Thread: Splitting field of x^p-x-a, a neq b^p-b

1. Splitting field of x^p-x-a, a neq b^p-b

Hi again guys. I have the following assignment:

Let k be a field of characteristic p and a in k an element which cannot be written b^p-b for any b in k. Compute the Galois group of the splitting field of f(x)=x^p-x-a.

In a previous thread here you helped me to determine that the above polynomial either splits in k or is irreducible there. It is shown here Math Forum - Ask Dr. Math how to do. The ansatz is simply what you guys taught me before:

Suppose [; \alpha ;] is a root, i.e. [; \alpha^p - \alpha - a = 0 ;]. Then so is [; \alpha+j ;], because [; (\alpha+j)^p-(\alpha+j)-a = \alpha^p - \alpha + j^p - j - a = \alpha^p - \alpha - a = 0 ;], hence yada yada.

My problem is with two things:
1) Alright, suppose [; \alpha ;] is a root... well, it's not. Wouldn't it follow from Fermat's little theorem that [; \alpha^p - \alpha - a = -a ;], which is not zero?

2) Fine, let's accept that [; \alpha ;] is indeed a root, that my objection in (1) is wrong. Then why is j^p-j = 0? Is it because j belongs to k while [; \alpha ;] does not?

(Bonus question: Determine all intermediate fields of [; Q(\sqrt{2+\sqrt{2}}) ;]. The intermediate fields correspond to the subgroups of the Galois group, in this case the cyclic group on four elements, so wouldn't there be only one intermediate field, corresponding to the (up to isomorphism) unique group of two elements, namely [; Q(\sqrt{2}) ;] ?)

2. Re: Splitting field of x^p-x-a, a neq b^p-b

Originally Posted by spudwish
My problem is with two things:
1) Alright, suppose [; \alpha ;] is a root... well, it's not. Wouldn't it follow from Fermat's little theorem that [; \alpha^p - \alpha - a = -a ;], which is not zero?
Fermat's little theorem only holds for integers. This root $\displaystyle \alpha$ is something that lives in the splitting field of $\displaystyle f(x)$, and is probably not an integer, or even an element of Z/pZ. However, since you know there exists a splitting field for every polynomial f(x), you can assume the existence of a root.

2) Fine, let's accept that [; \alpha ;] is indeed a root, that my objection in (1) is wrong. Then why is j^p-j = 0? Is it because j belongs to k while [; \alpha ;] does not?
In this instance j is necessary an integer (or more precisely, an element of Z/pZ). Fermat's little theorem holds in this case.

(Bonus question: Determine all intermediate fields of [; Q(\sqrt{2+\sqrt{2}}) ;]. The intermediate fields correspond to the subgroups of the Galois group, in this case the cyclic group on four elements, so wouldn't there be only one intermediate field, corresponding to the (up to isomorphism) unique group of two elements, namely [; Q(\sqrt{2}) ;] ?)
Yes.