Notation is not good because $\overline{\mathbb{F}}_p((t))$ is transcendental over $\mathbb{F}_p((t))$ Why $\overline{\mathbb{F}}_p((t))$ is transcendental over $\mathbb{F}_p((t))$・・・① ?
In this site, Maximal Unramified Extension of $\mathbb{F}_p((t))$
Notation $\overline{\mathbb{F}}_p((t))$ is not good and $\mathbb∪{F}_{p^n}((t))$ are welcomed because $\overline{\mathbb{F}}_p((t))$ is transcendental over $\mathbb{F}_p((t))$.
But why ① holds? I couldn't check ① by myself.
What element is transcendental element? And why?(In general, to prove transcedentality is difficult, is it managiblel in this case?)
Thank you in advance.
 A: *

*$Gal(\overline{\Bbb{F}_p}/\Bbb{F}_p)$ acts on $\overline{\Bbb{F}_p}((t))$


*it permutes the irreducible factors $\in \overline{\Bbb{F}_p}((t))[x]$ of a polynomial $\in \Bbb{F}_p((t))[x]$
so any element $\in \overline{\Bbb{F}_p}((t))$ with infinite orbit (ie. whose set of coefficients isn't contained in some finite field) will be transcendental  over $\Bbb{F}_p((t))$.

*

*$\Bbb{F}_{p^n}((t))$ is algebraic over $\Bbb{F}_p((t))$ (it is a $n$-dimensional $\Bbb{F}_p((t))$-vector space).

A: Let me first remark that any element of $\overline{\mathbb{F}_p}((t))$ which is algebraic over $\mathbb{F}_p((t))$ is a root of a polynomial with coefficients in $\mathbb{F}_p[[t]]$ whose leading coefficient is a power of $t$.  Indeed, it is the root of a monic polynomial with coefficients in $\mathbb{F}_p((t))$, and then you can multiply by a power of $t$ to make all the coefficients be in $\mathbb{F}_p[[t]]$.
Now topologize $\overline{\mathbb{F}_p}[[t]]$ by identifying it with $\overline{\mathbb{F}_p}^\mathbb{N}$ with the product topology, giving $\overline{\mathbb{F}_p}$ the discrete topology.  This topology is induced by a complete metric, and in particular satisfies the Baire category theorem.  Now let $A_{k,l}\subset\overline{\mathbb{F}_p}[[t]]$ be the set of all elements that are a root of a polynomial with coefficients in $\mathbb{F}_p[[t]]$ of degree $k$ and leading coefficient $t^l$.  I claim that $A_{k,l}$ is closed in $\overline{\mathbb{F}_p}[[t]]$.
Indeed, suppose we have a sequence $(a_n)$ in $A_{k,l}$ converging to an element $a\in\overline{\mathbb{F}_p}[[t]]$, with $a_n$ being a root of the polynomial $t^lx^k+c_{n,k-1}x^{k-1}+\dots+c_{n,0}$ with coefficients $c_{n,m}\in\mathbb{F}_p[[t]]$.  Since $\mathbb{F}_p[[t]]$ is compact (it is just $\mathbb{F}_p^{\mathbb{N}}$ with the product topology and $\mathbb{F}_p$ is finite), we can pass to a subsequence and assume that for each $m$ the coefficient sequence $(c_{n,m})$ converges to some $c_m\in\mathbb{F}_p[[t]]$.  By continuity of the ring operations, then, $a$ will be a root of the polynomial $t^lx^k+c_{k-1}x^{k-1}+\dots c_0$, so $a\in A_{k,l}$.
I further claim that $A_{k,l}$ has empty interior.  To prove this, consider a basic open subset $U\subseteq\mathbb{F}_p[[t]]$ consisting of all power series whose first $n$ coefficients are some fixed elements $a_0,\dots,a_{n-1}\in\overline{\mathbb{F}_p}$.  Let $\mathbb{F}_q$ be the subfield of $\overline{\mathbb{F}_p}$ generated by $a_0,\dots,a_{n-1}$ and consider an element of the form $s=at^n+a_{n-1}t^{n-1}+\dots+a_0\in U$ for some $a\in\overline{\mathbb{F}_p}$.  Suppose $s\in A_k$, so $x$ is a root of some degree $k$ polynomial $p(x)$ with coefficients in $\mathbb{F}_p[[t]]$.  Then $a$ is a root of the polynomial $q(x)=p(xt^n+a_{n-1}t^{n-1}+\dots+a_0)$ which has degree $k$ and coefficients in $\mathbb{F}_q[[t]]$.  But since $a\in\overline{\mathbb{F}_p}$, this implies $a$ is a root of a nonzero polynomial of degree at most $k$ with coefficients in $\mathbb{F}_q$ (consider $q(x)$ as a power series in $t$ with coefficients in $\mathbb{F}_q[x]$; then at least one coefficient is nonzero and all of them have $a$ as a root).  So, there are only finitely many such $a$, and in particular there is some choice of $a\in\overline{\mathbb{F}_p}$ such that $s$ will not be in $A_{k,l}$, so $A_{k,l}$ does not contain $U$.
Thus each $A_{k,l}$ is a closed subset of $\overline{\mathbb{F}_p}[[t]]$ with empty interior.  By the Baire category theorem, $\bigcup_k A_{k,l}$ is not all of $\overline{\mathbb{F}_p}[[t]]$.  Thus there is an element of $\overline{\mathbb{F}_p}[[t]]$ that is not algebraic over $\mathbb{F}_p((t))$.
(Probably this argument is more complicated than necessary and there is a more "hands-on algebra" way to pick the coefficients of a power series one-by-one to avoid being a root of a polynomial of any degree.  In particular, I would expect that if $K$ is an infinite extension of $k$, then $K((t))$ is not algebraic over $k((t))$, but this argument only proves that in the case that $k$ is finite, so that $k[[t]]$ is compact.)
