Does a local ring with a separably closed residue field admit a universal homeomorphism from a local ring with algebraically closed residue field? Let $A$ be a local ring with separably closed residue field. Under what conditions does there exist a local ring $B$ with an algebraically closed residue field and a local homomorphism $A\rightarrow B$ inducing a universal homeomorphism on spectra?
 A: You can always do this when $A$ has equicharacteristic (i.e. contains a field). For equicharacteristic $0$, there is nothing to prove, and for equicharacteristic $p > 0$ you can take the perfection¹
$$B = \underset{n}{\operatorname{colim}} F^n_* A.$$
Indeed, each map $F \colon F^n_* A \to F^{n+1}_* A$ is a universal homeomorphism, so the same goes for $A \to B$. In particular, $B$ is local and $A \to B$ is local. Moreover, the Frobenius on $B/\mathfrak m_B$ is surjective since the same goes for $B$, and $A/\mathfrak m_A \to B/\mathfrak m_B$ is algebraic since $A \to B$ is integral [Tag 04DF]. So we conclude that $B/\mathfrak m_B$ is algebraically closed since it is algebraic over a separably closed field as well as perfect. $\square$
On the other hand, if $A$ is normal of mixed characteristic $(0,p)$, then this is possible if and only if $A/\mathfrak m_A$ is already algebraically closed. Indeed, a universal homeomorphism $A \to B$ is integral by [Tag 04DF] and radicial by [Tag 01S4], so $\operatorname{Frac}(A) \to \operatorname{Frac}(B)$ is purely inseparable (algebraic). Since $\operatorname{char} A = 0$, this forces $\operatorname{Frac}(A) \stackrel\sim\to \operatorname{Frac}(B)$, hence by normality $A \stackrel\sim\to B$. $\square$
The remaining case where $A$ is non-normal of mixed characteristic (for example, it could have characteristic $p^n$ for some $n$) is mysterious to me. Even for $A$ integral of mixed characteristic $(0,p)$, I find it hard to picture how you can fit a local ring $B$ between $A$ and $\operatorname{Frac} A$ whose residue field is the algebraic closure of $A/\mathfrak m_A$. Better ask Brian Conrad or something...

¹ Note on notation: $F^n_*A$ is a compromise between writing $A^{1/p^n}$, which is not quite right when $A$ is not a domain, and denoting all of the objects in the colimit by $A$, which is confusing. The notation $F^n_*A$ refers to the affine morphism $F^n \colon \operatorname{Spec} A \to \operatorname{Spec} A$, where we think of an $A$-algebra $B$ as the quasi-coherent sheaf of algebras on $\operatorname{Spec} A$ given by $f_* B$ (or really $f_* \mathcal O_{\operatorname{Spec} B}$) for $f \colon \operatorname{Spec} B \to \operatorname{Spec} A$ induced by $A \to B$.
