Finite fields, vector spaces, and linear dependence

[vorpal]

Hey all

Comments

[synchcola]

If we take b = a^(n(j-i) - i), the relation is Tr[a^(n(j-i))] = s Tr[a^((n+1)(j-i))].
Use that relation over and over to get Tr[a^(n(j-i))] = s^(-n). So Tr[(a^(j-i) - (1/s))^n] = 0.

But if the trace of every power is zero, that means that some power of it is zero... I think? Even for finite fields?
So (a^(j-i) - (1/s))^n = 0 for some integer n.

[synchcola]

Wait, Tr[1] is the dimension of the vector space, so... you better use some element b with nonzero trace.

Tr[b a^(n(j-i) - i) a^i] = Tr[b a^((n+1)(j-i) - i) a^j]
so
Tr[b a^(n(j-i))] = s^(-n) Tr[b], etc.