Reduced Cartier multiples on a normal local surface

The minimal primes of $(a)$ are minimal in $\mathrm{Supp}(A\mathrm{/}aA)$, hence associated by [F4] and finite by [F3]. By [F12] each has height at most one; since $A$ is a domain and $a\mathrm{\ne }0$, each has height exactly one. Any height-one prime containing $a$ is one of these minimal primes, so list them as ${\mathfrak{p}}_1,\dots ,{\mathfrak{p}}_r$. Choose nonzero $u_i\in {\mathfrak{p}}_i$ and put $f={\prod }_i{u}_i\mathrm{\ne }0$, so $f\in {\bigcap }_i{\mathfrak{p}}_i\subseteq \mathfrak{m}$. The minimal primes of $(f)$ are finite by [F3, F4] and have height one by [F12]. Since $\mathfrak{m}$ is not contained in any of them, finite prime avoidance [F7] gives $g\in \mathfrak{m}$ outside their union. If a prime contains $(f,g)$, it contains a minimal prime of $(f)$ and strictly contains it, so in the two-dimensional local ring it must be $\mathfrak{m}$. Thus $(f,g)$ is $\mathfrak{m}$-primary, and ${\mathfrak{m}}^u\subseteq (f,g)$ for some $u\ge 1$.

For every $h\in {\mathfrak{m}}^{u+1}$ the ideal $(f+h,g)$ equals $(f,g)$: writing $h=\alpha f+\beta g$ with $\alpha ,\beta \in \mathfrak{m}$ gives $f+h=(1+\alpha )f+\beta g$, and $1+\alpha$ is a unit, so the two ideals coincide. Define the finite positive integer $\mathrm{\ell }:={\mathrm{length}}_{A}A\mathrm{/}(f+h,g)={\mathrm{length}}_{A}A\mathrm{/}(f,g);$ it is independent of $h$.

For each $h\in {\mathfrak{m}}^{u+1}$ put $Q_h=A\mathrm{/}(f+h)$. By step 1.2, $(f+h,g)$ is $\mathfrak{m}$-primary, so $f+h$ is nonzero; normality gives $A$ the $S_2$ property and hence Cohen--Macaulayness by [F11]. The element $f+h$ is regular in the domain $A$, so [F6] shows that $Q_h$ is a one-dimensional Cohen--Macaulay local ring. Its element $g$ is a parameter and therefore regular. Take a prime filtration of $Q_h$ itself, with factors $A\mathrm{/}{\mathfrak{P}}_i$. For each factor put ${\chi }_g(A\mathrm{/}{\mathfrak{P}}_i):=\mathrm{length}((A\mathrm{/}{\mathfrak{P}}_i)\mathrm{/}g(A\mathrm{/}{\mathfrak{P}}_i))-\mathrm{length}(0{:}_{A\mathrm{/}{\mathfrak{P}}_i}g)\mathrm{.}$ The kernel and cokernel have finite length, and the snake-lemma sequence for multiplication by $g$ shows that ${\chi }_g$ is additive along the filtration. If $\dim A\mathrm{/}{\mathfrak{P}}_i=0$, multiplication by $g$ is an endomorphism of a finite-length module, so its kernel and cokernel have the same length and ${\chi }_g=0$. If $\dim A\mathrm{/}{\mathfrak{P}}_i=1$, then ${\mathfrak{P}}_i$ is a minimal prime of $Q_h$ and $g\mathrm{\notin }{\mathfrak{P}}_i$; hence $g$ is a nonzero divisor on the domain $A\mathrm{/}{\mathfrak{P}}_i$ and ${\chi }_g={\mathrm{length}}_{A}A\mathrm{/}({\mathfrak{P}}_i,g)\ge 1$. Since $g$ is regular on $Q_h$, ${\chi }_g(Q_h)={\mathrm{length}}_{Q_h}{Q}_h\mathrm{/}g{Q}_h=\mathrm{\ell }$. Every minimal prime of $Q_h$ occurs as a factor in the filtration after localizing at that prime. Thus the number of one-dimensional factors, and hence the number of minimal primes of $Q_h$, is at most $\mathrm{\ell }$.

