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An ordinary node has two smooth branches

Example

In C2 with coordinates (x,y), let

f=y2−x2(1+x)=y2−x2−x3.

Near the origin the reduced curve X=Z(f) is the union of the two smooth branches y=±x1+x, whose equations are the Weierstrass polynomials y∓xs(x) for the holomorphic unit square root s of 1+x with s(0)=1. The branches meet only at the origin, where they have distinct tangent directions (1,±1), and the origin is the only singular point of X near 0; the germ is the ordinary node. Each branch carries the convergent parametrisation t↦(t,±ts(t)) with first coordinate t.

Facts & Assumptions

Given: The germ f=y2−x2(1+x)∈OC2,0 and its zero germ X=(Z(f),0).

[F1]

Holomorphic implicit function theorem: if F is holomorphic near (x0,w0) with F(x0,w0)=0 and ∂wF(x0,w0)≠0, then on a product of polydiscs around (x0,w0) the zero set of F is the graph w=φ(x) of a unique holomorphic function φ with φ(x0)=w0 (The holomorphic implicit function theorem).

[F2]

A Weierstrass polynomial of degree d in the last variable is monic of degree d with lower coefficients in the preceding germ ring vanishing at the base point; a germ regular in the last variable of order d is a unit times such a polynomial (Weierstrass polynomials in the last variable, Weierstrass preparation theorem).

[F3]

If f=gh in the germ ring and f is regular in the last variable, then g,h are regular and the product of their Weierstrass polynomials is the Weierstrass polynomial of f; in particular the degrees add, so a Weierstrass polynomial of degree 1 is not a product of two nonunits (Prepared factorizations correspond to germ factorizations).

[F4]

For a reduced germ f with factorisation f=u q1⋯qr into pairwise nonassociate irreducibles, the zero germ is Z(f)=⋃iZ(qi), the germs Z(qi) are exactly the irreducible components of Z(f), pairwise distinct and pairwise incomparable, and f is a reduced germ exactly when it is not divisible by the square of an irreducible; an irreducible germ is reduced (Finite unique irreducible components of a hypersurface germ, Irreducible hypersurface germs and their components, Reduced holomorphic germ for a hypersurface).

[F5]

A point q of a reduced hypersurface germ is regular exactly when the differential of a local reduced equation is nonzero at q, equivalently exactly when the germ is a holomorphic hypersurface graph near q; for a reduced germ h the vanishing ideal of Z(h) is (h) (Regular and singular points of an analytic hypersurface, The vanishing ideal of a reduced hypersurface germ is principal).

[F6]

The germ ring is a unique factorisation domain: factorisations into irreducibles are unique up to order and associates, and a nonzero nonunit is reduced exactly when all exponents in its factorisation equal 1 (The ring of holomorphic germs is a UFD, Unique factorisation domain, Irreducible and prime elements of an integral domain).

Proof technique: direct — construct the unit square root with the implicit function theorem, split the equation into its two degree-one Weierstrass factors, and locate the singular point with the gradient criterion.

Verification

1.1givenF1construct

Apply [F1] to F(x,w)=w2−(1+x) at the point (x0,w0)=(0,1): F(0,1)=0 and ∂wF(0,1)=2≠0, so there is a holomorphic function s on a neighbourhood of 0 with s(0)=1 and s(x)2=1+x identically. In particular s is a unit of OC,0 and s(x)≠0 for ∣x∣ small.

2.1givenF2F3algebra

Put g±:=y∓xs(x), the two germs determined by the unit s of step 1.1; each is a monic polynomial of degree 1 in y with coefficient ∓xs(x) in OC,0 vanishing at x=0. Each g± is irreducible in OC2,0. Suppose g+=gh with g,h nonunits. Since g+ is regular in y of order 1, [F3] makes g and h regular with Weierstrass polynomials G,H of positive degrees satisfying Wg+=GH; but Wg+=g+ has degree 1, while deg⁡G+deg⁡H≥2, a contradiction. The same argument applies to g−.

