HDB1 said:
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}.##