Is this field extension normal?

[PsychonautQQ]

Homework Statement

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

Homework Equations

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?

 

[fresh_42]

Homework Statement

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

Homework Equations

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]

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.

 

[fresh_42]

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.