Splitting Field Questions - Abstract Algebra Exam

[buzzmath]

I'm studying for my abstract algebra exam and I had a question and was wondering if anyone could help me out. If I'm given a polynomial is there a method to find the splitting field of the polynomial and the degree of that field? My book basically just gives the definition of a splitting field and there are really no examples. Thanks

 

[buzzmath]

I think you would first have to find all the zeros of the polynomial. But if the zeros aren't in the field that the polynomial is defined in is there any way to see these? I've seen an example of this using complex numbers, using Demoirves thm I think but I'm not really sure how that was done or if that is a standard way to find all the zeros. I think after you find all the zeros you can just take the basis of the field with the basis of the zeros and that is the splitting field? Are there exactly n zeros for an n degree polynomial? And then the degree of the splitting field is the dimension of this basis?
Thanks

 

[HallsofIvy]

Yes, you have to find all zeros of the polynomial. The field here is the set of rational numbers. Surely, you can determine whether a number is a rational number or not?

I don't know what you mean by "the basis of the field with the basis of the zeros". The splitting field is the smallest field containing all rational numbers and all zeros and the polynomial. It is true that such a field can be written as a vector space over the rational numbers- that may be the "basis" you are referring to. And yes, the dimension of that basis is the degree of the splitting field.

Each zero of the polynomial will be a "basis" element as long as it is "independent" of previously used zeros: as long as it cannot be written as a linear combination of those zeros.

For example, the zeros of the polynomial x²-2= 0 are \sqrt{2} and -\sqrt{2}. They are not 'independent' since one is -2 times the other. The smallest field containing all rational numbers and the zeros of x²-2 must contain any rational number a and \sqrt{2}. Since a field is closed under multiplication, it must also include numbers of the form a\sqrt{2}. Since a field is closed under addition, it must include numbers of the form a\sqrt{2}+ b. It's easy to show that sums and products of such things can be written in the same form. Of course, in a field, every non-zero number has an multiplicative inverse:
\frac{1}{a\sqrt{2}+b}= \frac{a\sqrt{2}}{2a^2- b^2}- \frac{b}{2a^2- b}
(and 2a²- b cannot be 0 because \sqrt{2} is not a rational number). That is, every number in that field can be written in the form a\sqrt{2}+ b(1), a vector space over the rational numbers with basis \{1, \sqrt{2}\} so the degree is 2.

 

[buzzmath]

Thanks, the concepts are a little clearer to me now. I am still have trouble with certain problems. For example how would I find the zeros of x^6 -1? I feel like I should know this but I don't know of a way to do it.

Also, my teacher ran through this problem really quickly in class. If p is prime, prove that the splitting field over F, the rational numbers, of the polynomial x^p - 1 is of degree p-1? he wrote x^p -1 =(x-1)(x^p-1 +...+1) since 1 is a rational we just need to find the splitting field of the other polynomial. so write a1, ...,a(p-1) as the roots of this polynomial. there are p-1 of them because the degree of the polynomial is p-1? [F(a1):F]= p-1 can we assume this? then x^p -1 =0 so x^p =1 so x = e^((i*2*k*pi)/p) where k = 0,...,p-1 how and why did you do this? then [F(a1, a2):F]=[F(a1,a2);F(a1)][F(a1):F] show this equals 1*(p-1). then to complete the proof show {a1,...,a(p-1)} is a cyclic group with generator a1. I have no idea how to do that but I don't need to know group theory. Although, I am curious. then the proof is done. why exactly?

Thanks for any help

 

[HallsofIvy]

x⁶-1= 0 means that x is a 6th root of unity. Those can be found by putting the number into "polar form". The roots are 1, \omega_6, \omega_6^2, \omega_6^3,\omega_6^4, and \omega_6^5.
Of course, \omega_6 is the "principal 6th root of unity":
e^{\frac{2\pi}{6}}= e^{\frac{\pi}{3}}= cos(\frac{\pi}{3})+ i sin(\frac{\pi}{3}).