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Galois fields

  1. Mar 1, 2008 #1
    [SOLVED] galois fields

    1. The problem statement, all variables and given/known data
    Let [itex]\bar{\mathbb{Z}}_p[/itex] be the algebraic closure of [itex]\mathbb{Z}_p[/itex] and let K be the subset of [itex]\bar{\mathbb{Z}}_p[/itex] consisting of all of the zeros of [itex]x^{p^n} - x[/itex] in [itex]\bar{\mathbb{Z}}_p[/itex]. My book proved that the subset K is actually a subfield of [itex]\bar{\mathbb{Z}}_p[/itex] and that it contains p^n elements. Then, out of nowhere, it said that K contains Z_p and provided absolutely no justification for that claim. Can someone fill me in on why that is so obvious?

    2. Relevant equations

    3. The attempt at a solution
  2. jcsd
  3. Mar 1, 2008 #2


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    Because it's a field, it's going to contain 1, and as a result it will also contain 1+1, 1+1+1, ...
  4. Mar 1, 2008 #3


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    Maybe a few general words are in order here, because you seem to be having trouble with things of this sort.

    If F is a field, what is the most basic thing we can say about its elements? We know it contains 0 and 1. Consequently, it's going to contain 1+1, 1+1+1, and so on. If this process stops generating new elements at some point, then we can find two distinct positive integers n and m such that 1+1+...+1 (n times) = 1+1+...+1 (m times). From this we conclude that there is a smallest positive integer k such that 1+1+...+1 (k times) = 0 (for instance, if n>m, then 1+1+...+1 (n-m times) will give us zero). This implies that a copy of Z/kZ sits inside of F. In this case we say that F has characteristic k. On the other hand, if the process keeps generating new elements, then F is going to contain a copy of the nonnegative integers, and hence all the integers (if n is in F, then -n had better be in F too). In this case we say that F has characteristic zero.

    So far we have only used the group structure of F, and not the field structure. Let's amend this now. If F contains a copy of Z, then for each n in F, 1/n is also in F. This implies that F contains a copy of Q (because m/n = 1/n + ... + 1/n (m times)). This is the most basic field of characteristic zero, and any field of characteristic zero is going to contain a copy of Q. Conversely, no field having characteristic other than zero can contain a copy of Q.

    For the other case, when Z/kZ sits in F, we must have that k is a prime, for otherwise the field F contains zero divisors. This is because if p is a prime divisor of k and k!=p, then in Z/kZ, k/p and p are nonzero, and (k/p)*p=k=0. Conversely, it's easy to see that Z/pZ is a field for every prime p. Thus we conclude that a field can only have prime nonzero characteristic, and this happens iff F contains a copy of Z/pZ.

    That's the long-winded version. The compact way to see all this is to consider the ring homomorphism f:Z->F given by f(n)=1+...+1 (n times). By the first isomorphism theorem, Z/kerf is going to be isomorphic to a subring of F, and thus must be an integral domain (because F is a field). This implies that kerf is a prime ideal of Z, and hence is either 0 or equal to pZ for some prime p. In the first case F will contain a copy of Z and hence of Q, and in the second case F will contain a copy of Z/pZ.

    Hopefully this clears things up for you a bit.
    Last edited: Mar 1, 2008
  5. Mar 1, 2008 #4
    I see, thanks.
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