Some true or false Field extension theory questions

[PsychonautQQ]

Homework Statement

Let S= {e^2*i*pi/n for all n in the natural numbers} and let F=Q

Is F:Q
1) algebraic?
2) finite?
3) simple?
4)separable?

Homework Equations

The Attempt at a Solution

1) Every element in S is a root of x^n-1 and every element of a in Q is a root of x-a, and thus I think somehow that means that the whole extension is algebraic is (i.e all the basis elements are algebraic and obviously all the elements of the base field are algebraic). Does this argument work? If not what would?

2) No. Every p in the natural numbers will be a basis that is linearly independent of all the elements before it, and there are infinite primes, so this will not be a finite extension.

3) I'm not sure. I want to say there was some theorem that algebraic extensions can ultimately be seen as simple but I'm not sure.

4) Yes, this extension is over Q where char(Q) = 0 so the extension is separable.

 

[fresh_42]

Homework Statement

Let S= {e^2*i*pi/n for all n in the natural numbers} and let F=Q

Is F:Q

I guess you mean $F=\mathbb{Q}(S)$.

1) algebraic?
2) finite?
3) simple?
4)separable?

Homework Equations

The Attempt at a Solution

1) Every element in S is a root of x^n-1 and every element of a in Q is a root of x-a, and thus I think somehow that means that the whole extension is algebraic is (i.e all the basis elements are algebraic and obviously all the elements of the base field are algebraic). Does this argument work? If not what would?

Yes. Although you don't need to mention the elements of $\mathbb{Q}$ itself. E.g. all real numbers are trivially algebraic over $\mathbb{R}$, even if not over $\mathbb{Q}$. Algebraic over itself is trivial, but your argument is correct.

2) No. Every p in the natural numbers will be a basis that is linearly independent of all the elements before it, and there are infinite primes, so this will not be a finite extension.

Yes. I don't think they are already a basis, because their powers until $p-1$ need to be included as well. However, they are linear independent and this is sufficient here.

3) I'm not sure. I want to say there was some theorem that algebraic extensions can ultimately be seen as simple but I'm not sure.

How could this be done with infinitely many elements in $S$? Read the theorem again, I'm sure it states something like: Every finite algebraic extension ... As far as I remember, the argument here is to simply multiply all minimal polynomials.

4) Yes, this extension is over Q where char(Q) = 0 so the extension is separable.

Not 100% sure whether separability applies to infinite extensions, too, but you're right. No multiple roots.