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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.

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What part of your answer are you uncertain about? – Chris Eagle Oct 24 '12 at 9:25
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.

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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

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