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Square-free reduction of a holomorphic equation

Statement

Let $n \geq 1$ , let $p \in C^{n}$ and let $f \in O_{C^{n}, p}$ be a nonzero nonunit. Then $f$ admits a square-free reduction: there are pairwise nonassociate irreducible germs $q_{1}, \dots, q_{r}$ and a unit $u$ with

$f = u q_{1}^{e_{1}} \dots q_{r}^{e_{r}}, e_{i} \geq 1, f_{r e d} : = q_{1} \dots q_{r},$

where $f_{r e d}$ is reduced in the sense of Reduced holomorphic germ for a hypersurface (no irreducible germ divides it twice). The associate class of $f_{r e d}$ depends only on $f$ : any other factorisation of $f$ into pairwise nonassociate irreducibles produces a product associate to $f_{r e d}$ . Moreover, on a neighbourhood of $p$ on which representatives of $f$ and $f_{r e d}$ are both defined, the two zero sets coincide:

$Z (f_{r e d}) = Z (f) .$

Facts & Assumptions

Given: A nonzero nonunit germ $f \in O_{C^{n}, p}$ .

[F1]

A nonzero nonunit germ is reduced when no irreducible element divides it twice; the zero and unit germs are excluded from hypersurface equations (Reduced holomorphic germ for a hypersurface).

[F2]

The holomorphic germ ring $O_{C^{n}, p}$ is a unique factorisation domain, hence an integral domain in which factorisations into irreducibles exist and are unique up to order and associates (The ring of holomorphic germs is a UFD, Unique factorisation domain).

[F3]

A nonzero nonunit of a unique factorisation domain has a factorisation $f = u q_{1}^{e_{1}} \dots q_{r}^{e_{r}}$ with $u$ a unit, the $q_{i}$ irreducible and pairwise nonassociate, and $r \geq 1$ ; the multiset of associate classes of the $q_{i}$ and the exponents are determined by $f$ (Unique factorisation domain).

Proof technique: direct — choose the UFD factorisation, drop repeated factors, and compare zero sets.

Proof

1.1givenF3

By [F3] choose a factorisation $f = u q_{1}^{e_{1}} \dots q_{r}^{e_{r}}$ with $u$ a unit, the $q_{i}$ pairwise nonassociate irreducible germs, $e_{i} \geq 1$ and $r \geq 1$ , and set $f_{r e d} : = q_{1} \dots q_{r}$ .

2.1step 1.1F2

The germ $f_{r e d}$ is a nonzero nonunit: it is a product of the nonunits $q_{i}$ in the domain of [F2], and a product of germs one of which is a nonunit cannot be a unit, while it is nonzero because a domain has no zero divisors and the $q_{i} \neq 0$ .

2.2step 1.1F3

The associate class of $f_{r e d}$ depends only on $f$ : if $f = v r_{1}^{g_{1}} \dots r_{s}^{g_{s}}$ is another factorisation into pairwise nonassociate irreducibles, then by uniqueness in [F3] the multiset ${(q_{i}, e_{i})}$ of associate classes with exponents equals ${(r_{j}, g_{j})}$ ; hence the set of associate classes occurring, and therefore the product $f_{r e d} = q_{1} \dots q_{r}$ up to a unit, is the same for the two factorisations.

3.1step 1.1step 2.1F1F3

No irreducible germ divides $f_{r e d}$ twice. Let $q$ be irreducible with $q^{2} ∣ f_{r e d}$ . Then $q ∣ q_{1} \dots q_{r}$ , and factoring the quotient $q_{1} \dots q_{r} / q$ into irreducibles exhibits $q \cdot (quotient)$ and $q_{1} \dots q_{r}$ as two irreducible factorisations of the same element; by uniqueness in [F3], $q$ is associate to one of the $q_{i}$ , say $q_{1}$ . But then $q_{1}^{2} ∣ f_{r e d}$ , so writing $f_{r e d} / q_{1}^{2}$ as a product of irreducibles and comparing with $q_{1} \dots q_{r}$ shows that the associate class of $q_{1}$ occurs at least twice among the classes of $q_{1}, \dots, q_{r}$ , contradicting their pairwise nonassociateness. Hence $f_{r e d}$ is reduced by [F1].

4.1step 1.1step 3.1F1

For the zero sets, put $k : = \max_{i} e_{i}$ and write the identities $f = f_{r e d} \cdot u \prod_{i} q_{i}^{e_{i} - 1}$ and $f_{r e d}^{k} = (u^{- 1} \prod_{i} q_{i}^{k - e_{i}}) f$ in the germ ring; after shrinking to a neighbourhood on which representatives of $f$ and $f_{r e d}$ are both defined, the first identity gives $Z (f_{r e d}) \subseteq Z (f)$ , and the second gives $Z (f) \subseteq Z (f_{r e d})$ , since a point with $f (x) = 0$ has $f_{r e d} (x)^{k} = 0$ and $C$ has no nilpotents. Therefore $Z (f_{r e d}) = Z (f)$ on that neighbourhood.

5.1step 3.1step 2.2step 4.1∎

Steps 3.1, 2.2 and 4.1 establish all three asserted properties of the square-free reduction $f_{r e d} = q_{1} \dots q_{r}$ .

Depends on

Used by

Complex-analytic hypersurface germ and its reduced equation Definition
Irreducible hypersurface germs and their components Definition
Regular and singular points of an analytic hypersurface Definition
A nonreduced equation can hide a smooth hypersurface Example
The coordinate axes form a reduced crossing Example
The vanishing ideal of a reduced hypersurface germ is principal Lemma
Finite unique irreducible components of a hypersurface germ Theorem

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Sources

Jiří Lebl, Tasty Bits of Several Complex Variables, Chapter 6 §§6.1–6.7 (standard reference, not scraped)
Jean-Pierre Demailly, Complex Analytic and Differential Geometry, Chapter II §§2, 4 and 6 (standard reference, not scraped)