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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