Total quotient ring and normalisation of a reduced plane curve germ

Fix a reduced complex-analytic plane curve germ $X$ at a point $p\in {\mathbb{C}}^2$, that is, a hypersurface germ in ${\mathbb{C}}^2$ given by an equation whose square-free reduction is itself (Complex-analytic hypersurface germ and its reduced equation). Write $I_p(X)$ for its vanishing ideal and define the local ring of the curve germ

$A:={\mathcal{O}}_{{\mathbb{C}}^2,p}\mathrm{/}{I}_p(X)\mathrm{.}$

By the principal vanishing-ideal lemma, choosing a reduced defining equation $f$ of $X$ gives $I_p(X)=(f)$ and hence $A={\mathcal{O}}_{{\mathbb{C}}^2,p}\mathrm{/}(f)$; in particular $A$ is the same ring for every reduced defining equation of $X$ (The vanishing ideal of a reduced hypersurface germ is principal).

Let

$S:=\{a\in A:a\text{ is a nonzerodivisor of }A\}$

be the set of nonzerodivisors. Then $S$ is a multiplicative subset of $A$ (Multiplicative subsets and the localisation $S^{-1}R$ as equivalence classes of fractions): $1\in S$, and if $s,t\in S$ and $stx=0$ for some $x\in A$, then $t(sx)=0$, so $sx=0$ because $t$ is a nonzerodivisor and then $x=0$ because $s$ is one; thus $st\in S$.

The total quotient ring of the curve germ is the localisation

$Q(A):=S^{-1}A,$

with its localisation map $A\to Q(A)$, $a\mapsto a\mathrm{/}1$. Since $S$ consists of the nonzerodivisors, this map is injective and every nonzerodivisor of $A$ becomes a unit in $Q(A)$; the ring $Q(A)$ is the largest localisation of $A$ in which the map is injective.

The normalisation of $A$ is the integral closure of $A$ in $Q(A)$: the set of elements of $Q(A)$ that are integral over $A$, i.e. roots of monic polynomials with coefficients in the image of $A$ (Integral elements over a commutative ring and algebraic integers, Integral closure in an extension ring and integrally closed domains). Explicitly, writing the localisation map as an inclusion,

$\bar{A}=\{b\in Q(A):b^k+a_1{b}^{k-1}+\cdots +a_k=0\text{ for some }k\ge 1\text{ and }{a}_i\in A\}\mathrm{.}$

The curve germ $X$ is normal when $A=\bar{A}$, that is, when $A$ is integrally closed in $Q(A)$.

Well-definedness. The ring $A$, and therefore the set $S$, the ring $Q(A)$ and the normalisation $\bar{A}$, depend only on the set germ $X$: the vanishing ideal $I_p(X)$ is attached to $X$, and the principal vanishing-ideal lemma identifies it with $(f)$ for every reduced defining equation $f$, so no choice of equation enters. The translation convention of Reduced holomorphic germ for a hypersurface identifies ${\mathcal{O}}_{{\mathbb{C}}^2,p}$ with the germ ring at the origin and transports the whole construction.

One branch and several branches. The ring $A$ is a domain exactly when the ideal $(f)=I_p(X)$ is a prime ideal of ${\mathcal{O}}_{{\mathbb{C}}^2,p}$. When $A$ is a domain, $Q(A)=\mathrm{Frac}(A)$ is its fraction field and the normalisation is the integral closure of $A$ in that fraction field, in agreement with Integral closure in an extension ring and integrally closed domains. When $X$ has several branches, $A$ has zero divisors, so no fraction field of $A$ exists; this is exactly why the ambient ring for integrality is the total quotient ring $Q(A)$, obtained by inverting precisely the nonzerodivisors. The product description of $Q(A)$ in terms of the branches of $X$, and the identification of the normalisation with the product of the normalisations of the branches, are proved in the next result on this page.

Nonzerodivisors and the localisation map. An element $a\in A$ is a nonzerodivisor exactly when the multiplication map $x\mapsto ax$ is injective, and this is the property that makes the localisation map $A\to S^{-1}A$ injective: $x\mathrm{/}1=0$ in $S^{-1}A$ means $ux=0$ for some $u\in S$, and then $x=0$ because $u$ is a nonzerodivisor. Thus $Q(A)$ contains $A$, and by the arithmetic of the localisation every $s\in S$ becomes a unit there (Multiplicative subsets and the localisation $S^{-1}R$ as equivalence classes of fractions); this is the ambient ring in which integrality is tested in the normalisation definition above.