Consider the following problem part of my abstract algebra course assignment:
Let F be an algebraic closure of $\mathbb{Z}_{p}$ (p prime) then prove that F is algebraic and Galois over $\mathbb{Z}_p$.
F is algebraic closure and so all roots of irreducible polynomials in $\mathbb{Z}_p$[x] are in F and but to prove that F is algebraic i need to choose any element say p which is in F and then prove that p is a root of element in $\mathbb{Z}_p $[x] but as F is given as algebraically closed so it is algebraic over $\mathbb{Z}_p$ .
But The problem is that I am not able to prove it to be Galois over $\mathbb{Z}_p$. I need to prove that Fixed field of $Aut_{\mathbb{Z}_p} F $ is $\mathbb{Z_p}$ itself. SO, If I choose an automorphism $f\in Aut_{\mathbb{Z}_p} F $ I need to prove that only for x $\in \mathbb{Z_p}$ we have $f(x) =x $ . So , For any superfield K" of $\mathbb{Z}_p$ I need to show that there exists atleast 1 x $f(x)\neq x$ .
But I am unable to form ideas on how should I proceed towards it. So, do u mind giving any advice.
Thank you!
I assume that $\mathbb{Z}_p$ refers to the field with $p$ elements; I'll use the more common notation $\mathbb{F}_p$.
An algebraic closure of a field $K$ is algebraic over $K$ by definition (it is unique up to isomorphisms fixing $K$, but it's irrelevant here).
Let $F_0$ be the fixed field of $\operatorname{Aut}(F)$.
The Frobenius homomorphism $\varphi\colon F\to F$, $\varphi(a)=a^p$ is an automorphism of $F$, because $x^p-b$ has a root for every $b\in F$. Since $F_0$ is the fixed field, $\varphi(a)=a$ for every $a\in F_0$. Thus $F_0$ consists of roots of $x^p-x$ and so it has at most $p$ elements; on the other hand, it contains $\mathbb{F}_p$.