1

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.

Chris Brooks
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spaceman_spiff
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  • Eisenstein's criterion. For b, its an 8th degree monic – Yunus Syed Apr 12 '17 at 17:51
  • @YunusSyed How do you apply Eisensteins criterion for $\mathbb{Q}(\sqrt[3]{7})$ ? , isnt it only for $\mathbb{Z}$ and there by for $\mathbb{Q}$ ? What are the necessary modifications ? – spaceman_spiff Apr 12 '17 at 17:57
  • For b): maybe you can try to prove that $x^8-7$ is an irreducible polynomial over $\mathbb{Q}(\sqrt[3]{7})$. – Fabian Apr 12 '17 at 18:04
  • @Fabian how thats the part I am not sure about , in the prev comment it is mentioned Eisensteins criterion, but how to do you apply it to this field – spaceman_spiff Apr 12 '17 at 18:07
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    Hint for (b): the degrees $[\mathbf Q(\sqrt[8]{7}): \mathbf Q]$ and $[\mathbf Q(\sqrt[3]{7}): \mathbf Q]$ are coprime so what can you say about $[\mathbf Q(\sqrt[8]{7}, \sqrt[3]{7}) : \mathbf Q(\sqrt[3]{7})]$? – Alex Macedo Apr 12 '17 at 18:19
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    Yes, your proof for part (a) is great, btw. – Greg Martin Apr 12 '17 at 18:41

1 Answers1

2

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 $\mathbf Q$.

Alex Macedo
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