prime subfield

**dori1123** · March 7th 2009, 11:07 AM

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

**ThePerfectHacker** · March 7th 2009, 02:04 PM

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, $\sigma (\alpha/\beta) = \sigma(\alpha)/\sigma(\beta) = \alpha/\beta$ .

**dori1123** · March 7th 2009, 05:35 PM

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 $\{(n\cdot 1)^{-1}| n \in \mathbb{Z}\}$ ?

**ThePerfectHacker** · March 7th 2009, 05:50 PM

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?

**dori1123** · March 7th 2009, 06:23 PM

Yes.

**ThePerfectHacker** · March 7th 2009, 06:58 PM

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.

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