20080603, 10:06  #1 
Dec 2003
Hopefully Near M48
1758_{10} Posts 
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. 
20080603, 10:35  #2 
Dec 2003
Hopefully Near M48
2×3×293 Posts 
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 
20080603, 10:59  #3 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
29·211 Posts 
Take the log[sub]p[/sub] and what do you get?

20080603, 11:53  #4 
Nov 2003
2^{2}·5·373 Posts 

20080603, 17:14  #5 
Dec 2003
Hopefully Near M48
2×3×293 Posts 
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 ? Last fiddled with by jinydu on 20080603 at 17:15 
20080603, 17:44  #6 
Undefined
"The unspeakable one"
Jun 2006
My evil lair
13747_{8} Posts 

20080604, 00:49  #7 
Dec 2003
Hopefully Near M48
2·3·293 Posts 
I've got the answer now (I had additional help); thanks.
Last fiddled with by jinydu on 20080604 at 00:49 
20080604, 00:56  #8  
Nov 2003
16444_{8} Posts 
Quote:
Allow me to rephrase. What are the roots of a^n1? a^m1? If a^n1 divides a^m1 then the roots of the former are also roots of the latter...... 'nuff said? 

20080726, 02:39  #9 
Oct 2006
vomit_frame_pointer
360_{10} Posts 
This appears in some elementary number theory book  Apostol's Introduction to Analytic Number Theory, 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. Last fiddled with by FactorEyes on 20080726 at 03:30 
20080729, 18:24  #10 
Aug 2002
Ann Arbor, MI
433 Posts 
Well for that problem, you can use the fact that (a,b)=(a,ab) and (a,b)=1 =>(a,bc)=(a,c) to get (WLOG m>=n)
(a^m1,a^n1)=(a^m1,a^ma^n)=(a^m1,a^n(a^(mn)1))=(a^m1,a^(mn)1) [as (a^m1,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. Last fiddled with by Kevin on 20080729 at 18:29 
20080806, 19:17  #11  
Oct 2007
linköping, sweden
2^{2}×5 Posts 
Quote:
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^n1) Writing m=qn+r, 0<=r<n, and using p^n congr. 1 repeatedly we arrive at p^r congr. 1 (mod p^n1), forcing r=0. As I used to say in my lectures on Algebra: How do we prove divisibility? Perform the division! 

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