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If $L/K$ and $F/L$ are Galois extensions inside $\mathbb{C}$, must $F/K$ be a Galois extension?

Spook
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2 Answers2

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Take the extension $$\mathbb{Q}\subset\mathbb{Q}(\sqrt{2}))\subset\mathbb{Q}(\sqrt[4]{2})$$ Then each of the intermediate steps are Galois as they are of degree two, but the total degree 4 extension $\mathbb{Q}\subset\mathbb{Q}(\sqrt[4]{2})$ is not Galois as some of the roots of the minimal polynomial of $\sqrt[4]{2}$ over $\mathbb{Q}$ is not in $\mathbb{Q}(\sqrt[4]{2})$.

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Consider the extension $\mathbb Q\subset\mathbb Q(\sqrt[4]{2})\subset \mathbb Q(\sqrt[4]{2},i) $. You have that $\mathbb Q(\sqrt[4]{2})/\mathbb Q$ is not Galois since it is not normal. Yu have to enlarge $\mathbb Q(\sqrt[4]{2})$ over $\mathbb Q$ in order to get Galois extension.

Reader
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  • But the question is: If $L/K$ and $F/L$ are Galois extensions inside $\mathbb{C}$, must $F/K$ be a Galois extension? Rather than whether a Galois extension can live inside a bigger Galois extension. – Spook Oct 29 '12 at 23:11