1. ## prime subfield

Let $F$ be a field and let $E$ be the prime subfield of $F$. Prove that every automorphism of $F$ fixes $E$.

2. Originally Posted by dori1123
Let $F$ be a field and let $E$ be the prime subfield of $F$. Prove that every automorphism of $F$ fixes $E$.
If $\sigma: F\to F$ is an automorphism then $\sigma (1) = 1$.
Let $n\cdot 1 = \underbrace{1 + 1 + ... 1}_{n\text{ times }}$ for $n>0$.
If $n=0$ then let $n\cdot 1 = 0$ and if $n<0$ then $n\cdot 1 = - [(-n)\cdot 1]$.

Now let $D = \{ n\cdot 1 | n \in \mathbb{Z} \}$ then $\sigma (x) = x$ for all $x\in D$.
To see this notice $\sigma (1+...+1) = \sigma (1) + ... + \sigma(1) = n\cdot \sigma(1) = n\cdot 1$.
This is because $\sigma (a+b) = \sigma(a) + \sigma(b)$ and $\sigma (-a) = -\sigma(a)$ since $\sigma$ is automorphism.

The prime subfield is $E = \{ \alpha/\beta | \alpha,\beta \in D , \beta \not = 0 \}$.
Notice that $\sigma (x) = x$ for all $x\in E$.
To see this notice, $\sigma (\alpha/\beta) = \sigma(\alpha)/\sigma(\beta) = \alpha/\beta$.

3. Originally Posted by ThePerfectHacker
The prime subfield is $E = \{ \alpha/\beta | \alpha,\beta \in D , \beta \not = 0 \}$.
Notice that $\sigma (x) = x$ for all $x\in E$.
To see this notice, $\sigma (\alpha/\beta) = \sigma(\alpha)/\sigma(\beta) = \alpha/\beta$.
I understand the first two cases. But why is that the third case? How about $\{(n\cdot 1)^{-1}| n \in \mathbb{Z}\}$?

4. Originally Posted by dori1123
I understand the first two cases. But why is that the third case? How about $\{(n\cdot 1)^{-1}| n \in \mathbb{Z}\}$?
I have no idea what you are trying to ask. What are you asking? Are you asking why $E$ is the prime subfield in the way I defined it?

5. Yes.

6. Originally Posted by dori1123
Yes.
The prime subfield is the intersection of all subfields i.e. it is the smallest subfield. If $E$ is a subfield then it means $1\in E$ but then $n\cdot 1\in E$ because $E$ is closed under addition and it has $0$ and it has all additive inverses. Therefore $D$ must be contained in $E$. However, $E$ is a field and so $\alpha/\beta \in E$ for all $\alpha,\beta \in D,\beta\not =0$. Therefore, the set of fraction is contained in $E$, but this set of fractions is itself a field and so $E$ is in fact the set of all these fractions that we can form.