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Finite Fields and ring homomorphisms HELP!

  1. Jan 12, 2010 #1
    1. The problem statement, all variables and given/known data

    Assuming the mapping Z --> F defined by n --> n * 1F = 1F + ... + 1F (n times) is a ring homomorphism, show that its kernel is of the form pZ, for some prime number p. Therefore infer that F contains a copy of the finite field Z/pZ.

    Also prove now that F is a finite dim vector space over Z/pZ; if this dim. is denoted d, then show that F has exactly p^d elements.

    I know that the kernel of a ring homomorphism is defined as ker(f) = {a in Z : f(a) = 0}
    but I am still having trouble exactly where to go from this...it appears that the only element of Z s.t. f(a) = 0, is 0 which would map to 0 * 1F = 0. But how is this of the form pZ, for some prime p??

    Any help or push in the right direction would be great...thanks.
     
  2. jcsd
  3. Jan 12, 2010 #2
    If [tex]R[/tex] and [tex]S[/tex] are rings, the kernel of a ring homomorphism [tex]\phi: R \to S[/tex] is an ideal of [tex]R[/tex]. What are the ideals of [tex]\mathbb{Z}[/tex]?

    Of those ideals, only some of them can be the kernel of a homomorphism [tex]\mu: \mathbb{Z} \to F[/tex] given by [tex]\mu(n) = n \cdot 1_F[/tex]. The others are incompatible with one of your hypotheses.
     
  4. Jan 13, 2010 #3
    Okay I get that the all of the ideals of Z are of the form mZ for some integer m, but im still not sure how n*1F implies that the kernel must be a prime ideal of Z?
     
  5. Jan 13, 2010 #4
    The important thing here is that [tex]F[/tex] is a field (or at least an integral domain). If the kernel of [tex]\mu[/tex] is [tex](m) = m\mathbb{Z}[/tex], and [tex]m = rs[/tex] is composite, think about the equation [tex]\mu(m) = \mu(r)\mu(s)[/tex].
     
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