Abstract Algebra: Extension Fields & Complex nth Roots

[Doom of Doom]

Two problems from my abstract algebra class...


1)
Let K be the algebraic closure of a field F and suppose E is a field such that  F F \subseteq E \subseteq K. Then is K the algebraic closure of E?

2)
Let n be a natural number with n\geq2, and suppose that \omega is a complex nth root on unity. Is there a formula for \left[\mathbb{Q}(\omega) : \mathbb{Q}\right] ?


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To 1), I must be missing something really silly, because it seems to me like it is obviously the case that K is also the algebraic closure of E, and that the proof should be easy. But I simply can't think of anything.


To 2) I would say no, but I am not exactly sure that I understand the question. For example, if n=8, then e^{i2\pi/8} is a complex 8th root of unity such that \left[\mathbb{Q}(e^{i2\pi/8}) : \mathbb{Q}\right]=4. However, i is also an 8th root of unity, but \left[\mathbb{Q}(i) : \mathbb{Q}\right]=2. Thus, for a given n, there is not necessarily a formula. Does this sound right?

 

[rasmhop]

1) Well for K to be the algebraic closure of E you must show:
a) K/E is algebraic.
b) If g(x) \in E[x], then g(x) splits completely in K. (HINT: Remember that K is an algebraic closure of itself so h(x) \in K[x] imply that h(x) splits completely in K).

2) Well \omega is a specific nth root of unity so it's acceptable for your formula to behave differently when given e^{i2\pi/8} and when given e^{i\pi/8}. You should probably look for a formula of the form:
\left[\mathbb{Q}\left(e^{ik\pi/n}\right) \, : \, \mathbb{Q} \right] = f(n,k)
so the formula can depend on both n and k, not just n.

 

[Doom of Doom]

Ok, for 2), I've got the formula

<br /> \left[\mathbb{Q}\left(e^{ik\pi/n}\right) \, : \, \mathbb{Q} \right] = \phi\left(\frac{n}{\gcd(n,k)}\right)<br />,

where phi is Euler's totient funciton.

Is that right?

Now I just have to prove it...