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 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.
 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-
May 15th, 2011, 06:59 AM   #3
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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?

 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 cant solve the quintic by origami?
May 15th, 2011, 09:06 AM   #5
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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 cant 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-

May 15th, 2011, 11:47 PM   #6
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Joined: Nov 2010

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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: $f/gcd(f',f)$), or is there
some other reason?

 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-
 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
 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-

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