$\DeclareMathOperator\trdeg{trdeg} \DeclareMathOperator\Frac{Frac} $ Let $k$ be an algebraically closed field of characteristic $0$ and $K$ a function field over $k$ of $\trdeg_k(K)=1$ and let $(A, \mathfrak{m}_A)$ a discrete valuation ring of rank one such that $k \subset A \subset K$ and the fraction field $\Frac(A)$ of $A$ is $K$. Denote by $\kappa_A= A/\mathfrak{m}_A$ the residue field.
Thinking geometrically $K$ can be recognized as a function field of a curve $X$ over $k$ and $A$ a stalk $\mathcal{O}_{X,x}$ with residue field $\kappa_A= \kappa(x) = \mathcal{O}_{X,x}/\mathfrak{m}_x$ at a regular point $x \in X$.

Let denote by $\widehat{K^A}$ the completion of $K$ for $A$, which is of course the same as the fraction field $\Frac(\widehat{A_{\mathfrak{m}_A}})$ of the $\mathfrak{m}_A$-completion of $A$ with respect it's maximal ideal $\mathfrak{m}_A$. Then $\widehat{K_A} \cong \kappa_A((T))$. (Note, that here we exploited that $\kappa_A $contains algebraically closed $k$, otherwise that's not true, eg $\mathbb{Q}_p = \Frac(\mathbb{Z}_p) \not \cong \mathbb{F}_p((T))$)

Let $\widehat{K^A}_{alg}$ an algebraic closure of $\widehat{K^A}$ and $\widehat{K^A}_{unr}$ the maximal unramified extension of $\widehat{K^A}$ in $\widehat{K^A}_{alg}$.

Question: In E. Peyre's "Unramified cohomology and rationality problems", p 3 is stated that $\widehat{K^A}_{unr}$ is isomorphic to algebraic closure $\kappa_A^s((T))_{alg}$ of $\kappa_A((T))$ in $\kappa_A^s((T))$ (where $\kappa_A^s$ is the separable closure of $\kappa_A^s$).

I not understand this statement. Isn't every element of $\kappa_A^s((T))$ already algebraic over $\kappa_A((T))$? So it make no sense for me to talk about "algebraic closure $\kappa_A^s((T))_{alg}$ of $\kappa_A((T))$ in $\kappa_A^s((T))$", or do I missing something. What the author means here and why is it true?

Moreover, why we have

$$ \widehat{K^A}_{alg} \cong \varinjlim_n \kappa_A^s((T^{1/n}))_{alg} $$

? The point which confuses be in this equality is why we should take the limit over $\kappa_A^s((T^{1/n}))_{alg}$ and not $\kappa_A^s((T^{1/n})) $? So why is this additional taking algebraic closure neccessary to exhaust all algebraic elements of $ \widehat{K^A}_{alg}$?