Artin's Lemma: Let G be a finite group of automorphisms of a field E, and F=EG. Then [E:F] is finite and no larger than |G|.
Proof: Let n=|G|, and let m be any integer greater than n. We shall show that any m elements x1,x2,...,xm of E are linearly dependent over F. This will prove that [E:F] cannot exceed n.
Let V be the E-vector space of solutions (a1,a2,...,am) of the n simultaneous linear equations e(x1)a1+e(x2)a2+...+e(xm)am=0 for each element e of G. [Sorry, I must use "e" instead of Jacobson's Greek letter "eta", which I can't get in HTML.] Since m>n, this space V has positive dimension. We shall show that it contains a nonzero vector all of whose coordinates are in F. The e=1 equation will then yield the F-linear dependence a1x1+a2x2+...+amxm=0 on the xj.
In fact, we shall show that a minimal-weight nonzero vector in V is necessarily proportional to one with all coordinates in F. Here the weight of a vector (a1,a2,...,am) is its number of nonzero coordinates, m-#(j:aj=0). To do this, we shall use the fact that if (a1,a2,...,am) is in V then so is (e(a1),e(a2),...,e(am)) for each e in G. In fact, we show that in every nonzero E-vector subspace of Em that is invariant under G, a minimal-weight vector is proportional to one in Fm.
Suppose b=(b1,b2,...,bm) is a nonzero vector of minimal weight. Then some bj is nonzero. Without loss of generality, we may assume that bj=1 (replace b by b/bj). For each e in G, the vector b'=b-e(b) then has weight strictly less than the weight of b, because b' has zero coordinates wherever b does, and its j-th coordinate vanishes where that of b does not. Hence b' must be the zero vector. Thus bk=e(bk) for each k in [1,m] and each e in G. Therefore each bk is in EG=F, and we are done. QED
Once we find that in fact [E:F]=|G|, we also obtain "independence of characters": the elements of G are E-linearly independent as functions from E to E. To see this, suppose on the contrary that we had elements ce of E, not all zero, such that the sum of cee over e in G was the zero function on E. Again let F=EG, and set n=[E:F]=|G|. Let (x1,x2,...,xm) be a basis for E over F. Consider the square matrix M over E with n rows [e(x1),e(x2),...,e(xn)] (one row for each element e of G). Then cM=0 where c is the row vector (ce). Hence M is degenerate, and thus has a nontrivial column kernel. But then Artin gives us a nonzero element (a1,...,an) of the column kernel all of whose entries are in F. Taking e=1, we again obtain an F-linear dependence on our basis vectors xi, reaching the desired contradiction. QED