Let $\displaystyle F$ be a field and let $\displaystyle E$ be the prime subfield of $\displaystyle F$. Prove that every automorphism of $\displaystyle F$ fixes $\displaystyle E$.
If $\displaystyle \sigma: F\to F$ is an automorphism then $\displaystyle \sigma (1) = 1$.
Let $\displaystyle n\cdot 1 = \underbrace{1 + 1 + ... 1}_{n\text{ times }}$ for $\displaystyle n>0$.
If $\displaystyle n=0$ then let $\displaystyle n\cdot 1 = 0$ and if $\displaystyle n<0$ then $\displaystyle n\cdot 1 = - [(-n)\cdot 1]$.
Now let $\displaystyle D = \{ n\cdot 1 | n \in \mathbb{Z} \}$ then $\displaystyle \sigma (x) = x$ for all $\displaystyle x\in D$.
To see this notice $\displaystyle \sigma (1+...+1) = \sigma (1) + ... + \sigma(1) = n\cdot \sigma(1) = n\cdot 1$.
This is because $\displaystyle \sigma (a+b) = \sigma(a) + \sigma(b)$ and $\displaystyle \sigma (-a) = -\sigma(a)$ since $\displaystyle \sigma$ is automorphism.
The prime subfield is $\displaystyle E = \{ \alpha/\beta | \alpha,\beta \in D , \beta \not = 0 \}$.
Notice that $\displaystyle \sigma (x) = x$ for all $\displaystyle x\in E$.
To see this notice, $\displaystyle \sigma (\alpha/\beta) = \sigma(\alpha)/\sigma(\beta) = \alpha/\beta$.
The prime subfield is the intersection of all subfields i.e. it is the smallest subfield. If $\displaystyle E$ is a subfield then it means $\displaystyle 1\in E$ but then $\displaystyle n\cdot 1\in E$ because $\displaystyle E$ is closed under addition and it has $\displaystyle 0$ and it has all additive inverses. Therefore $\displaystyle D$ must be contained in $\displaystyle E$. However, $\displaystyle E$ is a field and so $\displaystyle \alpha/\beta \in E$ for all $\displaystyle \alpha,\beta \in D,\beta\not =0$. Therefore, the set of fraction is contained in $\displaystyle E$, but this set of fractions is itself a field and so $\displaystyle E$ is in fact the set of all these fractions that we can form.