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Finite fields

  1. Feb 26, 2005 #1
    let F= { 0=a1, a2,a3,a4.....an} be a finite field. show that
    (1+a2)(1+a3).........(1+an) = 0.

    when n is odd, it's simple since 1 belongs to F. then odd number of elements are left( they're distinct from 0 and 1). at least one of them, say x,must have itself as its inverse. x^2 = 1 implies x=-1 ( since x != 1)...so the result is true.

    for n even and the case where none of them are inverses of themselves, can somebody suggest a solution???
     
  2. jcsd
  3. Feb 26, 2005 #2

    matt grime

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    One of the a_r must be 1, which is distinct from 0, so one of the terms is 2, which is zero in char 2, and wlog a_2 = -1 in any other field.

    It's a field. If elements multiply together to give 0 one of them must be zero, ie, 1+a_2=0 after relabelling. a_2= -1, the inverse of 1 which isn't zero and hence must be in the list you gave.
     
  4. Feb 26, 2005 #3
     
  5. Feb 26, 2005 #4

    matt grime

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    You get that the product is zero if and only if one of the factors is zero, since it is a field?

    So it suffices to show that one of the 1+a's is zero.

    But this is true since 1 has an additive inverse, -1 (which equals 1 in a field of char 2)
     
  6. Feb 27, 2005 #5
    thanks...that makes it clear...
    but here's another question :redface: does this mean i could have done this problem without considering separate cases???
     
  7. Feb 27, 2005 #6

    matt grime

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    Yes, absolutely. In any field there is a non-zero element, x, such that 1+x=0. The result is also true in any finite ring with unit, any finite domain too.
     
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