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An irreducible plane curve gives a connected punctured covering

Statement

Let W∈OC2,0 be a Weierstrass polynomial of degree m≥1 in the variable y (Weierstrass polynomials in the last variable), so that

W(x,y)=ym+am−1(x)ym−1+⋯+a0(x),aj∈OC,0, aj(0)=0,

and assume that W is reduced (Reduced holomorphic germ for a hypersurface) and irreducible in OC2,0. Then there is ε>0 such that, writing D∗:={x∈C:0<∣x∣<ε}, the zero set Z(W)∩(D∗×C) is a connected m-sheeted unramified covering of D∗; moreover every zero tends to the origin over the base point:

(xn,yn)∈Z(W), xn→0⟹yn→0.

Facts & Assumptions

Given: A reduced irreducible Weierstrass polynomial W of degree m≥1 in y.

[F1]

W is monic of degree m in y with coefficients in OC,0 vanishing at the origin, so W(0,y)=ym and W is regular in y of order m (Weierstrass polynomials in the last variable).

[F2]

For any r>0, after shrinking the coefficient disc V one has ∑j<m∣aj(x)∣rj−m<1 for x∈V, since all aj(0)=0. If ∣y∣≥r, then ∑j<m∣aj(x)yj∣<∣y∣m, so W(x,y)≠0. Thus every root of every slice over V lies in D={∣y∣<r}; this estimate uses only [F1].

[F3]

Since W itself is a reduced prepared polynomial, DW=Disc⁡y(W) is a nonzero germ (Reduced preparation has nonzero discriminant). For x0∈V, DW(x0)=0 exactly when the slice has a repeated root (The discriminant is ∏i<j(αi−αj)2 and vanishes exactly when a monic polynomial has a repeated root). On a root-containing representative this is the branch set of the fixed projection, as in Discriminant and branch set of a fixed Weierstrass projection.

[F4]

A monic polynomial of degree m over C has exactly m roots counted with multiplicity; hence for x0 with DW(x0)≠0 the slice W(x0,⋅) has exactly m distinct roots, all of them simple (A complex polynomial of degree n has exactly n roots counted with multiplicity, The discriminant is ∏i<j(αi−αj)2 and vanishes exactly when a monic polynomial has a repeated root).

[F5]

A nonzero holomorphic germ of one variable has finite order: either it is a unit or it equals xku with k≥1 and u a unit; consequently its zeros near 0 are isolated, and only x=0 can be a zero of the germ (The order of a zero is the exponent in its local holomorphic factorization).

[F6]

If x0∈V and τ is a simple root of W(x0,⋅), then near (x0,τ) the zero set of W is the graph of the unique holomorphic function φ with W(x,φ(x))=0 and φ(x0)=τ, by the implicit function theorem applied to ∂yW(x0,τ)≠0 (The holomorphic implicit function theorem).

[F7]

A holomorphic function on a punctured disc that is bounded extends holomorphically across the puncture (Characterizations of removable singularities).

[F8]

A holomorphic function on a connected open set in several variables that vanishes on a nonempty open subset vanishes identically (A holomorphic function vanishing on a nonempty open subset of a domain vanishes identically).

[F9]

If W=W1W2 with W1 and W2 Weierstrass polynomials of positive degree, then this gives a nontrivial factorization in OC2,0; this is the implication needed below (Prepared factorizations correspond to germ factorizations).

[F10]

A covering map has fibres whose points lie in pairwise disjoint sheets, each mapped homeomorphically onto the same evenly covered open set (Covering maps, evenly covered neighbourhoods, fibres, sheets, and trivial coverings).

Proof technique: contradiction — separate the covering into two open-and-closed parts, form the monic product of the roots in one part, and read a nontrivial Weierstrass factorisation of W.

Proof

1.1givenF1F2F3F4F5F6

Choose a disc V and radius r>0 as in [F2], and shrink V to {∣x∣<ε} so that DW(x)≠0 on its punctured part D∗, using [F3] and [F5]. By [F4] each slice over D∗ has exactly m distinct simple roots, all in D by [F2]. At any base point [F6] supplies a holomorphic graph through each root. Intersect the finitely many base neighbourhoods and shrink until these graphs stay in D and are pairwise disjoint. They exhaust each fibre, since a degree-m polynomial has at most m roots by [F4]. Thus they give an evenly covered neighbourhood with m holomorphic sheets. This proves directly, in the fixed coordinates, that Z(W)∩(D∗×C)=Z(W)∩(D∗×D) is an m-sheeted unramified covering.

