Let $R$ be a discrete valuation ring with a uniformizer $\pi$ and $(A, \mathfrak{m}_A)$ a local $R$-algebra that is essentially of finite type (i.e., is a localization of a finite type $R$-algebra), such that the image of $\pi$ is in $\mathfrak{m}_A$, and such that $A/\mathfrak{m}_A$ is a finite extension of $R/\pi$. Choose $x_1, \dotsc, x_n \in \mathfrak{m}_A$ such that $(\pi, x_1, \dotsc, x_n) = \mathfrak{m}_A$ in $A$ and let $f\colon R[X_1, \ldots, X_n] \rightarrow A$ be the $R$-algebra morphism satisfying $f(X_i) = x_i$. Let $B$ be the localization of $R[X_1, \ldots, X_n]$ at $(\pi, X_1, \dotsc, X_n)$ and let $g\colon B \rightarrow A$ be the morphism induced by $f$. Does $g$ make $A$ a finite type $B$-algebra?