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Math Help - Field arithmetic,

  1. #1
    Junior Member
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    Field arithmetic,

    Hi,

    I am reading a book about Reimann in which Field theory is introduced.
    I undertand the calculation of a clock arithmetic F_4 but I am struggling to understand the calculations for addition and multiplication of a proper field with 4 elements. (Power of a prime).

    Why is 1+3=2, in the clock arithmetic it is zero which I understand.
    Why is 3+3=0, in the clock it's 2 which makes sense.

    The multiplication table is equally confusing.

    Thanks
    Regards
    Craig.
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  2. #2
    Member alunw's Avatar
    Joined
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    It's not helpful for a book to denote elements of F4 by 0,1,2,3. Neither the addition nor multiplication in F4 are the same as arithmetic modulo 4, which does not give rise to a field because 2 is a zero divisor (2*2=0).
    The first two elements of F4 can be sensibly denoted by 0 and 1 since that is what they are. The other two elements are roots of some polynomial that does not have a root in F2, most likely x^2+x+1=0. Your book is unhelpfully denoting the two roots of this polynomial by 2 and 3. F4 has characteristic 2 so for any element x+x=0 hence the 3+3=0.
    Since we then have 3+0=3 3+3=0 and 3+2=1 (this is forced by the polynomial the two new elements are roots of) we must also have 3+1=2.
    The construction of F4 from F2 is a bit like that of the complex numbers where i is invented to make x^2+1=0 have a solution, but there is a huge difference: any polynomial has roots in C, so we don't need to add any more numbers once we have added i, but in F4 it is still easy to come up with polynomials that don't have roots. The same sort of thing happens with all the other finite fields as well.
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