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

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

where phi is Euler's totient funciton.

Is that right?

Now I just have to prove it...