Integral closure in the algebraic closure of $p$-adic numbers

Question

Let $p$ be a prime number and let $\overline{\mathbb{Q}}_p$ be a fixed algebraic closure of the $p$-adic numbers $\mathbb{Q}_p$. It is well know that the ring of integers of $\mathbb{Q}_p$ is the ring of $p$-adic integers $\mathbb{Z}_p$ and that for any finite extension $L$ of $\mathbb{Q}_p$(WLOG, we assume $L$ in the fixed algebraic closure), the ring of integers $\mathcal{O}_L$ is the integral closure of $\mathbb{Z}_p$ in $L$. But is there some extension of this property for an infinite extension? In other words

Is the ring of integers $\mathcal{O}_{\overline{\mathbb{Q}}_p}$of $\overline{\mathbb{Q}}_p$ the integral closure of $\mathbb{Z}_p$ in $\overline{\mathbb{Q}}_p$?

If no:

What is the difference between the two rings?

Answer 1

Let the integral closure of $\mathbb Z_p$ be $A$. We hope to check $\mathcal O_{\overline{\mathbb Q}_p}=A$.

To check $\mathcal O_{\overline{\mathbb Q}_p}\subseteq A$: let $x\in\mathcal O_{\overline{\mathbb Q}_p}$. There exists a finite extension $L\subset\overline{\mathbb Q}_p$ such that $x\in L$, so $x\in\mathcal O_L$. But now $x$ is in the integral closure of $\mathbb Z_p$ in $L$, hence $x\in A$.

To check $A\subset\mathcal O_{\overline{\mathbb Q}_p}$: let $x\in A$. Then there exists some equation $x^n+a_{n-1}x^{n-1}+\cdots+a_0=0$ where $a_i\in\mathbb Z_p$. But if $L\subset\overline{\mathbb Q}_p$ is the splitting field of this polynomial then $x\in \mathcal O_L\subset \mathcal O_{\overline{\mathbb Q}_p}$.