Suppose I have a local field $K$ with discrete valuation $v$, and we denote $k$ to be its residue field. Furthermore, suppose $K$ and $k$ to be perfect. I am having trouble understanding the following two things.
1) How do we know that the algebraic closure $\bar{K}$ is also a local field?
2) And how does the Galois group for $\bar{K}/K$ act on $\bar{k}$, the residue field of $\bar{K}$.
Any comments or explanation is appreciated. Thank you very much.
PS Please ignore the first one as it is not true as explained in the comment.
The algebraic closure is not a local field since its residue field is not finite.
Here’s what is going on. We can define a valuation $\overline v$ on $\overline K$ as follows: for each $x\in \overline K$, $x$ must lie in some finite extension $L/K$. If $v_L$ is the unique extension of $v$ to $L$, we define $\overline v(x)$ to be $v_L(x)$. This gives a valuation $v:\overline K\setminus\{0\}\to \mathbb Q$, which is not discrete, which is another way of seeing that $\overline K$ is not a local field.
Let $\overline{\mathcal O}$ denote the ring of integers of $\overline K$ - i.e. the ring of elements of non-negative valuation. Then $\overline O$ is still a local ring, with unique maximal $\overline{\mathfrak p}$ consisting of the elements with positive valuation.
The Galois group $\mathrm{Gal}(\overline K/K)$ acts on $\overline {\mathcal O}$ and fixes $\overline{\mathfrak p}$. Hence, it acts on the residue field $\overline k = \overline{\mathcal O}/\overline{\mathfrak p}$.