# Finding splitting field and Galois group questions

1. Dec 5, 2016

### PsychonautQQ

1. The problem statement, all variables and given/known data
2 questions here:

1)
Let g(x) = x^6 - 10 be a polynomial in Q(c) where c is a primitive 6th root of unity. Find a splitting field for this polynomial and determine it's Galois group

2)
let f(x) = x^3 + x^2 + 2 with coefficients in $F_3$. Find a splitting field K for this polynomial and determine it's Galois group

2. Relevant equations

3. The attempt at a solution
Question 1)
If we adjoin b = the 6th real root of 10 to Q(c) then we have a splitting field. Furthermore, the roots of f will be {b, bc, bc^2, bc^3, bc^4, bc^5}. The Galois group will permute the roots whilst fixing every element of Q(c). So I believe this galois group will be isomorphic to $(Z_6,+)$ where the generator will take b-->bc and fix all elements of Q(c).

Question 2)
I'm having trouble constructing a splitting field for this polynomial. If I assume r is a root and then divide f(x) by x-r I don't seem to arrive at anything helpful. Perhaps I should consider that the multiplicative group in F_7(r) will be cyclic of order 26 (since the degree of r is 3 and 3*9 = 27) so r^27=r, and then use this knowledge to see if any powers of r are also roots. Does this make sense? Does anyone else have any other insights to make this simpler?

2. Dec 7, 2016

### Staff: Mentor

I think this is correct. Only one remark: you can say $g(x)$ is a polynomial over $\mathbb{Q}(c)$ or in $\mathbb{Q}(c)[x]$.
I don't understand why you consider $\mathbb{F}_7$ here or even $\mathbb{Z}_{26}$.
If you divided $f(x)=x^3+x^2+2$ by $(x-r)$ with a root $r$ such that $r^3+r^2+2=0$ then you can also do the rest and split $f(x)$ completely. If I made no mistake, then you will need $\sqrt{r+1}$ in the splitting field. Together with $r^2$ this looks like a four-dimensional extension of $\mathbb{F}_3$ but I didn't check linear dependencies.

3. Dec 8, 2016

### PsychonautQQ

I divided by x-r and then solved the resulting quadratic and calculated that it has a discriminant -3r^2 - 2r + 1. it appears our splitting fields will be a bit different.

4. Dec 8, 2016

### Staff: Mentor

I don't think so. $-3=0$ and $-2=1$ in $\mathbb{F}_3$.

5. Dec 8, 2016

### PsychonautQQ

Ahhh we are such genius's! lol