^d=1$ and any eigenvalue of $g$ must be a $d$-th root of unity. Since distinct field homomorphisms $1,g,\ldots, g^{d-1}$ are linearly independent over $F$, it follows that $X^d-1\in F[X]$ is the minimal polynomial of $g$. Since $F$ contains $n$-th roots of unity, we may assume that $g$ is diagonalizable over $F$. Moreover, the diagonal matrix $D_g$ corresponding to $g$ has all $d$-th roots of unity appearing on its diagonal entries.

By Galois theory, the field $K^g=\{x\in K| gx=x\}$ is a subfield of $K$ with extension degree $n/d$ over $F$. Then there is a basis $\{y_1,\ldots, y_{n/d}\}$ of $K^g$ over $F$. For each $d$-th root of unity $\zeta$, take a eigenvector $x\in K$ of $g$ so that $g x= \zeta x$. Then $\{xy_1, \ldots, xy_{n/d}\}$ forms a set of $n/d$ linearly independent eigenvectors. Thus, each $d$-th root of unity appears on exactly $n/d$ diagonal entries of $D_g$, and it follows that $(X^d-1)^{n/d}$ is the characteristic polynomial of $g$. Since the coefficient of $X^{n-1}$ in $(X^d-1)^{n/d}$ is zero, we have $\mathrm{tr}(g)=0$.

Hence, the character of the representation is the regular representation.

Answer 2

Here is a different approach which also avoids the normal basis theorem. The idea is that we can compute the trace of a linear map after extending scalars to any field extension of the ground field. So, we extend scalars to $K$ itself; Galois theory produces an isomorphism

$$K \otimes_F K \cong \prod_{g \in G} K$$

where the map from the LHS to the RHS is given by sending $x \otimes y$ to the function $g \mapsto g(x) y$. With this choice of map, the Galois action of $G$ on the left copy of $K$ is isomorphic to the regular representation $K[G]$, so it has the same character as the regular representation after extension of scalars to $K$, and hence also before extension of scalars.

This isomorphism can even be used to prove the normal basis theorem itself although I haven't seen this argument anywhere.