Lemma 115.4.12. Let A be a Noetherian local normal domain of dimension 2. For f \in \mathfrak m nonzero denote \text{div}(f) = \sum n_ i (\mathfrak p_ i) the divisor associated to f on the punctured spectrum of A. We set |f| = \sum n_ i. There exist integers N and M such that |f + g| \leq M for all g \in \mathfrak m^ N.
Proof. Pick h \in \mathfrak m such that f, h is a regular sequence in A (this follows from Algebra, Lemmas 10.157.4 and 10.72.7). We will prove the lemma with M = \text{length}_ A(A/(f, h)) and with N any integer such that \mathfrak m^ N \subset (f, h). Such an integer N exists because \sqrt{(f, h)} = \mathfrak m. Note that M = \text{length}_ A(A/(f + g, h)) for all g \in \mathfrak m^ N because (f, h) = (f + g, h). This moreover implies that f + g, h is a regular sequence in A too, see Algebra, Lemma 10.104.2. Now suppose that \text{div}(f + g ) = \sum m_ j (\mathfrak q_ j). Then consider the map
where \mathfrak q_ j^{(m_ j)} is the symbolic power, see Algebra, Section 10.64. Since A is normal, we see that A_{\mathfrak q_ i} is a discrete valuation ring and hence
Since V(f + g, h) = \{ \mathfrak m\} this implies that c becomes an isomorphism on inverting h (small detail omitted). Since h is a nonzerodivisor on A/(f + g) we see that the length of A/(f + g, h) equals the Herbrand quotient e_ A(A/(f + g), 0, h) as defined in Chow Homology, Section 42.2. Similarly the length of A/(h, \mathfrak q_ j^{(m_ j)}) equals e_ A(A/\mathfrak q_ j^{(m_ j)}, 0, h). Then we have
The equalities follow from Chow Homology, Lemmas 42.2.3 and 42.2.4 using in particular that the cokernel of c has finite length as discussed above. It is straightforward to prove that e_ A(\mathfrak q^{(m)}/\mathfrak q^{(m + 1)}, 0, h) is at least 1 by Nakayama's lemma. This finishes the proof of the lemma. \square
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