Automorphism of Z_p(t) and its invariant field

\mathbf{Problem:} Let E=\mathbb{Z}_p(t) where t is transcedental over \mathbb{Z}_p. Let G be the group of automorphism generated by the automorphism of E such that t\to t+1. Determine F=\text{Inv}G and [E:F].

\mathbf{Proof:} Suppose \sigma is the automorphism of E that t\to t+1.  Group G is generated by \sigma. It is easy to verify that G is the cyclic group \{1,\sigma,\cdots,\sigma^{p-1}\}. So If F=\text{Inv}G, then [E:F]=|G|=p.

Let’s determine F. Note that \mathbb{Z}_p(u) where u=t^p-t is invariant under \sigma, because a^p=a, for \forall \,a\in \mathbb{Z}_p.  So \mathbb{Z}_p(u)\subset F.

On the other hand, \mathbb{Z}_p(u) is a field of characteristic p and E is a splitting field of x^p-x-u over \mathbb{Z}_p(u). x^p-x-u has no root in \mathbb{Z}_p(u), thus it is irreducible on \mathbb{Z}_p(u)[x] by a previous conclusion. So [E:\mathbb{Z}_p(u)]=p=[E:F], which means F=\mathbb{Z}_p(u) where u=t^p-t.

\text{Q.E.D}\hfill \square

\mathbf{Problem:} What if G is replaced by the group of automorphism such that t\to at+b, a,b\in\mathbb{Z}_p, a\neq 0

If a=1, b=0, then F=E and [E:F]=1

If a=1, b\neq 0, then the conclusion is the same with previous problem.

If a\neq 1,0, |G|=p-1=[E:F], F=\mathbb{Z}_p(u) where \displaystyle u=\prod\limits_{i=1}^{p-1}\sigma^{i}(t).

\mathbf{Remark:} Jacobson, Algebra I p243.

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