Just for this post, I’ll specialize my definition of “character” to the case where my representation is 1-dimensional. This means that , so a the trace of each element of this map (i.e. the character afforded by this representation) is just a homomorphism to .
A famous theorem of Dedekind says that every list of distinct characters of a group, G, into some field, k, is linearly independent. I’m not going to prove this, since I proved something very similar when proving Hilbert’s Theorem 90 (in fact, I may have done it already). Anyway, it is just your standard induction and contradiction combination. There is a slick trick in there if you attempt to do it yourself, though, so beware.
I’d like to quickly prove a nice important and useful result that seems to escape the standard curriculum. It is called the Normal Basis Theorem. It basically says that any Galois field extension, if interpreted as a vector space, has as a basis the roots of a single irreducible polynomial over the base field. I’ll explain a little better when we actually get there.
First let’s set up a lemma: If is Galois with and are a basis for L over k, then is an L-basis for .
Proof: Let . So the lemma claims that . Suppose not. Then there exists some (the dual space) such that vanishes on some element of S.
Since this is finite dimensional there is some , so that we have the representation (the standard dot product). So if vanishes on S, then for all i. i.e. for all i, contradicting Dedekind, since some was assumed non-zero.
Alright. Time to prove the big theorem now. Every Galois extension has a normal basis.
Proof: Assume notation from the lemma. We want to construct a normal basis for . If , then are linearly dependent over k. So there are some not-all-zero constants such that . Now by applying we get that
Note that is an nxn matrix. And that . The sum above tells us that . We want to choose such that we force all to be zero. This will be the case if is invertible.
Let be a basis for L over k. So any . i.e. we have that
Now let . Let . Then by the Lemma, we have some constants in L such that .
But component-wise we see this is .
i.e. we have that which shows that it is not equivalently 0.
Suppose k is infinite. Then there are such that .
Thus are linearly independent and hence span L, so they are the normal basis.
Unfortunately, we made that assumption that k was infinite for a quick step. And in fact, there is no easy remedy to get the finite case. It is still true there, but we need to take an entirely new approach. I’m not sure if I’m going to do this next time.