Galois Theory - Fixed Field of F and Definition of Aut(K/F) ....

[Math Amateur]

I am reading Dummit and Foote, Chapter 14 - Galois Theory.

I am currently studying Section 14.2 : The Fundamental Theorem of Galois Theory ... ...

I need some help with Corollary 10 of Section 14.2 ... ... and the definition of \(\displaystyle \text{Aut}(K/F)\) ... ...

Corollary 10 reads as follows:
View attachment 6666
Now the Definition of \(\displaystyle \text{Aut}(K/F)\) is as follows:
https://www.physicsforums.com/attachments/6667
Now in Corollary 10 we read the following:

" ... ... Then

\(\displaystyle | \text{Aut}(K/F) | \ \le \ [ K \ : \ F ] \)with equality if and only if F is the fixed field of \(\displaystyle \text{Aut}(K/F)\) ... ... "
My question is as follows:

Given the definition of \(\displaystyle \text{Aut}(K/F)\) shown above, isn't \(\displaystyle F\) guaranteed to be the fixed field of \(\displaystyle \text{Aut}(K/F)\) ... ... ?
Hope someone can resolve this problem/issue ...

Help will be much appreciated ...

Peter===========================================================================================
The above post will be easier to follow if readers understand D&F's definition of a Galois Extension and a Galois Group ... so I am providing the definition as follows ... ... :

https://www.physicsforums.com/attachments/6668

 

[Euge]

Hi Peter,

Let $L$ be a field, and let $H$ be a subgroup of the automorphism group $\operatorname{Aut}{L}$. The fixed field of $H$, usually denoted $L^H$, is defined as the set of all $x\in L$ such that $\sigma x = x$ for all $\sigma \in H$, i.e., the set of all elements of $L$ that are left fixed by $H$.

In your situation, $L = K$ and $H = \operatorname{Aut}(K/F)$. By definition of $H$, $H$ fixes all elements of $F$, so $K^H \supset F$. The definition does not say that the only elements fixed by $H$ are the elements of $F$. As an example, consider the extension field $\Bbb Q(\sqrt[3]{2})/\Bbb Q$. The only automorphism of $\Bbb Q(\sqrt[3]{2})$ is the identity, so the fixed field of $\operatorname{Aut}(\Bbb Q(\sqrt[3]{2})/\Bbb Q$ is $\Bbb Q(\sqrt[3]{2})$, which is different from $\Bbb Q$. In particular, $\Bbb Q(\sqrt[3]{2})$ is not a Galois extension of $\Bbb Q$.

 

[Math Amateur]

Hi Peter,

Let $L$ be a field, and let $H$ be a subgroup of the automorphism group $\operatorname{Aut}{L}$. The fixed field of $H$, usually denoted $L^H$, is defined as the set of all $x\in L$ such that $\sigma x = x$ for all $\sigma \in H$, i.e., the set of all elements of $L$ that are left fixed by $H$.

In your situation, $L = K$ and $H = \operatorname{Aut}(K/F)$. By definition of $H$, $H$ fixes all elements of $F$, so $K^H \supset F$. The definition does not say that the only elements fixed by $H$ are the elements of $F$. As an example, consider the extension field $\Bbb Q(\sqrt[3]{2})/\Bbb Q$. The only automorphism of $\Bbb Q(\sqrt[3]{2})$ is the identity, so the fixed field of $\operatorname{Aut}(\Bbb Q(\sqrt[3]{2})/\Bbb Q$ is $\Bbb Q(\sqrt[3]{2})$, which is different from $\Bbb Q$. In particular, $\Bbb Q(\sqrt[3]{2})$ is not a Galois extension of $\Bbb Q$.

Thanks Euge ... that helped in a big way ...

Most grateful for your help and support ...Peter