Find the monic polynomial of $\sqrt[8]{7}$ over each of the following fields

Question

Find the monic polynomial of $\sqrt[8]{7}$ over each of the following fields

(a) $\mathbb{Q}(\sqrt{7})$

(b) $\mathbb{Q}(\sqrt[3]{7})$

My attempt:

(a) I can guess the answer to be $X^4 - \sqrt{7}$, but I am not sure how to prove it is irreducible over $\mathbb{Q}(\sqrt{7})$. I have tried to use the tower law as follows

$$[\mathbb{Q}(\sqrt[8]{7}): \mathbb{Q}] = [\mathbb{Q}(\sqrt[8]{7}): \mathbb{Q}(\sqrt{7})][\mathbb{Q}(\sqrt{7}): \mathbb{Q}]$$

i.e. $$8 = [\mathbb{Q}(\sqrt[8]{7}): \mathbb{Q}(\sqrt{7})]2$$

Is this the right way to prove the that $X^4 - \sqrt{7}$ is the required irreducible ?

(b) Here I have no clue and would like some hints to finish to the proof.

Answer 1

In general, if you have a candidate $f(x)$ for the minimal polynomial of $\sqrt[8]{7}$ over an extension $K$ of the rationals but you don't know how to check if $f(x)$ is really irreducible, you might want to consider the diagram

\begin{matrix} && K(\sqrt[8]{7})& \\ &\huge\diagup & & \huge\diagdown \\ \mathbf Q(\sqrt[8]{7})& & & & K\\ &\huge\diagdown & & \huge\diagup \\ &&\mathbf Q \end{matrix}

to compute $[K(\sqrt[8]{7}):K] = \deg f$.

Your solution for (a) does exactly that and is perfect. Can you do that for (b)?

You might want to show that:

Exercise: If $L/\mathbf Q$ and $K/\mathbf Q$ are extensions of degree $m$ and $n$ with $\gcd(m,n) = 1$, then the compositum $LK$ has degree $mn$ over $\mathbf Q$.