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✓ 2 results · all verified · 2 also independently AI-judged
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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 z00z11−z01z10, so the image is the quadric surface V+(z00z11−z01z10)⊆P3. On z00≠0, 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 Z0Z2−Z12=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 p01p23−p02p13+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 dim⁡V=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(x−y)⊆Ak1×Ak1.

Counterexample

1.1givenalgebra

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

2.1step 1.1

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

3.1step 2.1∎

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.

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 Pn−1; for fixed [x] its fibre is the projective space of hyperplanes through [x], also Pn−1 when n≥1. 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.

1.1givenalgebra

As an R-algebra, C≅R[t]/(t2+1), so C⊗RC≅C[t]/(t2+1).

2.1step 1.1algebra

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

3.1step 2.1∎

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.

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 s2−itiu3−jvj(0≤i≤2, 0≤j≤3). 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