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Math Help - group of invertible elements

  1. #1
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    group 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
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  2. #2
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    Re: group of invertible elements

    hint: it suffices to prove that the map r→r2 is injective (why?)

    since K is of characteristic 2, show r2 - s2 = (r - s)2.
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  3. #3
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    Re: 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?
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  4. #4
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    Re: group of invertible elements

    fix r in K*, if there is ONE solution to x2 - 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) = x2 + (a - a)x + -a2 = x2 - r.

    so the mapping a→a2 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)

    = a2b2.

    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:

    a2b2 = (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.
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  5. #5
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    Re: 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?
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  6. #6
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    Re: group of invertible elements

    yes.
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