Lemma 55.14.4. Let $R$ be a discrete valuation ring with fraction field $K$ and residue field $k$. Assume $X \to \mathop{\mathrm{Spec}}(R)$ is at-worst-nodal of relative dimension $1$ over $R$. Let $X \to X'$ be the contraction of an exceptional curve $E \subset X$ of the first kind. Then $X'$ is at-worst-nodal of relative dimension $1$ over $R$.

**Proof.**
Namely, let $x' \in X'$ be the image of $E$. Then the only issue is to see that $X' \to \mathop{\mathrm{Spec}}(R)$ is at-worst-nodal of relative dimension $1$ in a neighbourhood of $x'$. The closed fibre of $X \to \mathop{\mathrm{Spec}}(R)$ is reduced, hence $\pi \in R$ vanishes to order $1$ on $E$. This immediately implies that $\pi $ viewed as an element of $\mathfrak m_{x'} \subset \mathcal{O}_{X', x'}$ but is not in $\mathfrak m_{x'}^2$. Since $\mathcal{O}_{X', x'}$ is regular of dimension $2$ (by definition of contractions in Resolution of Surfaces, Section 54.16), this implies that $\mathcal{O}_{X'_ k, x'}$ is regular of dimension $1$ (Algebra, Lemma 10.106.3). On the other hand, the curve $E$ has to meet at least one other component, say $C$ of the closed fibre $X_ k$. Say $x \in E \cap C$. Then $x$ is a node of the special fibre $X_ k$ and hence $\kappa (x)/k$ is finite separable, see Algebraic Curves, Lemma 53.19.7. Since $x \mapsto x'$ we conclude that $\kappa (x')/k$ is finite separable. By Algebra, Lemma 10.140.5 we conclude that $X'_ k \to \mathop{\mathrm{Spec}}(k)$ is smooth in an open neighbourhood of $x'$. Combined with flatness, this proves that $X' \to \mathop{\mathrm{Spec}}(R)$ is smooth in a neighbourhood of $x'$ (Morphisms, Lemma 29.34.14). This finishes the proof as a smooth morphism of relative dimension $1$ is at-worst-nodal of relative dimension $1$ (Algebraic Curves, Lemma 53.20.3).
$\square$

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