# Finding splitting field and Galois group questions

• PsychonautQQ
In summary, we discussed two questions regarding finding splitting fields and determining the Galois group of given polynomials. For the first question, we found that adjoing the 6th real root of 10 to Q(c) would give us a splitting field and that the Galois group would be isomorphic to (Z_6,+). For the second question, we divided the polynomial by a root and found that the splitting field would involve the root and its square root, resulting in a four-dimensional extension of F_3.
PsychonautQQ

## Homework Statement

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

## 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?

PsychonautQQ said:

## Homework Statement

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

## 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).
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]##.
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?
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.

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.

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

PsychonautQQ
Ahhh we are such genius's! lol

fresh_42

## What is a splitting field?

A splitting field is a field extension of a given field in which a polynomial can be completely factored into linear factors. It is the smallest field that contains all the roots of the polynomial.

## How do you find the splitting field of a polynomial?

To find the splitting field of a polynomial, you need to first factor the polynomial into irreducible factors. Then, you can construct a field extension by adjoining all the roots of the polynomial to the original field. This extension will be the splitting field of the polynomial.

## What is a Galois group?

A Galois group is a group of automorphisms of a field extension that leave the base field fixed. It describes the symmetry of the roots of a polynomial in the splitting field and provides information about the solvability of the polynomial by radicals.

## How do you find the Galois group of a polynomial?

To find the Galois group of a polynomial, you need to first find the splitting field of the polynomial. Then, you can identify the automorphisms that fix the base field and form a group. This group will be the Galois group of the polynomial.

## What can the Galois group tell us about a polynomial?

The Galois group can tell us whether a polynomial is solvable by radicals, which means it can be expressed in terms of the coefficients using only the basic arithmetic operations and taking roots. It also provides information about the number of distinct roots of the polynomial in the splitting field.

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