The perturbation set is nonempty: choose nonzero elements from the finitely many height-one primes ${\mathfrak{p}}_i$, multiply them, and then multiply by a nonzero element of ${\mathfrak{m}}^{u+1}$. Choose $h\in {\mathfrak{m}}^{u+1}\cap {\mathfrak{p}}_1\cap \cdots \cap {\mathfrak{p}}_r$ for which the number $s$ of minimal primes of $Q_h$ is maximal; this is possible because the set of possible counts is a nonempty finite set of integers bounded by $\mathrm{\ell }$. Replace $f$ by $f^{\mathrm{\prime }}=f+h$, and let ${\mathfrak{q}}_1,\dots ,{\mathfrak{q}}_s$ be the height-one primes of $A\mathrm{/}{f}^{\mathrm{\prime }}A$. Every original ${\mathfrak{p}}_i$ is one of these: $f^{\mathrm{\prime }}\in {\mathfrak{p}}_i$, so a minimal prime of $f^{\mathrm{\prime }}A$ lies inside ${\mathfrak{p}}_i$, and both have height one.

For each $j$ construct $a_j\in A$ with $v_{{\mathfrak{q}}_j}(a_j)=1$ and $a_j\mathrm{\notin }{\mathfrak{q}}_i$ for $i\mathrm{\ne }j$: the local ring $A_{{\mathfrak{q}}_j}$ is a discrete valuation ring, so choose $b_j\in {\mathfrak{q}}_j$ of valuation one, choose $c_j\in {\mathfrak{q}}_j$ outside all other ${\mathfrak{q}}_i$ by finite prime avoidance, and if $c_j$ does not have valuation one take $d_j$ in the product of the other ${\mathfrak{q}}_i$ and outside ${\mathfrak{q}}_j$; then $a_j=c_j^2+b_j{d}_j$ has valuation one at ${\mathfrak{q}}_j$ and is a unit modulo every other ${\mathfrak{q}}_i$.

Let $h^{\mathrm{\prime }}$ be the product of $a_j^2$ over the indices with $v_{{\mathfrak{q}}_j}(f^{\mathrm{\prime }})=1$ and of $a_j$ over the indices with $v_{{\mathfrak{q}}_j}(f^{\mathrm{\prime }})>1$, multiplied by an element of ${\mathfrak{m}}^{u+1}$ lying outside every ${\mathfrak{q}}_j$, which exists by finite prime avoidance. Then $h^{\mathrm{\prime }}\in {\mathfrak{m}}^{u+1}$ and, because every original ${\mathfrak{p}}_i$ occurs among the ${\mathfrak{q}}_j$ by step 3.1, $h^{\mathrm{\prime }}\in {\bigcap }_i{\mathfrak{p}}_i$. The element $c:=f^{\mathrm{\prime }}+h^{\mathrm{\prime }}$ has valuation exactly one at each ${\mathfrak{q}}_j$. No new height-one prime occurs: such a prime would be an additional minimal prime of $A\mathrm{/}(f^{\mathrm{\prime }}+h^{\mathrm{\prime }})$, contradicting the maximal $s$ in step 3.1 because $h^{\mathrm{\prime }}$ belongs to the same perturbation set.

We have $c\in \mathfrak{m}$; it is nonzero because $v_{{\mathfrak{q}}_j}(c)=1$ for each $j$. Normality gives $A$ the $S_2$ property by [F11], so $A$ is Cohen--Macaulay of dimension two; since $A$ is a domain, $c$ is regular. Thus $A\mathrm{/}cA$ is a one-dimensional Cohen--Macaulay local ring by [F6]. Its associated primes are therefore exactly its minimal primes ${\mathfrak{q}}_1,\dots ,{\mathfrak{q}}_s$; each localization $A_{{\mathfrak{q}}_j}\mathrm{/}c{A}_{{\mathfrak{q}}_j}$ is a field because the valuation of $c$ there is one. By detection of a ring at its associated primes, $A\mathrm{/}cA$ is reduced.

The height-one primes ${\mathfrak{p}}_i$ containing $a$ all occur among the ${\mathfrak{q}}_j$, because $f^{\mathrm{\prime }}\in {\bigcap }_i{\mathfrak{p}}_i$ shows ${\mathfrak{p}}_i\supseteq f^{\mathrm{\prime }}A$ and hence contains a minimal prime of $f^{\mathrm{\prime }}A$, and both have height one. Choosing $n\ge {\max }_i{v}_{{\mathfrak{p}}_i}(a)$ makes $c^n\mathrm{/}a$ have nonnegative valuation at every height-one prime of $A$; by the $S_2$ intersection property of the normal domain $A$, an element of $\mathrm{Frac}(A)$ with nonnegative valuation at every height-one prime lies in $A$. Hence $a\mid c^n$.

The element $c\in \mathfrak{m}$ has $A\mathrm{/}cA$ reduced by step 6.1 and $a\mid c^n$ by step 7.1; the Axiom of Choice and the Axiom of Dependent Choice are inherited from the cited commutative-algebra suppliers, with no further choice used.