Why is $H^0(X_n, \mathcal{O}_{X_n})$ a local artin ring.

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Let $V$ be regular, proper and of dimension 2 over $S = $ spec $R$, for $R$ a complete discrete valuation ring with uniformizer $t$, maximal ideal $\mathfrak{m}$, and algebraically closed residue field $k$, and let $X$ denote its closed fiber, which we assume to be connected. Let $X_N = V\times_S $ spec $(R/t^{N+1}R)$. I want to show that $H^0(X_{N}, \mathcal{O}_{X_{N}}^*)$ is the group of units in a local Artin ring with residue field $k$.

My initial attempt went as follows:

My goal seems to be to show that $\mathcal{O}_{X_N}(X_N)$ is a local Artin ring with residue field $k$. To this end, I was thinking that by some sort of Künneth formula we should have$$\mathcal{O}_{X_N}(X_N) = H^0(X_N, \mathcal{O}_{X_N})\cong H^0(V,\mathcal{O}_V)\otimes_{R} H^0\left(\text{spec }(R/\mathfrak{m}^{N+1}), \mathcal{O}_{\text{spec} (R/\mathfrak{m}^{N+1})}\right)$$$$= H^0(V,\mathcal{O}_V)\otimes_{R} (R/t^{N+1}R)\cong \mathcal{O}_V(V)/t^{N+1}\mathcal{O}_V(V)$$ and then writing $A = \mathcal{O}_V(V)$ and $B = A/t^NA$, we want to show that $B$ is a local Artin ring with residue field $k$. We see that $A/tA\cong \mathcal{O}_{X}(X) \cong k$ (this gives local and residue field $k$). To see that $B$ is an Artin ring, we note that it's Noetherian since it has a principal maximal ideal $tB$ which is nilpotent, and then it's also Artinian since it's of dimension $0$.