Homework Help: Abstract algebra proof involving prime numbers

1. Sep 10, 2010

christinamora

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. Sep 10, 2010

Dick

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.

3. Sep 10, 2010

christinamora

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
4. Sep 11, 2010

Dick

Now that looks right.