Let $K=F(\alpha)$. It strikes me that there are two possibilities:
$\alpha \in F$, in which case $K\cong F$
$\alpha\not\in F$, in which case $K$ is the splitting field of $(x-\alpha)$ (and hence Galois).
(I actually think that even #1 might be described as a Galois extension - $|G(F/F)|=[F:F]...
I have been looking at Galois extensions and Galois groups and have been wondering if all Galois extensions are simple. I don't think this is true.
For example with $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$ would be a simple extension of $\mathbb{Q}$.
With something li...
I have been looking at Galois extensions and Galois groups and have been wondering if all Galois extensions are simple. I don't think this is true.
For example with $\mathbb{Q}(\sqrt{2}, \sqrt{3}) = \mathbb{Q}(\sqrt{2} + \sqrt{3})$ would be a simple extension of $\mathbb{Q}$.
With something li...
Linear & Abstract algebra
Linear & Abstract algebra