Quartic polynomial

Sep 2012
764
53
British Columbia, Canada
Ok, I know I've been inactive for quite a while, so here's a little problem so that none of you forget me. :mrgreen:
Show that the polynomial \(\displaystyle x^4+2x^3-x-1\) is irreducible.
 

CRGreathouse

Forum Staff
Nov 2006
16,046
936
UTC -5
I suppose you won't let
Code:
polisirreducible(x^4+2*x^3-x-1)
count?

Let's see... here's one I don't think you'll have expected. First transform x -> x+1 to get x^4 + 6x^3 + 12x^2 + 9x + 1, then substitute x = 10 to get 17291. Note that this is prime, and that all coefficients are between 1 and 12. If they were between 0 and 9, Schur's theorem would tell us that the polynomial is irreducible. But recent work by Samuel Gross and Michael Filaseta shows that the same result holds with coefficients between 0 and 49598666989151226098104244512918. :D :D :D
 

mathbalarka

Math Team
Mar 2012
3,871
86
India, West Bengal
The question in the current form is incorrect. I can factorize that over \(\displaystyle \mathbb{F}_5\), for example, if you forgive me the cheek. :mrgreen:

PS : Ed, I was thinking that now might be the time to continue the Math Q&A. What do you think? And others?
 

agentredlum

Math Team
Jul 2011
3,372
234
North America, 42nd parallel
Here's an alternative approach.

By the Rational Root Theorem we know

\(\displaystyle P(x) \ = \ x^4 \ + \ 2x^3 \ - \ x \ - \ 1\)

Does not have any rational roots. This means P(x) cannot be reduced to anything that has a rational linear factor. There remains only one possibility to test , can it be reduced to the product of two quadratic factors with rational coefficients?

\(\displaystyle x^4 \ + \ 2x^3 \ - \ x \ - \ 1 \ = \ (x^2 \ + \ ax \ + \ 1 )(x^2 \ + \ bx \ - \ 1) \ =\)

\(\displaystyle x^4 \ + \ (b \ + \ a)x^3 \ + \ abx^2 \ + \ (b \ - \ a)x \ - \ 1\)

Itis clear that either a or b or both must be 0 otherwise we get a x^2 term which we don't want.

If both a , b = 0 will lead to

\(\displaystyle (x^2 \ + \ 1)(x^2 \ - \ 1)\)

contradicting RRT. WLOG let a = 0 , then we are forced to write

\(\displaystyle (x^2 \ + \ 1)(x^2 \ + \ bx \ - \ 1)\)

\(\displaystyle x^4 \ + \ bx^3 \ + \ bx \ - \ 1\)

Comparing to P(x) we get b = 2 and -1 , impossible.

Note: I'm not comfortable with this 'analysis'

:D
 

agentredlum

Math Team
Jul 2011
3,372
234
North America, 42nd parallel
mathbalarka said:
PS : Ed, I was thinking that now might be the time to continue the Math Q&A. What do you think? And others?
I agree.

:D
 

Hoempa

Math Team
Apr 2010
2,780
361
Yes, that would be nice. What sort of questions could we do?
 

mathbalarka

Math Team
Mar 2012
3,871
86
India, West Bengal
Hoempa said:
Yes, that would be nice. What sort of questions could we do?
Anything, more or less, that requires thinking more than throwing up previously digested results/tedious works. General form would be anything from logic (we should concentrate on this more than previous times) to combinatorics, from elementary number theory to solving solvable quintics. The question should be general and less advanced (for example, ask me what S x P1 is diffeomorphic to and you'll finding me learning differential topology throughout a month) and more than all one should think more in those.

I nominate Eddy as leader and set up a general guideline for posting problems and proposing solutions. PM all the ones who you think are interested in participating. Let them know through this post if they are interested by posting something like "I am in". Now, someone needs to call Eddy. Ed? Edward? Can you hear me?
 

mathbalarka

Math Team
Mar 2012
3,871
86
India, West Bengal
Hmm, in general I'd have said 'Patience, Daniel-San!' but this seems like a good idea. Let's use this thread as Area 51 and that as a Test Specimen for how well does a Q&A works out.
 

CRGreathouse

Forum Staff
Nov 2006
16,046
936
UTC -5
I figure we can just sort of fool around there until we come up with a real Q&A thread, however long that takes.