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Let $F \subset E$ be a Galois extension. Let $F \subset G \subset E$. What can we say about 'being Galois' for extensions $F \subset G$ and $G \subset E$?

I think for $G \subset E$ the answer is that it is Galois. To show Galois I need to show normal and separable. I have a theorem that says it will be normal so all thats left is to show it is separable. $F \subset E$ being separable means the irreducible polynomial of any element $u \in E$ over has no multiple roots in any further extension of $E$, but I'm not sure how this means any polynomial over $G$ would also have no multiple roots.

For $F \subset G$ I'm not even sure what the answer is. I thought no, but I see something later in my notes that makes me think it is. I don't know how to show normal, but for separability I have: The extension of $E$ over $F$ being separable means the irreducible polynomial of any element $u \in E$ over has no multiple roots in any further extension of $E$. If x is a root in $G$ then x will also be a root in an extension of $G$ so since there are no multiple roots in $E$ there are none in $G$ so it is separable.

rae306
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Vinny Chase
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    $G\subset E$ is indeed always Galois. $F\subset G$ needn't be; that's actually equivalent to the Galois group $Gal(E/G)$ being normal in $Gal(E/F)$. – Wojowu May 03 '18 at 20:37

2 Answers2

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See the fundamental theorem of Galois theory. As has been pointed out in the comments by Wojowu, $E/G$ is always Galois, but $G/F$ is Galois iff $Gal(E/G)\triangleleft Gal(E/F)$ in which case $Gal(G/F)\cong Gal(E/F)/Gal(E/G)$

Jonathan Dunay
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Hint: Consider $\mathbb Q \subseteq \mathbb Q(\sqrt[3]{2}) \subseteq \mathbb Q(\sqrt[3]{2},\omega)$, where $\omega$ is a primitive cubic root of unity.

lhf
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