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Field (Z_p +, *)

  1. Sep 5, 2011 #1
    1. The problem statement, all variables and given/known data

    I'm working on fields proof (intro level stuff) and I don't quite know how to interpret this field: (Zp, #, *) for prime p where [x]#[y]=[x+y] and [x]*[y]=[xy]

    For (Z3, #, *), it was the 3 element set {0, 1, 2} and for example, 2#2=1 because 4 mod 3 is 1, and 1#2=0 for the same reason.

    I'm supposed to show that this is a field but I can't figure out what the notation even means, what are the elements in the set, how do some of them operate on each other? That's what I need help with.

    2. Relevant equations

    A field means:

    Both operations are associative and commutative
    There is a identity element for #, and every element in the set has an inverse that gives #'s identity
    there is an identity element for * that is not the same as the identity for #, and every element except For the identity element for # has an inverse that gives the identity for *

    3. The attempt at a solution

    Zp ={2, 3, 5, 7, 11,....}? Then 13#2=15_p=? Also, I don't see any identity or inverses here.
     
  2. jcsd
  3. Sep 5, 2011 #2

    micromass

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    Hi Arcana!

    Do you know what cosets and quotient groups are?

    The set [itex]\mathbb{Z}_p[/itex] contains p elements, that is

    [tex]\mathbb{Z}_p=\{0,1,2,3,4,5,...,p-1\}[/tex]

    The rule is that p=0. So the numbers 0,1,2,...,p-1 are distinct but p=0.

    For example, in [itex]\mathbb{Z}_5[/itex], we have that 5=0. So 3+3=6=5+1=1 holds in [itex]\mathbb{Z}_5[/itex]

    Is that a bit clear??

    More rigourous perhaps. We define on the ring [itex]\mathbb{Z}[/itex] an equivalence relation as follows

    [itex]x\sim y ~\Leftrightarrow~p~\text{divides}~x-y[/itex]

    Then the quotient set w.r.t. that equivalence relation is [itex]\mathbb{Z}_p[/itex].

    Am I making sense??
     
  4. Sep 5, 2011 #3
    Z_p = {0,1,2,3,...,p-1}. The notation means that you are adding and multiplying mod p, just like you were doing in Z_3. The bracket notation comes into play because I guess you are really dealing with sets here. The elements of [x] are all integers that are congruent to x mod p. For example, with Z_3, [1] = {1, 4, 7, ...}. But don't let this give you troubles, you don't have to actually pay attention (much) to the fact that [1] is actually a set. As the two operations were defined, you can just operate as if they were integers mod p.


    Now, just check the field axioms.

    BTW, what book are you using?
     
  5. Sep 5, 2011 #4
    Thanks guys, I get it now.

    We haven't done rings yet micro, we're doing fields first because my prof hates having to rush through them at the end of the semester.


    We're using Durbin's Modern Algebra, but this was a handout. It seems like all my profs hate the books that the committees choose so they always disregard them and just give notes.
     
  6. Sep 5, 2011 #5
    What goes wrong if p is not prime?
     
  7. Sep 5, 2011 #6

    micromass

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    Well, nothing in the definition goes wrong. If p is not prime, then we can still define [itex]\mathbb{Z}_p[/itex]. But this will not be a field anymore.

    For example, in [itex]\mathbb{Z}_6[/itex], we will have 2*3=6=0. So in this case there are zero divisors. It turns out that [itex]\mathbb{Z}_p[/itex] always has zero divisors if p is nonprime.
     
  8. Sep 5, 2011 #7
    great, thanks! :)
     
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