2.2step 1.1F2algebra

With s as in step 1.1, f=y2−x2s(x)2=(y−xs(x))(y+xs(x)). The two factors g±:=y∓xs(x) are Weierstrass polynomials of degree 1 in y: each is monic of degree 1 with its coefficient ∓xs(x) lying in OC,0 and vanishing at x=0. The germ f is itself a Weierstrass polynomial of degree 2: it is monic of degree 2 in y, its coefficients −x2(1+x) vanish at the origin, and f(0,y)=y2, so it is regular in y of order 2 and is its own Weierstrass preparation by [F2].

2.3step 1.1algebra

The zero germs of g+ and g− are distinct: Z(g+) is the graph of φ+(x)=xs(x) and Z(g−) is the graph of φ−(x)=−xs(x) over the x-coordinate. Since s(x)≠0 for ∣x∣ small by step 1.1, the two graphs intersect exactly where xs(x)=0, that is, only at x=0, and for every small x≠0 the values xs(x) and −xs(x) differ; hence the two set germs at the origin are distinct, and neither is contained in the other.

3.1step 2.1step 2.3F4F6

The germ f is reduced, and X=Z(f) has exactly the two irreducible components Z(g+) and Z(g−). Indeed f=g+g− exhibits f as a product of two nonassociate irreducibles, each occurring once; by the uniqueness of factorisation in the UFD [F6], no irreducible germ divides f twice, so f is reduced, and the factorisation of a reduced germ into pairwise nonassociate irreducibles has all exponents one and is unique up to order and associates. Applying the decomposition statement [F4] to f=g+g− gives X=Z(g+)∪Z(g−) with Z(g+) and Z(g−) the two irreducible components of X.

3.2step 1.1step 2.2F5construct

Each branch is a holomorphic hypersurface graph and carries an injective convergent parametrisation. For ∣t∣ small, γ+(t)=(t, ts(t)) is holomorphic with γ+(Δ)=Z(g+)∩{∣x∣<ρ} for a suitable radius ρ>0, since y−xs(x)=0 is exactly the graph y=xs(x); similarly γ−(t)=(t,−ts(t)) has image Z(g−)∩{∣x∣<ρ}. Both maps are injective because their first coordinate is t, and both are restrictions of the holomorphic function s of step 1.1, hence convergent. Every point of each branch is therefore regular as a point of that branch by [F5]. At a point other than the origin, step 2.3 separates the two graphs, so X locally equals the branch through that point and is regular there. This does not assert regularity of their union at the origin.

4.1

The origin is the only singular point of X near 0. By step 3.1 the reduced defining germ of X is f, and

df=(−2x(1+x)−x2,  2y)=(−x(2+3x),  2y)

vanishes at the origin, so the origin is singular by [F5]. Conversely let q=(x,y)∈X with q≠0; by step 3.1 the point lies on Z(g+) or on Z(g−), and by step 2.3 that forces x≠0. If q=(x,xs(x)) with x≠0 small, then df(q)=(−x(2+3x), 2xs(x)) has 2xs(x)≠0 because x≠0 and s(x)≠0; the same computation with −xs(x) gives df(q)≠0 on the other branch. At such a point the other factor g∓ is nonvanishing, so f is a unit times the local graph equation g±; hence f is a local reduced equation there. Thus every point of X other than the origin is regular by [F5], and the two branches meet there with distinct tangent directions (1,±1), since s(0)=1 makes their linear parts y=±x. [step 3.1, step 3.2, F5, algebra]

5.1step 3.2step 4.1∎

Steps 2.1 to 4.1 establish the assertions: y2−x2(1+x) is a reduced equation of X whose irreducible components are the two smooth branches y=±x1+x described by the unit square root s of 1+x; the origin is their only intersection and the only singular point of the germ, with distinct tangent directions, so X is an ordinary node; and each branch carries the convergent parametrisation t↦(t,±ts(t)) with first coordinate t.

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