Alphabeta Math
Pipeline-generated
How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

2 results · all verified · 2 also independently AI-judged
Every result on this page is machine-checked by a proof checker and read in full and owner-audited; the judge is an additional, independent cross-model AI review of the proofs; all 2 also cleared it.

Products Segre and Veronese Embeddings and Grassmannians — Examples

1 · Prerequisites

2 · Summary

These coordinate calculations exercise the constructions on the preceding page. They are leaves: none supplies a premise for later material.

3 · Logical flowchart

4 · Definitions, theorems and proofs

None yet.

5 · Examples, counterexamples and false statements

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The Segre image of P1 times P1 is a quadric surface

Example

For ([s:t],[u:v])P1×P1, Segre gives [z00:z01:z10:z11]=[su:sv:tu:tv]. The sole 2×2 minor is z00z11z01z10, so the image is the quadric surface V+(z00z11z01z10)P3. On z000, its inverse sends [z] to ([z00:z10],[z00:z01]); the other three charts give the analogous formulas.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The quadratic Veronese image of P1 is a plane conic

Example

The quadratic Veronese map is [s:t][s2:st:t2]=[Z0:Z1:Z2]. Its image has equation Z0Z2Z12=0. The line aZ0+bZ1+cZ2=0 pulls back to the binary quadratic as2+bst+ct2=0, illustrating the hyperplane-section correspondence.

ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Lines in P3 and the Klein quadric Gr(2,4)

Example

A line in P3 is a 2-plane in k4. With Plucker coordinates p01,p02,p03,p12,p13,p23, the unique Plucker relation is p01p23p02p13+p03p12=0. Thus Gr(2,4) is the Klein quadric in P5.

ExampleConstruction: AI-generatedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The boundary Grassmannians Gr(0,V) and Gr(dim V,V)

Example

If dimV=n, the only 0-plane is {0} and the only n-plane is V, so both Gr(0,V) and Gr(n,V) are one point. Their Plucker spaces are P(Λ0V)=P(k) and P(ΛnV) respectively, again one point. There is no r-dimensional subspace when r<0 or r>n, so the stipulated Grassmannian is empty in those cases.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The product of Zariski topologies is too coarse on A1 times A1

Statement refuted

For every field k, the Zariski topology of Ak1×Ak1 equals the product of the two factor Zariski topologies.

Facts & Assumptions

Given: An infinite field k and the diagonal D=V(xy)Ak1×Ak1.

Counterexample

1.1

The affine-product coordinate ring is k[x,y], so D is Zariski closed.

givenalgebra
2.1

Each factor has the cofinite Zariski topology. For a point (a,b) off D, every basic product neighbourhood U×V has UV. Thus it contains (c,c)D for some cUV.

step 1.1
3.1

Hence no product-topology neighbourhood of (a,b) lies in the complement of D. The complement is not open, so D is not closed in the product topology, although it is closed in the affine-product Zariski topology. This refutes the statement.

step 2.1
ExampleConstruction: Literature-sourcedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

The incidence variety of a point and a hyperplane

Example

In Pn×(Pn), write a point as [x0::xn] and a hyperplane as [a0::an]. Incidence is the closed bilinear equation i=0naixi=0. For fixed [a] its fibre is that hyperplane Pn1; for fixed [x] its fibre is the projective space of hyperplanes through [x], also Pn1 when n1. When n=0, the unique point of P0 lies on no hyperplane: the incidence locus and both fibres are empty.

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

Tensor products of domains need not be domains over a nonclosed field

Statement refuted

The tensor product of two domains over every field is a domain.

Counterexample

Given: The domains C over the non-algebraically-closed field R.

1.1

As an R-algebra, CR[t]/(t2+1), so CRCC[t]/(t2+1).

givenalgebra
2.1

In C[t], t2+1=(ti)(t+i) with distinct factors, so the Chinese remainder calculation gives C[t]/(t2+1)C×C.

step 1.1algebra
3.1

The two nonzero coordinate idempotents have zero product, so this tensor product is not a domain. This refutes the statement and explains the base-field hypothesis in the affine product theorem.

step 2.1
ExampleConstruction: AI-generatedVerification: AI-adaptedjudge pass (gpt-5.6-terra)audited 2026-09-07Open item page →

A bidegree (2,3) Segre-Veronese embedding

Example

For ([s:t],[u:v])P1×P1, the bidegree (2,3) coordinates are the twelve products s2itiu3jvj(0i2, 0j3). Scaling the two representatives by λ,μ scales every coordinate by λ2μ3. Up to the target coordinate permutation determined by the chosen Veronese monomial orderings, this is σ(ν1,2×ν1,3), hence is the stated closed embedding.

Sources