The Two Artin-Schreier Theorems

Proof. We prove by the following steps: y 1 Start with induction on n, for n = 1 the result is trivial. y 2 Let g be such that σ1(g) ̸= σ2(g) and consider ∑ aiσi(gx) = 0 and ∑ aiσ1(g)σi(x) = 0 y 3 Cancel one summand by showing a2 = 0 and eventually show all ai = 0 One can prove by induction. Let a1σ1 ≡ 0 then since σ1 does not map to 0 ∈ K one must have a1 = 0. Suppose now that for any linear combination of n − 1 characters (n ∈ N) one has linear independence as described in the Theorem. Suppose now that ∑n i=1 aiσi ≡ 0 for some ai ∈ K and characters σi : G → K∗ with i = 1, . . . , n. Since the characters are mutually different there is a g ∈ G such that σ1(g) ̸= σ2(g). For all x ∈ G, one has σ1(g) ∑n i=1 aiσi(x) = 0 and ∑n i=1 aiσi(g)σi(x) = 0. Subtracting the two gives