The Stacks project

Example 53.23.1 (Contracting a rational bridge). 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 bridge will be an irreducible component $C \subset X$ (viewed as an integral closed subscheme) with the following properties

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

  2. the scheme theoretic intersection $C \cap X'$ has degree $2$ over $H^0(C, \mathcal{O}_ C)$, and

  3. $C$ has genus zero.

Set $k' = H^0(C, \mathcal{O}_ C)$ and $k'' = H^0(C \cap X', \mathcal{O}_{C \cap X'})$. Then $k'$ is a field (Varieties, Lemma 33.9.3) and $\dim _{k'}(k'') = 2$. Since there are at least two irreducible components of $X$ passing through each point of $C \cap X'$, we conclude these points are nodes of $X$ and smooth points on both $C$ and $X'$ (Lemma 53.19.17). Hence $k'/k$ is a finite separable extension of fields and $k''/k'$ is either a degree $2$ separable extension of fields or $k'' = k' \times k'$ (Lemma 53.19.7). By Section 53.14 there exists a pushout

\[ \xymatrix{ C \cap X' \ar[r] \ar[d] & X' \ar[d]^ a \\ \mathop{\mathrm{Spec}}(k') \ar[r] & Y } \]

with many good properties (all of which we will use below without further mention). Let $y \in Y$ be the image of $\mathop{\mathrm{Spec}}(k') \to Y$. Then

\[ \mathcal{O}_{Y, y}^\wedge \cong k'[[s, t]]/(st) \quad \text{or}\quad \mathcal{O}_{Y, y}^\wedge \cong \{ f \in k''[[s]] : f(0) \in k'\} \]

depending on whether $C \cap X'$ has $2$ or $1$ points. This follows from Lemma 53.14.1 and the fact that $\mathcal{O}_{X', p} \cong \kappa (p)[[t]]$ for $p \in C \cap X'$ by More on Algebra, Lemma 15.38.4. Thus we see that $y \in Y$ is a node, see Lemmas 53.19.7 and 53.19.4 and in particular the discussion of Case II in the proof of (2) $\Rightarrow $ (1) in Lemma 53.19.4. Thus the singularities of $Y$ are at-worst-nodal.

We can extend the commutative diagram above to a diagram

\[ \xymatrix{ C \cap X' \ar[r] \ar[d] & X' \ar[d]^ a \ar[r] & X \ar[ld]^ c & C \ar[ld] \ar[l] \\ \mathop{\mathrm{Spec}}(k') \ar[r] & Y & \mathop{\mathrm{Spec}}(k') \ar[l] } \]

where the two lower horizontal arrows are the same. Namely, $X$ is the scheme theoretic union of $X'$ and $C$ (thus a pushout by Morphisms, Lemma 29.4.6) and the morphisms $C \to Y$ and $X' \to Y$ agree on $C \cap X'$. Finally, we claim that

\[ c_*\mathcal{O}_ X = \mathcal{O}_ Y \quad \text{and}\quad R^1c_*\mathcal{O}_ X = 0 \]

To see this use the exact sequence

\[ 0 \to \mathcal{O}_ X \to \mathcal{O}_ C \oplus \mathcal{O}_{X'} \to \mathcal{O}_{C \cap X'} \to 0 \]

of Morphisms, Lemma 29.4.6. The long exact sequence of higher direct images is

\[ 0 \to c_*\mathcal{O}_ X \to c_*\mathcal{O}_ C \oplus c_*\mathcal{O}_{X'} \to c_*\mathcal{O}_{C \cap X'} \to R^1c_*\mathcal{O}_ X \to R^1c_*\mathcal{O}_ C \oplus R^1c_*\mathcal{O}_{X'} \]

Since $c|_{X'} = a$ is affine we see that $R^1c_*\mathcal{O}_{X'} = 0$. Since $c|_ C$ factors as $C \to \mathop{\mathrm{Spec}}(k') \to X$ and since $C$ has genus zero, we find that $R^1c_*\mathcal{O}_ C = 0$. Since $\mathcal{O}_{X'} \to \mathcal{O}_{C \cap X'}$ is surjective and since $c|_{X'}$ is affine, we see that $c_*\mathcal{O}_{X'} \to c_*\mathcal{O}_{C \cap X'}$ is surjective. This proves that $R^1c_*\mathcal{O}_ X = 0$. Finally, we have $\mathcal{O}_ Y = c_*\mathcal{O}_ X$ by the exact sequence and the description of the structure sheaf of the pushout in More on Morphisms, Proposition 37.67.3.

All of this means that $Y$ is also a proper scheme over $k$ having dimension $1$ and $H^0(Y, \mathcal{O}_ Y) = k$ whose singularities are at-worst-nodal (Lemma 53.19.17) and that $Y$ has the same genus as $X$. We will say $c : X \to Y$ is the contraction of a rational bridge.

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