Vaguely, an arithmetic surface is an algebraic curve with coefficients in some "ring of integers".

More formally it is an integral normal scheme $X$ of dimension 2 of finite type over a Dedekind domain $R$ with some additional properties over its reduction modulo the primes of $R$ (this reductions being the same as the fibers of the morphism $X\rightarrow \text{Spec } R$). You can see the actual definition at the Wikipedia page on arithmetic surfaces. As this question is local I think there is no harm in assume $R$ a DVR.

Also, if I understand correctly, given a variety $V$ and a point $p\in V$, the completion $\widehat{\mathcal{O}}_{V,p}$ of the local ring at $p$ is a way to understand the singularity of the variety at $p$. For example when the point $p$ is smooth this completion is isomorphic to $k[[x_1,\dots,x_n]]$ for $n=\dim V$ as was already discussed here.

Now take $x\in X$ a closed point in the arithmetic surface. My idea is that $\widehat{\mathcal{O}}_{X,x}$ is something that is determined on the regularity of $x$ both as a point in $X$ and as a point in its special fiber $X_s$ (where $s$ is the closed point of $\text{Spec }R$). So, if we fix the regularity type of $x$ at $X$ and at $X_s$ then $\widehat{\mathcal{O}}_{X,x}$ should also be fixed.

Thinking in this way I want to know if the following is true:

Let $R$ be a DVR with algebraically closed residue field $k$ and $X/R$ be an algebraic surface with semistable reduction. Then for all point $x\in X_s$ regular in $X$ we have either $$\widehat{\mathcal{O}}_{X,x}\cong R[[T]] \ \ \text{ or } \ \ \widehat{\mathcal{O}}_{X,x}\cong R[[S,T]]/(ST-\pi)$$ for an uniformizer $\pi$ of $R$.

Here semistable reduction means that the special fiber is reduced and the only singularities of it are nodes. The points with local completion $R[[T]]$ should be the ones with smooth reduction and the ones with local completion $R[[S,T]]/(ST-\pi)$ should be the nodes.

I think this is equivalent to ask if, étale-locally every semistable algebraic surface can be modeled as $R[S,T]/(ST-\pi)$.

Edit: Added the hypothesis of algebraically closed as discussed in the comments.

Vaguely, an arithmetic surface is an algebraic curve with coefficients in some "ring of integers".

More formally it is an integral normal scheme $\mathfrak{X}$ of dimension 2 of finite type over a Dedekind domain $R$ with some additional properties over its reduction modulo the primes of $R$ (this reductions being the same as the fibers of the morphism $\mathfrak{X}\rightarrow \text{Spec } R$). You can see the actual definition at the Wikipedia page on arithmetic surfaces. As this question is local I think there is no harm in assume $R$ a DVR.

Also, if I understand correctly, given a variety $V$ and a closed point $p\in V$, the completion $\widehat{\mathcal{O}}_{V,p}$ of the local ring at $p$ is a way to understand the singularity of the variety at $p$. For example when the point $p$ is smooth this completion is isomorphic to $k[[x_1,\dots,x_n]]$ for $n=\dim V$ as was already discussed here.

Now take $x\in \mathfrak{X}$ a closed point in the arithmetic surface. My heuristic is that $\widehat{\mathcal{O}}_{\mathfrak{X},x}$ is something that is determined by the regularity of $x$ both as a point in $\mathfrak{X}$ and as a point in its special fiber $\mathfrak{X}_s$ (where $s$ is the closed point of $\text{Spec }R$). So, if we fix the regularity type of $x$ at $\mathfrak{X}$ (say, regular) and at $\mathfrak{X}_s$ (say, smooth or normal crossing) then $\widehat{\mathcal{O}}_{\mathfrak{X},x}$ should also be fixed.

Thinking in this way I want to know if the following is true:

Let $R$ be a DVR with algebraically closed residue field $k$ and $\mathfrak{X}/R$ be an algebraic surface with semistable reduction. Then for all point $x\in \mathfrak{X}_s$ regular in $\mathfrak{X}$ we have either $$\widehat{\mathcal{O}}_{\mathfrak{X},x}\cong R[[T]] \ \ \text{ or } \ \ \widehat{\mathcal{O}}_{\mathfrak{X},x}\cong R[[S,T]]/(ST-\pi)$$ for an uniformizer $\pi$ of $R$.

Here semistable reduction means that the special fiber is reduced and the only singularities of it are nodes. The points with local completion $R[[T]]$ should be the ones with smooth reduction and the ones with local completion $R[[S,T]]/(ST-\pi)$ should be the nodes.

