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Abstract algebra proof involving prime numbers

  1. Sep 10, 2010 #1
    The question states prove,
    If p is prime and p | a^n then p^n | a^n

    I am pretty sure I have i just may need someone to help clean it up.

    There are two relevant theorems i have for this.
    the first says p is prime if and if p has the property that if p | ab then p | a or p | b

    the second one is that if p is prime and p | a1a2a3.....an, then p must divide one of the a_i.

    so for the proof i am assuming p | a^n which i can rewrite as

    p | a*a*a.......an-1*an. so this is saying p(q) = a*a*a....an-1*an for some integer q.

    now if I look at p^n | a^n thats the same as

    p*p*p.....pn-1*pn | a*a*a.....an-1*an

    well p(p*p*p....pn-1*p) | a*a*a....an-1*an

    is that the way to go?
    Or maybe before when i had that p(q) = a*a*a....an-1*an for some integer q.

    just set q = p^n-1 so that p(q) = p^n.

    I feel like the later way should do it.
    Is this right?
  2. jcsd
  3. Sep 10, 2010 #2


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    It's actually hard to know whether you have it or not. Because that's pretty unreadable. If you know that "if p is prime and p | a1a2a3.....an then p must divide one of the a_i" and you apply that to p | a^n=a*a*a*...a (n times), what do you conclude about the divisibility of a by p? It's just the special case where all of the a_i are equal to a.
  4. Sep 10, 2010 #3
    ok so it is just a special case, so if p | a^n then the fact that p should divide one of the a_i means simply p|a, since every a_i is a.

    so from p |a^n implies p |a .

    so if p divides a we have that p(q) = a for some integer q.

    since we have an equation, i can raise both sides to the n power,
    so now i have p^n(q^n) = a^n which implies p^n | a^n.

    that looks like it should be good right?
    Last edited: Sep 10, 2010
  5. Sep 11, 2010 #4


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    Now that looks right.
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