# Is this field extension normal?

• PsychonautQQ
In summary: We don't say ##n## is primitive or not. A root of ##x^n-1## can be primitive or not: ##\exp(\frac{1}{n}\,\cdot\,2\pi\,i)^1## is, ##\exp(\frac{1}{n}\,\cdot\,2\pi\,i)^0## is not. ##c## is primitive simply means, that you can hop around the circle starting at ##c^0## by steps of ##\cdot c## and hit all other roots before you hit one twice. Like ##(c^0,c^3,c^2,c^1)## hops in
PsychonautQQ

## Homework Statement

Consider the field extension Q(c):Q, where c is a primitive nth root of unity. Is this extension normal?

## The Attempt at a Solution

I believe this polynomial splits in x^n - 1,where it's n roots are exactly the powers of c. Thus this extension is normal because Q(c) is a finite splitting field of the polynomial x^n-1 over Q. Is this correct?

PsychonautQQ said:

## Homework Statement

Consider the field extension Q(c):Q, where c is a primitive nth root of unity. Is this extension normal?

## The Attempt at a Solution

I believe this polynomial splits in x^n - 1,where it's n roots are exactly the powers of c. Thus this extension is normal because Q(c) is a finite splitting field of the polynomial x^n-1 over Q. Is this correct?
Yes. Only that ##x^n-1## is divided by the minimal polynomial of ##\mathbb{Q}(c)##. "splits in ##x^n-1##" doesn't make sense. And you get less than ##n## factors, since ##(x-1)## is one factor (eventually with others) which you don't need.
You already know, which powers of ##c## are primitive, so you have two ways to write down the minimal polynomial: either by all the factors that give you all the necessary roots in terms of powers of ##c##, or explicitly with the roots in ##\mathbb{C}##. Both ways show you the decomposition. The way using complex numbers has the advantage, that one easily sees, that and which powers of ##c## are involved.

PsychonautQQ
fresh_42 said:
Yes. Only that ##x^n-1## is divided by the minimal polynomial of ##\mathbb{Q}(c)##. "splits in ##x^n-1##" doesn't make sense. And you get less than ##n## factors, since ##(x-1)## is one factor (eventually with others) which you don't need.
You already know, which powers of ##c## are primitive, so you have two ways to write down the minimal polynomial: either by all the factors that give you all the necessary roots in terms of powers of ##c##, or explicitly with the roots in ##\mathbb{C}##. Both ways show you the decomposition. The way using complex numbers has the advantage, that one easily sees, that and which powers of ##c## are involved.
My bad, I should have said x^n - 1 splits in Q(c). x^n-1 is divided by the minimal polynomial of Q(c)? The minimal polynomial of Q(c) will not b the nth cyclotomic polynomial because n is not primitive, do I need to know more about it's minimal polynomial. But the nth cyclotomic polynomial will be a factor of the minimal polynomial in c because all of it's roots are powers of c.

PsychonautQQ said:
My bad, I should have said x^n - 1 splits in Q(c). x^n-1 is divided by the minimal polynomial of Q(c)? The minimal polynomial of Q(c) will not b the nth cyclotomic polynomial because n is not primitive, do I need to know more about it's minimal polynomial. But the nth cyclotomic polynomial will be a factor of the minimal polynomial in c because all of it's roots are powers of c.
Well the ##n-##th roots of unity are all complex numbers, because at latest ##x^n-1## splits in ##\mathbb{C}##.
You know, that they divide the unit circle in equidistant parts. Since ##1^n-1=0\, , \,(x-1)\,|\,(x^n-1)## and our first point is ##1##. The others are therefore ##\exp(\frac{k}{n}\,\cdot\,2\pi\,i)=\exp(\frac{1}{n}\,\cdot\,2\pi\,i)^k=c^k## with ##k=1,\ldots,n-1##. Split. You know from previous posts, that you need a primitive root ##c## as given to get all others as a power of it.

I'm not sure how you define the cyclotomic polynomial. In my books, it is the (irreducible) minimal polynomial (which of course has to divide ##x^n-1##).

We don't say ##n## is primitive or not. A root of ##x^n-1## can be primitive or not: ##\exp(\frac{1}{n}\,\cdot\,2\pi\,i)^1## is, ##\exp(\frac{1}{n}\,\cdot\,2\pi\,i)^0## is not. ##c## is primitive simply means, that you can hop around the circle starting at ##c^0## by steps of ##\cdot c## and hit all other roots before you hit one twice. Like ##(c^0,c^3,c^2,c^1)## hops in steps of ##c^3## and hits all of the ##4-##th roots of unity, whereas ##(c^0,c^2)## in steps of ##c^2## does not.

PsychonautQQ

## 1. What is a field extension?

A field extension is a mathematical concept in abstract algebra where one field is contained within another field. It extends the original field by adding new elements to it.

## 2. What does it mean for a field extension to be normal?

A field extension is considered normal if it satisfies a specific condition known as the normality condition. This condition ensures that the extension field is "nice" and behaves well under certain operations, such as taking roots of polynomials.

## 3. How can I determine if a field extension is normal?

There are several methods for determining if a field extension is normal. One way is to check if every irreducible polynomial in the original field has all of its roots in the extension field. Another way is to check if the extension field is the splitting field of a polynomial over the original field.

## 4. Why is it important for a field extension to be normal?

A normal field extension has many useful properties, such as being a Galois extension and having a well-defined Galois group. This makes it easier to study and understand the structure and behavior of the extension field. Additionally, normal field extensions play a crucial role in many areas of mathematics, including number theory and algebraic geometry.

## 5. What are some examples of normal field extensions?

Some examples of normal field extensions include finite extensions of the rational numbers, such as the complex numbers, as well as algebraic extensions of finite fields. Additionally, any extension of a perfect field is normal.

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