I think this is equivalent to ask if, étale-locally every semistable algebraic surface can be modeled as $R[S,T]/(ST-\pi)$.

Edit: Added the hypothesis of algebraically closed as discussed in the comments.

Vaguely, an arithmetic surface is an algebraic curve with coefficients in some "ring of integers".

More formally it is an integral normal scheme $X$ of dimension 2 of finite type over a Dedekind domain $R$ with some additional properties over its reduction modulo the primes of $R$ (this reductions being the same as the fibers of the morphism $X\rightarrow \text{Spec } R$). You can see the actual definition at the Wikipedia page on arithmetic surfaces. As this question is local I think there is no harm in assume $R$ a DVR.

Also, if I understand correctly, given a variety $V$ and a point $p\in V$, the completion $\widehat{\mathcal{O}}_{V,p}$ of the local ring at $p$ is a way to understand the singularity of the variety at $p$. For example when the point $p$ is smooth this completion is isomorphic to $k[[x_1,\dots,x_n]]$ for $n=\dim V$ as was already discussed here.

Now take $x\in X$ a closed point in the arithmetic surface. My idea is that $\widehat{\mathcal{O}}_{X,x}$ is something that is determined on the regularity of $x$ both as a point in $X$ and as a point in its special fiber $X_s$ (where $s$ is the closed point of $\text{Spec }R$). So, if we fix the regularity type of $x$ at $X$ and at $X_s$ then $\widehat{\mathcal{O}}_{X,x}$ should also be fixed.

Thinking in this way I want to know if the following is true:

Let $R$ be a DVR with algebraically closed residue field $k$ and $X/R$ be an algebraic surface with semistable reduction. Then for all point $x\in X_s$ regular in $X$ we have either $$\widehat{\mathcal{O}}_{X,x}\cong R[[T]] \ \ \text{ or } \ \ \widehat{\mathcal{O}}_{X,x}\cong R[[S,T]]/(ST-\pi)$$ for an uniformizer $\pi$ of $R$.

Here semistable reduction means that the special fiber is reduced and the only singularities of it are nodes. The points with local completion $R[[T]]$ should be the ones with smooth reduction and the ones with local completion $R[[S,T]]/(ST-\pi)$ should be the nodes.

Edit: Added the hypothesis of algebraically closed as discussed in the comments.

Vaguely, an arithmetic surface is an algebraic curve with coefficients in some "ring of integers".

More formally it is an integral normal scheme $X$ of dimension 2 of finite type over a Dedekind domain $R$ with some additional properties over its reduction modulo the primes of $R$ (this reductions being the same as the fibers of the morphism $X\rightarrow \text{Spec } R$). You can see the actual definition at the Wikipedia page on arithmetic surfaces. As this question is local I think there is no harm in assume $R$ a DVR.

Also, if I understand correctly, given a variety $V$ and a point $p\in V$, the completion $\widehat{\mathcal{O}}_{V,p}$ of the local ring at $p$ is a way to understand the singularity of the variety at $p$. For example when the point $p$ is smooth this completion is isomorphic to $k[[x_1,\dots,x_n]]$ for $n=\dim V$ as was already discussed here.

Now take $x\in X$ a closed point in the arithmetic surface. My idea is that $\widehat{\mathcal{O}}_{X,x}$ is something that is determined on the regularity of $x$ both as a point in $X$ and as a point in its special fiber $X_s$ (where $s$ is the closed point of $\text{Spec }R$). So, if we fix the regularity type of $x$ at $X$ and at $X_s$ then $\widehat{\mathcal{O}}_{X,x}$ should also be fixed.

Thinking in this way I want to know if the following is true:

Let $R$ be a DVR with algebraically closed residue field $k$ and $X/R$ be an algebraic surface with semistable reduction. Then for all point $x\in X_s$ regular in $X$ we have either $$\widehat{\mathcal{O}}_{X,x}\cong R[[T]] \ \ \text{ or } \ \ \widehat{\mathcal{O}}_{X,x}\cong R[[S,T]]/(ST-\pi)$$ for an uniformizer $\pi$ of $R$.

Here semistable reduction means that the special fiber is reduced and the only singularities of it are nodes. The points with local completion $R[[T]]$ should be the ones with smooth reduction and the ones with local completion $R[[S,T]]/(ST-\pi)$ should be the nodes.

I think this is equivalent to ask if, étale-locally every semistable algebraic surface can be modeled as $R[S,T]/(ST-\pi)$.

Edit: Added the hypothesis of algebraically closed as discussed in the comments.