2.1step 1.1F1F2F4

Every zero over D∗ lies in D by step 1.1. Let (xn,yn) be any sequence of zeros of the chosen representative with xn→0, allowing xn=0; for all sufficiently large n, xn∈V, and when xn=0 one has yn=0 by [F1]. Thus the tail of (yn) lies in the closed disc D‾. For any convergent subsequence ynk→y∞, continuity of the polynomial W on a neighbourhood of {0}×D‾ gives W(0,y∞)=lim⁡kW(xnk,ynk)=0, so [F1] gives y∞m=0 and y∞=0, even if the limit was initially allowed to lie on ∂D. If yn did not tend to 0, a subsequence bounded away from 0 would have a convergent subsequence in D‾ with nonzero limit, a contradiction. Hence yn→0.

3.1step 2.1assume-contra

Suppose for contradiction that the total space F:=Z(W)∩(D∗×D) is disconnected, so that F=F1⊔F2 with F1,F2 nonempty, open and closed in F.

4.1step 2.1step 3.1F10

The function k(x):=∣F1∩π−1(x)∣ is locally constant on D∗: if U⊆D∗ is a disc over which the covering trivialises with sheets V1,…,Vm, then each Vi is connected by [F10], and F1∩Vi is open and closed in Vi because F1 is open and closed in F; hence Vi⊆F1 or Vi∩F1=∅ for each i, so k is constant on U. As D∗ is connected, k is constant, say k(x)=k for all x∈D∗, and 1≤k≤m−1 because F1 and F2 are nonempty.

5.1step 4.1F3F4F6construct

On such a disc U the sheets of F1 are graphs of holomorphic functions φi:U→D by [F6]; define W1(x,y):=∏i∈I(y−φi(x)) on U×D, where I is the set of sheets contained in F1, so W1 is monic of degree k in y with holomorphic coefficients on U. For two discs U,U′ the definitions agree on U∩U′, because at each x∈U∩U′ both are the monic degree-k polynomial in y whose k roots are the distinct points of F1 over x by [F3] and [F4]; hence W1 is a well-defined holomorphic function on D∗×D, monic of degree k in y. Defining W2 in the same way from F2, we get a monic holomorphic function of degree m−k on D∗×D with k+(m−k)=m.

6.1step 5.1F3F4

For every x∈D∗, the two monic polynomials W1(x,⋅)W2(x,⋅) and W(x,⋅) in y have the same m distinct roots, hence are equal; therefore W=W1W2 on D∗×D.

6.2step 1.1step 5.1F7

The coefficients of W1 and W2 are, up to sign, the elementary symmetric functions of the corresponding root values; by step 1.1 all roots lie in the bounded disc D, so every coefficient is a bounded holomorphic function on the punctured disc D∗, and [F7] extends each coefficient holomorphically across the puncture.

7.1step 2.1step 6.2F1

The extended coefficients of W1 and W2 vanish at x=0: the roots of W1 and of W2 all tend to 0 as x→0 by step 2.1, and the elementary symmetric functions are continuous in the roots, so each coefficient has limit 0. Hence the extensions are Weierstrass polynomials of degrees k≥1 and m−k≥1.

8.1step 6.1step 7.1F8

The functions W,W1,W2 are holomorphic on the polydisc V×D and satisfy W=W1W2 on the nonempty open subset D∗×D, so by [F8] the identity holds on V×D; hence W=W1W2 in OC2,0 with W1,W2 Weierstrass polynomials of positive degree.

9.1step 8.1F9contradiction

Step 8.1 gives a factorization of W into positive-degree Weierstrass polynomials. The implication recorded in [F9] makes this a nontrivial germ factorization, contradicting the assumed irreducibility of W; therefore F is connected.

10.1step 1.1step 2.1step 9.1discharge-contradiction∎

By step 1.1 the local covering from step 9.1 is the full zero set Z(W)∩(D∗×C); it is connected and m-sheeted, unramified by step 1.1, and every sequence of its zeros whose base coordinates tend to 0 has fibre coordinates tending to 0 by step 2.1.

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