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Thread: finite fields of elements

  1. #1
    Mar 2009

    finite fields of elements

    Show that there are no finite fields of 10 elements.

    Any help with this problem would be appreciated!
    Last edited by mr fantastic; Apr 30th 2009 at 01:52 PM. Reason: restored question and closed thread.
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  2. #2
    Super Member Gamma's Avatar
    Dec 2008
    Iowa City, IA

    order 10

    first of all, all finite fields have order a power of a prime, $\displaystyle p^n$ with p prime and $\displaystyle n\in \mathbb{N}$. $\displaystyle 10=2*5$ so it is not a finite field.

    Alternatively, a field must be an abelian group under addition. There are 2 groups of order 10, and only one is an abelian group of order 10, the cyclic one. You can check with Sylow theorem or take my word for it. (If you are curious, the non abelian group of order 10 is the dihedral one). This is clearly not a field as $\displaystyle 2*5=10 \equiv 0$ and fields do not have zero divisors
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  3. #3
    Apr 2009
    I'd like to add a few remarks to Gamma's.

    Here is how it goes:
    Every field has at least two elements, the neutral elements with
    respect to addition (called zero or 0) and the neutral element with respect to
    multiplication (called one or 1).

    Now, take 1, and add it to itself (i.e., 1+1).
    If what you get is zero, then the field is said to have characteristic two
    (since you added one twice).
    If not, add another one. If that's zero, characteristic is three.
    In any case, since there are only finitely many elements in the field,
    and since you cannot have repeats in the sequence that you create
    unless you repeat with 0, you must eventually have that after
    p times (p an integer less or equal to the size of the field)
    you get 0. Then you can show that p is prime.
    If not, you would have p = a * b and hence a*b = p = 0
    but a, b not zero, i.e., a and b are zero divisors.

    So, now we know that finite fields have characteristic p a prime.
    (In fact, the integers mod p are fields). Now if the field isn't equal to
    the set Z_p = {0,1,1+1,1+1+1, ...,1+1+...(p-1) times}
    then at least we know that Z_p is
    a subfield. A subfield is just a subset that is in itself a field.
    So, every finite field has a Z_p as a subfield.

    Now, use linear algebra to verify that the field is a vector space over
    its subfield Z_p (use the field axioms to verify the vector space axioms).
    But if its a vector space over Z_p, it must be finite dimensional
    (after all, it is a finite field), and hence there is a basis of the field
    over the subfield Z_p, say b_1,b_2,...,b_h.
    That means that everything in the field is a linear combination
    of the b_i's with coefficients in Z_p.
    This gives p^h field elements.
    One can show (using an irreducible polynomial of degree h) that fields
    of order p^h exist (in fact, unique up to isomorphism).

    So, long story, in the end since 10 is not a prime power, there is no
    finite field of order 10.


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