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. $\square$

## Post a comment

Your email address will not be published. Required fields are marked.

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).

Unfortunately JavaScript is disabled in your browser, so the comment preview function will not work.

All contributions are licensed under the GNU Free Documentation License.

In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following input field. As a reminder, this is tag 0C5Z. Beware of the difference between the letter 'O' and the digit '0'.