TheoremStatement: AI-adaptedProof: AI-adaptedPipeline-generatedprecheck passjudge pass (gpt-6.1-sol)audited 2026-10-02

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Reduced hypersurface germs have pure codimension one

Statement

Assume the Axiom of Choice (The Axiom of Choice). Let $n \geq 1$ , let $p \in C^{n}$ and let $X$ be a nonempty reduced complex-analytic hypersurface germ at $p$ , with reduced defining germ $f$ . Then

$\dim_{p} X = n - 1$ , where the local dimension is the Krull dimension of the local ring $O_{C^{n}, p} / (f)$ (Local Krull dimension of a hypersurface germ);
every irreducible component of $X$ (Irreducible hypersurface germs and their components) has local dimension $n - 1$ and has a defining prime ideal of height one: writing the components as $Z (q_{i})$ with $q_{i}$ irreducible, the ideals $(q_{i})$ are prime and $ht ((q_{i})) = 1$ .

The statement concerns the principal ideal generated by a single reduced equation; it asserts nothing about arbitrary analytic ideals or set germs not cut out by one equation.

Facts & Assumptions

Given: The Axiom of Choice, a reduced nonzero nonunit germ $f$ at $p \in C^{n}$ , and its zero germ $X = Z (f)$ with irreducible factorisation $f = u q_{1} \dots q_{r}$ .

[F1]

The local dimension is $\dim_{p} X = \dim O_{C^{n}, p} / (f)$ for the reduced equation $f$ , and it equals the Krull dimension of that quotient (Local Krull dimension of a hypersurface germ, Krull dimension of a nonzero ring).

[F2]

The factorisation $f = u q_{1} \dots q_{r}$ exists with $u$ a unit and $q_{1}, \dots, q_{r}$ pairwise nonassociate irreducibles; $X = ⋃_{i} Z (q_{i})$ , each $Z (q_{i})$ is an irreducible hypersurface germ, and these are exactly the irreducible components of $X$ , uniquely determined with their number $r \geq 1$ (Finite unique irreducible components of a hypersurface germ, Irreducible hypersurface germs and their components).

[F3]

For an ideal $I$ with $R / I \neq 0$ , $\dim (R / I) = \sup {n : p_{0} ⊊ \dots ⊊ p_{n} is a strict chain of primes all containing I}$ (Dimension of a quotient via chains above an ideal).

[F4]

Each $q_{i}$ is a nonzero germ, so some invertible complex-linear map $T_{i}$ makes $T_{i} (q_{i})$ regular in the last variable of some order $d_{i} \geq 1$ ; then $T_{i} (q_{i}) = v_{i} W_{i}$ with $v_{i}$ a unit and $W_{i}$ a Weierstrass polynomial of degree $d_{i}$ , and $O_{C^{n}, 0} / (W_{i})$ is a finitely generated $O_{C^{n - 1}, 0}$ -module generated by $1, z_{n}, \dots, z_{n}^{d_{i} - 1}$ , with $O_{C^{0}, 0} = C$ when $n = 1$ (After a linear coordinate change, every nonzero germ is regular in the last variable, Weierstrass preparation theorem, A quotient by a Weierstrass polynomial is a finite module over the smaller germ ring).

[F5]

The induced map $O_{C^{n - 1}, 0} \to O_{C^{n}, 0} / (W_{i})$ is injective: if $h \in O_{C^{n - 1}, 0}$ lies in $(W_{i})$ , say $h = W_{i} g$ , then dividing $h$ by $W_{i}$ both as $h = 0 \cdot W_{i} + h$ and as $h = W_{i} g + 0$ and invoking uniqueness of the Weierstrass remainder forces $h = 0$ (Weierstrass division theorem).

[F6]

Assume the Axiom of Choice. For an injective integral extension $A \subseteq B$ of nonzero commutative rings one has $\dim A = \dim B$ ; a finite module extension is integral, so a ring that is a finitely generated module over a subring is integral over it (Injective integral extensions preserve Krull dimension, Integrality and finite-module characterizations for one element).

