# Why are Galois extensions mapped to themselves? [closed]

Let $L$ be the splitting field of a separable irreducible polynomial $f(x)\in\Bbb{Q}[x]$. Consider the Galois subfield $K\subset L$. Let $\sigma\in \text{Gal}(L/\Bbb{Q})$. As $K$ is a galois extension, $\sigma(K)=K$.

I don't understand why $K$ being a Galois extension implies $\sigma(K)=K$. Any help would be great.

## closed as off-topic by user26857, zz20s, Ben Sheller, user228113, Daniel W. FarlowMay 27 '16 at 0:37

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Since $K/F$ is a Galois extension, it is a normal extension, meaning that the minimal polynomial of any $a\in K$, $f_a$, splits over $K$.
Let $\sigma\in\mathrm{Gal}\left(L/F\right)$ and let $a\in K$. $\sigma\left(a\right)$ is a root of $f_a$, hence $\sigma\left(a\right)\in K$. This proves $\sigma\left(K\right)\subseteq K$, and thus $\sigma\left(K\right)=K$ (dimension or something similar).
• Instead of dimension as final step I'd suggest to consider $\sigma^{-1}$ – Hagen von Eitzen Jan 7 '16 at 7:02
• I'm sorry what is $K/L$? $K$ is a subfield of $L$. – fierydemon Jan 7 '16 at 7:04