My Math Forum  

Go Back   My Math Forum > College Math Forum > Abstract Algebra

Abstract Algebra Abstract Algebra Math Forum


Reply
 
LinkBack Thread Tools Display Modes
May 13th, 2011, 08:42 AM   #1
Member
 
Joined: Nov 2010

Posts: 53
Thanks: 0

Origami trouble

I am working with a proof (which I think is quite easy), but still I am a bit stuck
I want to show that one can use origami to solve polynomials of degree ? 4. We have the following thm:
Thm: Let ??C be algebraic over Q and let Q?L be a splittingfield of the minimalpolynomial of ? over Q. Then ? is an origami number (which means it is solvable by origami)
if and only if [L:Q]=(2^a)*(3^b), where a og b are integers bigger then 0.

As I understand it, the solution should present itself when I look at possible splitting fields. We know that there is a upper bound to the degree of the extensions:
Thm: Let f?F[x] be a polynomial of degree n>0, and let L be a splittingfield of f over F. Then [L:F]?n!

Which means that we have the upper bound [L:Q]?3! for cuberoots and [L:Q]?4! for quartroots. But I cannot see if this really helps us or not... I would really appreciate any help with this.
studentx is offline  
 
May 13th, 2011, 12:12 PM   #2
Senior Member
 
Joined: Jun 2010

Posts: 618
Thanks: 0

Re: Origami trouble

studentx,

I am a bit confused by your question, maybe I am not understanding it correctly. For if we have a general quartic polynomial (over the rationals), then the degree of the splitting field (degree of the extension) must divide 24, since the Galois group is a subgroup of the symmetric group on four letters, which has order 4! = 24. Now, if the degree does divide 24, then it must be of the form 2?¹3?². A similar analysis holds for cubics and quadratics.

-Ormkärr-
 is offline  
May 15th, 2011, 06:59 AM   #3
Member
 
Joined: Nov 2010

Posts: 53
Thanks: 0

Re: Origami trouble

Hi Ormkärr

Thanks for the help. I really like your approach. But one thing still confuse me. you said:
Quote:
if we have a general quartic polynomial...
, must not
the polynomial also be separable?
studentx is offline  
May 15th, 2011, 07:29 AM   #4
Member
 
Joined: Nov 2010

Posts: 53
Thanks: 0

Re: Origami trouble

By the way, a bit of toppic. It suddenly hit me that you cannot write 5! on the form 2?¹3?². Could you use this as an argument for the insolvability of the Quintic, or does it just
mean that you can`t solve the quintic by origami?
studentx is offline  
May 15th, 2011, 09:06 AM   #5
Senior Member
 
Joined: Jun 2010

Posts: 618
Thanks: 0

Re: Origami trouble

studentx,

I am glad that you are asking questions.

Quote:
Originally Posted by studentx
But one thing still confuse me. you said:
Quote:
if we have a general quartic polynomial...
, must not
the polynomial also be separable?
Technically, you are correct, but over the rationals it is not a problem.

Quote:
Originally Posted by studentx
By the way, a bit of toppic. It suddenly hit me that you cannot write 5! on the form 2?¹3?². Could you use this as an argument for the insolvability of the Quintic, or does it just
mean that you can`t solve the quintic by origami?
I wish! No, I think it just means you can't solve it by origami. The insolvability of quintics depends on much more subtle properties.

-Ormkärr-
 is offline  
May 15th, 2011, 11:47 PM   #6
Member
 
Joined: Nov 2010

Posts: 53
Thanks: 0

Re: Origami trouble

Hi Ormkärr

When you say
Quote:
over the rationals it is not a problem
, is this not a problem because we can get rid of multiple roots in fields with char 0 (f.ex: ), or is there
some other reason?
studentx is offline  
May 16th, 2011, 08:48 AM   #7
Senior Member
 
Joined: Jun 2010

Posts: 618
Thanks: 0

Re: Origami trouble

Yes, that is the reason. Over a field of characteristic zero, every irreducible polynomial is separable. Looking back over the thread, I realize that I have imposed the condition F = ? myself. I hope this is not a problem.

-Ormkärr-
 is offline  
May 18th, 2011, 02:41 AM   #8
Member
 
Joined: Nov 2010

Posts: 53
Thanks: 0

Re: Origami trouble

Hi!

F = Q is not a problem. So you can safely assume/impose that
studentx is offline  
May 18th, 2011, 05:18 AM   #9
Senior Member
 
Joined: Jun 2010

Posts: 618
Thanks: 0

Re: Origami trouble

studentx,

I just wanted to state, in case it wasn't clear, that even if f, of degree n, has repeated roots, it still has Galois group contained in Sn, although now the assertions of the Fundamental Theorem are no longer precise.

-Ormkärr-
 is offline  
Reply

  My Math Forum > College Math Forum > Abstract Algebra

Tags
origami, trouble



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Origami constructions studentx Abstract Algebra 6 June 14th, 2014 12:57 AM
Trouble w/ P.M.F. WWRtelescoping Algebra 2 March 2nd, 2014 12:49 AM
Having trouble converting units/trouble understanding Adrian Algebra 4 February 5th, 2012 02:00 PM
having trouble jordanshaw Calculus 8 October 13th, 2010 05:29 PM
Having Trouble with the Problem!!?? flxjones Calculus 1 November 15th, 2009 09:11 AM





Copyright © 2018 My Math Forum. All rights reserved.