Let $k$ be a field. I want to prove that the integral closure of $k[[t]]$ in algebraic closure of $k((t))$ (we call it $A$) is not a Noetherian ring.
Thank you for your help. Reference (a webpage, pdf, etc.) is also appreciated.
P.S
From the theory of extension of valuation, we can say $\operatorname{Spec} A$ is Noetherian as topological space. Thus we can construct an example of $\operatorname{Spec} A$ which is not noetherian as a scheme but Noetherian as a topological space.
If $k$ has characteristic $0$ then $$A=\bigcup_{n\ge 1} \overline{k}[[t^{1/n!}]], \qquad \overline{k((t))} = A[t^{-1}]$$ and $(t^{1/n!})$ is a strictly increasing sequence of ideals.
$Spec(A) = \{0\} \cup (\bigcup_{n\ge 1}(t^{1/n!}))$ is a 'Noetherian topological space'.
When $char(k)=p$ then $t^{1/(n+1)!-1/n!}$ is not integral over $k[[t]]$ thus $$(t^{1/n!})\subsetneq (t^{1/(n+1)!})\subsetneq A$$