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Galois groups over the rationals 
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#1
Apr3007, 05:09 PM

P: 308

1. The problem statement, all variables and given/known data
Construct a polynomial of degree 7 with rational coefficients whose Galois group over Q is S7 2. Relevant equations I need an irreducible polynomial of degree 7 that has exactly 2 nonreal roots. 3. The attempt at a solution I have just been using trial and error, graphing polynomials of degree 7, making sure that eisenstein's criterion is in force for the irreducible part. I just change up the coefficients and hope the graph crosses the x axis 5 times. I'm not having any luck. Any hints or pointers on how to construct this polynomial? Thanks, CC 


#2
Apr3007, 06:08 PM

HW Helper
P: 2,567

It's not hard to find a polynomial of degree n with any n given numbers as roots, nor is it hard to find a real one of even degree with no real roots. I don't know how you think this will give you the answer though.



#3
Apr3007, 06:51 PM

Sci Advisor
P: 2,340

Hi, happyg1,
Let me go out on a limb here and try to guess the answer to the question implicit in the reply by StatusX. My guess is that you are trying to use Theorem 4.16 in Jacobson, Basic Algebra to find a polynomial over Q which has Galois group [itex]S_7[/itex], as a homework problem for your course in modern algebra. This theorem states that if f is a polynomial of degree p, where p is prime, which is irreducible over Q and has exactly two nonreal roots over C (aha!), then the Galois group of f is [itex]S_p[/itex]. So your problem is reduced to finding an irreducible polynomial of degree seven which has precisely five real roots. But this is fairly elementary given that trial and error is likely to succeed. On scrap paper, you can use analytic geometry to concoct a family of polynomials with five real roots (i.e. plot x^7 and add terms to pull the graph up/down in just the right way to give five roots), then you can probably use your favorite sufficient condition for irreducibility to find a rational coefficient polynomial in this family which is irreducible, and in your writeup you can simply present your polynomial, prove it is irreducible over Q, use Sturm's theorem to carefully verify that it has precisely five real roots, and then use Theorem 4.16 to show that it must have Galois group [itex]S_7[/itex]. (I think I see a clever approach which uses a "coincidence of small groups", but never mind.) "Trial and error": actually, in a sense most irreducible polynomials of degree p have Galois group [itex]S_p[/itex]. The hard part is finding ones with smaller Galois groups, since they get rather rare rather quickly! OTH, every finite solvable group (maybe even every finite group) arises as the Galois group of some polynomial over Q. These topics are discussed in various algebra books. 


#4
Apr3007, 09:51 PM

P: 308

Galois groups over the rationals
I'm working from herstein section 5.8, but it's the same thing.
I've resorted to Matlab and I think I might have one that works...just need some clarification on eisenstein's criterion. The prime number has to divide every coefficient except the highest degree one...and p^2 does not divide the constant? am I there? I got a big 'ol polynomial. I hope i'm right about this 'cause I'm TIRED. CC 


#5
Mar410, 07:31 AM

P: 4

yeh, you have eisenstein's criterion correct.



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