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?
Best Answer
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.