p^n - 1 divides p^m - 1 ==> n divides m
 

This question seems almost perfect for this forum. I guess there's some kind of trick that I can't think of at the moment.

P.S. I found the answer to my previous question, on cyclic groups.

Hmm... If I replace p with x and ask the question for polynomials over Q, the result becomes much easier. An interesting fact, but I don't see how it implies the result.

Thanks

[spoiler]Take the log[sub]p[/sub] and what do you get?[/spoiler]

[QUOTE=jinydu;135060]This question seems almost perfect for this forum. I guess there's some kind of trick that I can't think of at the moment.

P.S. I found the answer to my previous question, on cyclic groups.[/QUOTE]

Think "splitting fields"

[QUOTE=R.D. Silverman;135068]Think "splitting fields"[/QUOTE]

I think I know what you mean. The original problem was to show that if the field of order p^n is a subfield of the field of order p^m, then n divides m. I reduced it to a problem on divisibility of integers, thinking it would make the problem easier, or at least more familiar. Is that not the case?

[QUOTE=retina;135064][spoiler]Take the log[sub]p[/sub] and what do you get?[/spoiler][/QUOTE]

I have no idea what you have in mind. What's nice about [tex]\log_p(p^n-1)[/tex]?

[QUOTE=jinydu;135088]I have no idea what you have in mind. What's nice about [tex]\log_p(p^n-1)[/tex]?[/QUOTE]What do you get if you write the two numbers in base p? Sorry if my hints are too obtuse, I didn't think you wanted the answer given directly.

I've got the answer now (I had additional help); thanks.

[QUOTE=jinydu;135088]I think I know what you mean. The original problem was to show that if the field of order p^n is a subfield of the field of order p^m, then n divides m. I reduced it to a problem on divisibility of integers, thinking it would make the problem easier, or at least more familiar. Is that not the case?



I have no idea what you have in mind. What's nice about [tex]\log_p(p^n-1)[/tex]?[/QUOTE]

No.

Allow me to rephrase. What are the roots of a^n-1? a^m-1?
If a^n-1 divides a^m-1 then the roots of the former are also roots
of the latter...... 'nuff said?

This appears in some elementary number theory book -- Apostol's [I]Introduction to Analytic Number Theory[/I], I think. So there's an elementary solution, which I found without appealing to field extensions. I have my notebook of solutions somewhere.

EDIT: The actual exercise is "If a > 1, then (a^m - 1, a^n - 1) = a^(m,n) - 1" -- here (r,s) is the g.c.d. of r and s.

Well for that problem, you can use the fact that (a,b)=(a,a-b) and (a,b)=1 =>(a,bc)=(a,c) to get (WLOG m>=n)

(a^m-1,a^n-1)=(a^m-1,a^m-a^n)=(a^m-1,a^n(a^(m-n)-1))=(a^m-1,a^(m-n)-1) [as (a^m-1,a^n)=1)]

Then by the Euclidean algorithm you can reduce one of the terms to a^((m,n))-1. I believe it's clear that it's a divisor (you can explicitly write out the quotient), and since any divisor of (a,b) can't be greater than a or b, this must be the greatest common divisor.

[QUOTE=jinydu;135088]I think I know what you mean. The original problem was to show that if the field of order p^n is a subfield of the field of order p^m, then n divides m. I reduced it to a problem on divisibility of integers, thinking it would make the problem easier, or at least more familiar. Is that not the case?



I have no idea what you have in mind. What's nice about [tex]\log_p(p^n-1)[/tex]?[/QUOTE]

Suppose E is a subfield of F, with p^n and p^m elements, respectively.
Let d denote the dimension of F as a vector space over E, and let B be a basis.
There are (p^n)^d linear combinations of B, with coefficients in E,
hence p^m=p^(nd), m= nd.

This is a special case of the multiplicativity of field degrees
(m and n are the degrees of E and F over the prime field).

For the divisibility question, we are assuming p^m congr. 1 (mod p^n-1)
Writing m=qn+r, 0<=r<n, and using p^n congr. 1 repeatedly
we arrive at p^r congr. 1 (mod p^n-1), forcing r=0.

As I used to say in my lectures on Algebra: How do we prove divisibility? Perform the division!