**Proof.**
Since $X \to \mathop{\mathrm{Spec}}(R)$ is proper, the $K$-rational point extends to a morphism $a : \mathop{\mathrm{Spec}}(R) \to X$ by the valuative criterion of properness (Morphisms, Lemma 29.42.1). Let $x \in X$ be the image under $a$ of the closed point of $\mathop{\mathrm{Spec}}(R)$. Then $a$ corresponds to an $R$-algebra homomorphism $\psi : \mathcal{O}_{X, x} \to R$ (see Schemes, Section 26.13). It follows that $\pi \not\in \mathfrak m_ x^2$ (since the image of $\pi $ in $R$ is not in $\mathfrak m_ R^2$). Hence $\mathcal{O}_{X_ k, x} = \mathcal{O}_{X, x}/\pi \mathcal{O}_{X, x}$ is regular (Algebra, Lemma 10.106.3). Then $X_ k \to \mathop{\mathrm{Spec}}(k)$ is smooth at $x$ by Algebra, Lemma 10.140.5. It follows that $x$ is contained in a unique irreducible component $C_ i$ of $X_ k$, that $\mathcal{O}_{C_ i, x} = \mathcal{O}_{X_ k, x}$, and that $m_ i = 1$. The fact that $C_ i$ has a $k$-rational point implies that the field $\kappa _ i = H^0(C_ i, \mathcal{O}_{C_ i})$ (Varieties, Lemma 33.26.2) is equal to $k$. This proves (1). We have $H^0(X_ k, \mathcal{O}_{X_ k}) = k$ because $H^0(X_ k, \mathcal{O}_{X_ k})$ is a field extension of $k$ (Lemma 55.9.9) which maps to $H^0(C_ i, \mathcal{O}_{C_ i}) = k$. The genus equality follows from Lemma 55.9.10.
$\square$

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