Is the Order of an Automorphism in a Field with Characteristic p Equal to p?

[HDB1]

Please, I have a question about automorphism:

Let $\mathbb{K}$ be a field, if $\operatorname{char}(\mathbb{K})=p$, then the order of automorphism $\phi$ is $p$, i.e. $\phi^p=\operatorname{id}$, where $i d$ is identity map.

Is that right? please, if yes, how we can prove it, and what will happen if $\operatorname{char}(\mathbb{K})=0$Thanks in advance,

 

[HDB1]

Dear @fresh_42 , if you could help, I would appreciate that,

 

[fresh_42]

Please, I have a question about automorphism:

Let $\mathbb{K}$ be a field, if $\operatorname{char}(\mathbb{K})=p$, then the order of automorphism $\phi$ is $p$, i.e. $\phi^p=\operatorname{id}$, where $i d$ is identity map.

Is that right? please, if yes, how we can prove it, and what will happen if $\operatorname{char}(\mathbb{K})=0$Thanks in advance,

This is not the case.

Consider $\mathbb{F}_3=\mathbb{Z}/3\mathbb{Z}=\{\bar 0\, , \,\bar 1\, , \,\bar 2\, , \,\}$ and the polynomial $f(x)=x^2+\bar1 \in \mathbb{F}_3[x].$ It has no zeros in $\mathbb{F}_3$ since $f(\bar 0)=\bar 1 \, , \,f(\bar 1)= \bar 2 \, , \,f(\bar 2)=\bar 2 .$ Therefore, it has no factors of degree $1,$ i.e. it is irreducible. If we add a zero $\mathrm{i}$ of $x^2+1$ to $\mathbb{F}_3,$ i.e. we build
$$
\mathbb{K}=\mathbb{F}_3[x]/\langle x^2+1 \rangle \cong \mathbb{F}_3[\mathrm{i}]
$$
then $\mathbb{F}_3 \subsetneq \mathbb{F}_3[\mathrm{i}]=\mathbb{K}$ is a proper field extension and $\operatorname{char}\mathbb{F}_3=\operatorname{char}\mathbb{K}=3.$

We define $\sigma (a+b\mathrm{i}):=a-b \mathrm{i}$ for all $a,b \in \mathbb{F}_3.$ Then $\sigma$ is an automorphism of $\mathbb{K}$ and $\sigma^2=\operatorname{id}_{\mathbb{K}},$ i.e. $\operatorname{ord}(\sigma)=2.$

However, and maybe this is what you meant, there is a certain automorphism for fields of finite characteristic. Say $\operatorname{char}\mathbb{K}=p.$ Then $x\longmapsto x^p$ is the Frobenius homomorphism which is an automorphism of $\mathbb{K}$ and the identity map on the prime field $\mathbb{F}_p$ of $\mathbb{K}.$