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