Following [this](https://math.stackexchange.com/questions/1860288/completion-and-algebraic-closure) question:


Given a valued field $K$, denote with $\bar{K}$ its algebraic closure and with $\hat{K}$ the completion. Then both $\hat{\bar{K}}$ and $\hat{\bar{\hat{K}}}$ are complete and algebraically closed. If $K=\mathbb{Q}$, these constructions always give $\mathbb{C}$ or $\mathbb{C}\_p$ for a prime $p$, i.e. both constructions give the same fields.


Is this true for any field $K$? Or is there a field $K$ with a valuation $v$ such that for no extensions of $v$ to $\bar{K}$ we have $\hat{\bar{K}}=\hat{\bar{\hat{K}}}$?



Completion and algebraic closure
First you have to observe that since all extensions of the valuation to $\bar{K}$ are conjugate, $\hat{\bar{K}}$ is well-defined up to (non-unique) isomorphism.  

Now, since $\hat{\bar{K}}$ is complete and $K$ is dense in $\hat{K}$, the inclusion $K\subset \hat{\bar{K}}$ extends continuously to $K\subset \hat{K}\subset \hat{\bar{K}}$ (in fact you can identify $\hat{K}$ with the closure of $K$ in $\hat{\bar{K}}$). Since $\hat{\bar{K}}$ is an algebraically closed extension of $\hat{K}$, it contains a unique copy of $\bar{\hat{K}}$, namely the algebraic closure of $\hat{K}$ in it. Moreover this copy obviously contains $\bar{K}$ which is dense in $\hat{\bar{K}}$, hence it is also dense. Since $\hat{\bar{K}}$ is complete, it is therefore isomorphic to the completion of $\bar{\hat{K}}$. 


