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

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- Nov 8th 2012, 03:31 AMMarcoRogroup of invertible elements
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 - Nov 8th 2012, 07:59 AMDevenoRe: group of invertible elements
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}. - Nov 8th 2012, 08:30 AMMarcoRoRe: group of invertible elements
Hi Deveno,

thank you for your reply, I think that I understood the idea.

In the same situation, if the characteristic is >2, then G seems to be a subgroup of K* with index 2. Could you also give a hint how to see this? - Nov 8th 2012, 08:58 AMDevenoRe: group of invertible elements
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. - Nov 8th 2012, 09:24 AMMarcoRoRe: group of invertible elements
What exactly is the grouphomomorphism a->a^2 you are applying the group isomorphy theorem to?

It is a map from K* to G, right? - Nov 8th 2012, 09:41 AMDevenoRe: group of invertible elements
yes.