[F7]

Assume the Axiom of Choice. $\dim O_{C^{m}, 0} = m$ for every $m \geq 0$ (Krull dimension of the holomorphic germ ring).

[F8]

Assume the Axiom of Choice. In a Noetherian commutative ring, if $x$ is a nonzerodivisor and $p$ is a prime ideal minimal over $(x)$ , then $ht (p) = 1$ (A minimal prime over a principal nonzerodivisor has height one). The germ ring is Noetherian (The ring of holomorphic germs is Noetherian) and a domain in which a nonzero irreducible germ is prime (The ring of holomorphic germs is a UFD, Irreducible holomorphic germs are prime).

Proof technique: direct — prove each branch quotient has dimension $n - 1$ by a finite integral extension, compare prime chains for the union, and apply the principal ideal height theorem.

Proof

1.1givenF1F2

By [F1] and [F2] the local dimension of $X$ is $\dim O_{C^{n}, p} / (f)$ , that of the component $Z (q_{i})$ is $\dim O_{C^{n}, p} / (q_{i})$ , and $X = ⋃_{i} Z (q_{i})$ with the $Z (q_{i})$ the irreducible components; note $r \geq 1$ .

2.1step 1.1F4F5F6F7

For every $i$ one has $\dim O_{C^{n}, p} / (q_{i}) = n - 1$ . By [F4] choose the linear coordinates $T_{i}$ , the degree $d_{i} \geq 1$ and the Weierstrass polynomial $W_{i}$ ; the ring automorphism induced by the invertible linear change carries $(q_{i})$ to $(T_{i} q_{i}) = (W_{i})$ , so $\dim O / (q_{i}) = \dim O / (W_{i})$ . The residue classes $1, \dots, z_{n}^{d_{i} - 1}$ generate $O / (W_{i})$ as an $O_{C^{n - 1}, 0}$ -module by [F4], and the structure map is injective by [F5]; hence $O_{C^{n - 1}, 0} \subseteq O / (W_{i})$ is an injective integral extension of nonzero rings by [F6], so $\dim O / (W_{i}) = \dim O_{C^{n - 1}, 0} = n - 1$ by [F6] and [F7].

2.2step 1.1F8

For every $i$ the ideal $(q_{i})$ is a prime ideal of height one. It is prime because $q_{i}$ is irreducible and irreducible germs are prime by [F8]; it is trivially minimal over itself, its generator $q_{i}$ is a nonzero nonzerodivisor since $O$ is a domain, and the ring is Noetherian by [F8]; therefore $ht ((q_{i})) = 1$ by the height-one corollary in [F8].

3.1step 1.1step 2.1F2F3F8

The local dimension of $X$ is $n - 1$ . Every prime ideal $p$ containing $(f)$ contains the product $q_{1} \dots q_{r}$ of the irreducible factors up to the unit $u$ , hence contains some $(q_{j})$ by primality of the $q_{j}$ in [F8]; consequently, in a strict chain of primes all containing $(f)$ , the smallest member already contains some $(q_{j})$ , so every member of the chain contains $(q_{j})$ and the chain is a chain of primes containing $(q_{j})$ . Conversely every chain of primes containing $(q_{j})$ contains $(f) \subseteq (q_{j})$ . By [F3] this gives $\dim O / (f) = \max_{i} \dim O / (q_{i}) = n - 1$ by step 2.1.

4.1step 3.1step 2.1step 2.2F1F6F7F8∎

Steps 1.1, 2.1, 2.2 and 3.1 establish all assertions: $\dim_{p} X = \dim O / (f) = n - 1$ ; each irreducible component $Z (q_{i})$ has local dimension $\dim O / (q_{i}) = n - 1$ and defining prime $(q_{i})$ of height one; and $r \geq 1$ with the components uniquely determined. The Axiom of Choice is used exactly through the integral-extension dimension preservation [F6], the numerical dimension of the germ ring [F7] and the height-one corollary [F8], as declared in the Statement; no choice is used beyond these.

Depends on

Used by

Singular locus of a reduced analytic hypersurface Theorem

Dependency tree · two levels

69 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

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)