Splitting field, polynomials

Question

Let $F\subset \mathbb C$ be the splitting field of $x^7-2$ over $\mathbb Q$ and let $z=e^{2\pi i/7}$ be a primitive seventh root of unity. What would be $[F:\mathbb Q(z)]$ and $[F:\mathbb Q(2^{1/7})]$? I believe $[F:\mathbb Q(z)]>[F:\mathbb Q(2^{1/7})]$. Am I right? Any help would be appreciated.

Answer 1

You can check easily $[F(u_1, u_2, ... , u_n):F]= F[(u_1,u_2,...u_n):F(u_1,...,u_{n-1})]...[F(u_1):F]$.

As boris pointed out in his answer, since $[\mathbb{Q}[2^\frac{1}{7}]:\mathbb{Q}] = 7, \ \ [\mathbb{Q}[z]:\mathbb{Q}] = 6 $ and their greatest common divisor is 1, $[\mathbb{Q}[2^\frac{1}{7}]:\mathbb{Q}] = 6*7 = 42$.

(This is an exercise that we can find in any text.

Precisely, If $u,v \in K$ are algebraic over $F$, with $[F(u):F]=m$ and $[F(v):F]=n$ and their GCD is 1, then $[F(u,v):F]=mn$.)

Thus your conclusion is correct.

Answer 2

We have $F = \mathbb{Q}[z, 2^{\frac{1}{7}}]$.

$[\mathbb{Q}[2^\frac{1}{7}]:\mathbb{Q}] = 7, \ \ [\mathbb{Q}[z]:\mathbb{Q}] = 6 $ and so $[F:\mathbb{Q}] = 6*7 = 42$.

Then you can calculate those quantities.