Example 53.22.1 (Contracting a rational tail). Let $k$ be a field. Let $X$ be a proper scheme over $k$ having dimension $1$ and $H^0(X, \mathcal{O}_ X) = k$. Assume the singularities of $X$ are at-worst-nodal. A *rational tail* will be an irreducible component $C \subset X$ (viewed as an integral closed subscheme) with the following properties

$X' \not= \emptyset $ where $X' \subset X$ is the scheme theoretic closure of $X \setminus C$,

the scheme theoretic interesection $C \cap X'$ is a single reduced point $x$,

$H^0(C, \mathcal{O}_ C)$ maps isomorphically to the residue field of $x$, and

$C$ has genus zero.

Since there are at least two irreducible components of $X$ passing through $x$, we conclude that $x$ is a node. Set $k' = H^0(C, \mathcal{O}_ C) = \kappa (x)$. Then $k'/k$ is a finite separable extension of fields (Lemma 53.19.7). There is a canonical morphism

inducing the identity on $X'$ and mapping $C$ to $x \in X'$ via the canonical morphism $C \to \mathop{\mathrm{Spec}}(k') = x$. This follows from Morphisms, Lemma 29.4.6 since $X$ is the scheme theoretic union of $C$ and $X'$ (as $X$ is reduced). Moreover, we claim that

To see this, denote $i_ C : C \to X$, $i_{X'} : X' \to X$ and $i_ x : x \to X$ the embeddings and use the exact sequence

of Morphisms, Lemma 29.4.6. Looking at the long exact sequence of higher direct images, it follows that it suffices to show $H^0(C, \mathcal{O}_ C) = k'$ and $H^1(C, \mathcal{O}_ C) = 0$ which follows from the assumptions. Observe that $X'$ is also a proper scheme over $k$, of dimension $1$ whose singularities are at-worst-nodal (Lemma 53.19.17) has $H^0(X', \mathcal{O}_{X'}) = k$, and $X'$ has the same genus as $X$. We will say $c : X \to X'$ is the *contraction of a rational tail*.

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