Trace/Norm of Field Extension vs Trace/Determinant of Linear Operators

Question

Dummit and Foote (3rd ed, page 582-3) defines the norm and trace of an element of a field extension as follows:

Let $K/F$ be any finite field extension, and let $\alpha\in K$. Let $L$ be a Galois extension of $F$ containing $K$, and let $H$ be the subgroup of $Gal(L/F)$ whose fixed field is $K$. (I'm assuming that everything has to be separable).

Define the norm of $\alpha$ to be $\displaystyle\prod_{\sigma} \sigma(\alpha)$, where the product is taken over all embeddings $\sigma$ of $K$ into an algebraic closure of $F$. Equivalently, one $\sigma$ is taken from each coset of $H$ in $Gal(L/F)$. In particular, if $K$ is also Galois, the product runs over all $\sigma\in Gal(K/F)$.

The trace is defined similarly, but with a sum instead of a product.

Now, say that $\alpha$ has degree $d$ over $F$, while $K$ has degree $n$ over $F$. It follows from the Tower Law that $d$ divides $n$. I want to prove that there are $d$ distinct "Galois conjugates" (i.e., possible images of $\alpha$ under automorphisms) and that each is repeated $n/d$ times in the product/sum. Now, since everything is separable, the $d$ conjugates are clearly the $d$ distinct roots of $\alpha$'s minimal polynomial. But I'm not quite sure how to prove that each appears the same number of times. It seems plausible because of symmetry, but how do I actually prove it?

After this, I also want to prove that, if one represents the $F$-linear operator "multiplication by $\alpha$" with an $n\times n$ matrix, then the trace of the matrix equals the trace of $\alpha$, and the determinant of the matrix equals the norm of $\alpha$.

Now, I can show that the matrix has the same minimal polynomial as $\alpha$. And since $\lambda$ is a root of both minimal polynomials iff it's an eigenvalue of the matrix. Now the issue I get stuck on is multiplicity: I don't seem to know anything about the characteristic polynomial, so I don't know how to show that each root/eigenvalue has multiplicity $n/d$, as would be required for the norm to equal the product of the eigenvalues and for the trace to equal the sum of the eigenvalues. I know that the other invariant factors must divide the minimal polynomial and each other in order, but this is possible without all the eigenvalues having the same multiplicity, right?

Answer 1

OK ill have another go at it, hopefully I understand it better.

Consider the set of $\sigma(\alpha)$ for $\sigma \in Gal L/F$, then

$ \sigma(\alpha)=\tau(\alpha)$ gives $ \tau^{-1}\sigma(\alpha)=\alpha$ so

$\tau^{-1}\sigma \in Gal(L/F(\alpha))$

This implies that there are $d$ many distinct $\sigma(\alpha)$ each occurring $l/d$ many times. ($l$ being the degree of $L$ over $F$. Now to move down to $K$ consider what happens if $\sigma \restriction K=\tau \restriction K$.

then $\tau^{-1}\sigma \in Gal(L/K)$ and so there are $l/n$ of these so we have $\frac{l}{d}/\frac{l}{n}=n/d$ repetitions.

This implies that $$ Tr_{K/F}(\alpha)= \frac{n}{d} \sigma(Tr_{F(\alpha)/F}(\alpha))$$ Similar for product.

For the definition it terms of linear transforms, note that if we chose as basis for $F(\alpha)$, $\{1, \alpha, \alpha^2, \ldots, \alpha^{d-1}\}$ then the matrix representation of multiplication by $\alpha$ in this basis is the companion matrix of the minimal polynomial. The general case follows by extensions.

Answer 2

So for the first place I got stuck, showing that each root appears $n/d$ times, it seems that, if you let $H'$ be the subgroup that fixes $F(\alpha)$, then since there are $n/d$ cosets of $H$ in each coset of $H'$, and since each coset of $H'$ fixes a root, each root appears $n/d$ times, once for each coset of $H$.

For the matrix thing, if $k_1,\ldots,k_{n/d}$ is a basis of $K$ over $F(\alpha)$, then since $1, \alpha,\ldots, \alpha^{d-1}$ is a basis for $F(\alpha)$ over $F$, a basis for $K$ over $F$ is given by the products $k_i\alpha^j$, where $1\leq k\leq n/d$ and $0\leq j\leq d-1$. Then, the matrix of "multiplication by $\alpha$" with respect to this basis is the direct sum of $n/d$ copies of the $d\times d$ companion matrix for the minimal polynomial of $\alpha$, from which the trace and determinant are easily seen to be as desired.