The Stacks project

Lemma 55.9.1. Let $X$ be a regular model of a smooth curve $C$ over $K$.

  1. the special fibre $X_ k$ is an effective Cartier divisor on $X$,

  2. each irreducible component $C_ i$ of $X_ k$ is an effective Cartier divisor on $X$,

  3. $X_ k = \sum m_ i C_ i$ (sum of effective Cartier divisors) where $m_ i$ is the multiplicity of $C_ i$ in $X_ k$,

  4. $\mathcal{O}_ X(X_ k) \cong \mathcal{O}_ X$.

Proof. Recall that $R$ is a discrete valuation ring with uniformizer $\pi $ and residue field $k = R/(\pi )$. Because $X \to \mathop{\mathrm{Spec}}(R)$ is flat, the element $\pi $ is a nonzerodivisor affine locally on $X$ (see More on Algebra, Lemma 15.22.11). Thus if $U = \mathop{\mathrm{Spec}}(A) \subset X$ is an affine open, then

\[ X_ K \cap U = U_ k = \mathop{\mathrm{Spec}}(A \otimes _ R k) = \mathop{\mathrm{Spec}}(A/\pi A) \]

and $\pi $ is a nonzerodivisor in $A$. Hence $X_ k = V(\pi )$ is an effective Cartier divisor by Divisors, Lemma 31.13.2. Hence (1) is true.

The discussion above shows that the pair $(\mathcal{O}_ X(X_ k), 1)$ is isomorphic to the pair $(\mathcal{O}_ X, \pi )$ which proves (4).

By Divisors, Lemma 31.15.11 there exist pairwise distinct integral effective Cartier divisors $D_ i \subset X$ and integers $a_ i \geq 0$ such that $X_ k = \sum a_ i D_ i$. We can throw out those divisors $D_ i$ such that $a_ i = 0$. Then it is clear (from the definition of addition of effective Cartier divisors) that $X_ k = \bigcup D_ i$ set theoretically. Thus $C_ i = D_ i$ are the irreducible components of $X_ k$ which proves (2). Let $\xi _ i$ be the generic point of $C_ i$. Then $\mathcal{O}_{X, \xi _ i}$ is a discrete valuation ring (Divisors, Lemma 31.15.4). The uniformizer $\pi _ i \in \mathcal{O}_{X, \xi _ i}$ is a local equation for $C_ i$ and the image of $\pi $ is a local equation for $X_ k$. Since $X_ k = \sum a_ i C_ i$ we see that $\pi $ and $\pi _ i^{a_ i}$ generate the same ideal in $\mathcal{O}_{X, \xi _ i}$. On the other hand, the multiplicity of $C_ i$ in $X_ k$ is

\[ m_ i = \text{length}_{\mathcal{O}_{C_ i, \xi _ i}} \mathcal{O}_{X_ k, \xi _ i} = \text{length}_{\mathcal{O}_{C_ i, \xi _ i}} \mathcal{O}_{X, \xi _ i}/(\pi ) = \text{length}_{\mathcal{O}_{C_ i, \xi _ i}} \mathcal{O}_{X, \xi _ i}/(\pi _ i^{a_ i}) = a_ i \]

See Chow Homology, Definition 42.9.2. Thus $a_ i = m_ i$ and (3) is proved. $\square$


Comments (1)

Comment #9890 by Doug Liu on

The first displayed equation in the proof, should be ?


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