hint: it suffices to prove that the map r→r^{2} is injective (why?)
since K is of characteristic 2, show r^{2} - s^{2} = (r - s)^{2}.
Hello!
Could anybody help me with the following problem?
Let K be a finite field and denote the (multipl.) group of invertible elements of K with K^*. We set G:={r^2:r is in K^*}.
How can one prove that G=K^* if K has characteristic 2?
Marco
fix r in K*, if there is ONE solution to x^{2} - r in K[x], call it a, then we have TWO (since K is a field there is at most two roots to a polynomial of degree 2), since -a is likewise a (distinct) solution (since char(K) ≠ 2, a ≠ -a) and:
(x - a)(x + a) = x^{2} + (a - a)x + -a^{2} = x^{2} - r.
so the mapping a→a^{2} sends both a and -a to the same image.
now, let's show that this map is a homomorphism:
(ab)^{2} = abab = a(ba)b = a(ab)b (fields have an abelian multiplicative group)
= a^{2}b^{2}.
and the kernel of this homomorphsim is {-1,1} (why?)
the question now is: is this homomorphism surjective onto the subgroup G of squares of K*? (the squares of K* IS a subgroup because K* is finite, and we have proven closure:
a^{2}b^{2} = (ab)^{2}).
clearly (?) it is, so the subgroup G of squares of K* is isomorphic to K*/{1,-1} which has order |K*|/2, so:
[K*:G] = |K*|/|G| = |K*|/(|K*|/2) = 2.