Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

This exercise is taken from Falko Lorenz's Algebra 5.2.

Why is $\sqrt[3]{2}$ not an element of $\mathbb{Q}(\sqrt[7]{5})$? Why is there no extension E of $\mathbb{R}$ such that $E:\mathbb{R}=3$.

My Progress: The first question I have answered as follows (of which I am not certain of my answer). Could anyone give me a hint on the second question?

Suppose $\sqrt[3]{2}$ is an element of $\mathbb{Q}(\sqrt[7] {5})$. Then $\mathbb{Q}(\sqrt[3]{2})\subset \mathbb{Q}(\sqrt[7]{5})$. This implies that $[\mathbb{Q}(\sqrt[7]{5}):\mathbb{Q}(\sqrt[3]{2})][\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=[\mathbb{Q}(\sqrt[7]{5}):\mathbb{Q}]$ But since $X^3-2$ is irreducible in $\mathbb{Q}[X]$ with root $\sqrt[3]{2}$, $[\mathbb{Q}(\sqrt[3]{2}):\mathbb{Q}]=3$ and similarly $X^7-5$ is irreducible in $\mathbb{Q}[X]$ with root $\sqrt[7]{5}$, $[\mathbb{Q}(\sqrt[7]{5}):\mathbb{Q}]=7$, contradiction.

share|improve this question
What part of your answer are you uncertain about? –  Chris Eagle Oct 24 '12 at 9:25
add comment

1 Answer

up vote 0 down vote accepted

The proof for the 1st question is Ok. Here is a hint for the 2nd question: Every cubic polynomial over $\mathbb{R}$ has a real root.

share|improve this answer
Ahh Ok. So Suppose E:R=3, then there must exist some minimal polynomial of degree 3 which has a zero in E. But then since every cubic polynomial over R has a real root, this implies E is in R. –  Chris Oct 24 '12 at 9:51
the end of your argument is not correct –  Martin Brandenburg Oct 24 '12 at 10:34
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.