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.

✓ 10 results · all verified · 1 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. The 9 not AI-judged were verified by owner audit (typically over a confirmed judge false positive), not failures.

Proj Projective Schemes Twisting Sheaves and Ampleness — Examples

1 · Prerequisites

2 · Summary

This companion page records examples and counterexamples for the Proj, twisting-sheaf and ampleness development on the main page. Every chart computation is carried out in its own item, including the empty, nilpotent, rank-zero and characteristic-two cases.

For the polynomial ring, the standard charts of Proj⁡ k[x0,…,xn] have rings k[x0/xi,…,xn/xi] and the overlap change is the displayed ratio of coordinates; for n=0 the space is a single point. The twist transitions on Pk1 are e1=tne0 with t=x1/x0, including negative n, where the homogeneous unit x1n is the frame. A nilpotent irrelevant ideal makes Proj⁡ empty even though the spectrum is not, and the unit section of OX has zero ideal OX and empty zero scheme. A nonzero homogeneous equation cuts the hypersurface Proj⁡(k[x]/(F))⊂Pkn with chart ring k[xa/xi]/(F/xideg⁡F).

Three items delimit the concepts. Γ(Pk1,O(−1))=0, so an invertible sheaf need not be globally generated; the second Veronese T=k[x2,xy,y2] has the same Proj⁡ as k[x,y] but is not isomorphic to it as a graded algebra, so Proj⁡ forgets the grading; and the structure sheaf is globally generated but not very ample, since no immersion into projective space pulls O(1) back to OPk1. The degree-two Veronese map exhibits the conic Z0Z2−Z12=0 as the scheme-theoretic image of Pk1, and the projective bundle of a trivial module gives PS(OSr)≅PSr−1 for r≥1, with the rank-zero case empty.

3 · Logical flowchart

4 · Definitions, theorems and proofs

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passjudge pass (gpt-6-sol)audited 2026-09-30Open item page →

Polynomial Proj charts

Example

Assume the Axiom of Choice (The Axiom of Choice). Let k be a field and n≥0, and give S=k[x0,…,xn] the total-degree grading with deg⁡xi=1. Then Proj⁡S=Pkn, the chart D+(xi) is Spec⁡k[x0/xi,…,xn/xi] with the variable xi/xi omitted (it equals 1), and on the overlap D+(xixj) the coordinate change between the i-th and j-th charts sends xaxj=xa/xixj/xi, so it is the transition formula xa(i)↦xa(j)/xi(j) of the published charts of Pkn (Relative projective space from standard charts). For n=0 the space is the one-point scheme Spec⁡k.

Facts & Assumptions

Given: The Axiom of Choice, A field k, an integer n≥0, the graded polynomial ring S=k[x0,…,xn] with deg⁡xi=1, and the scheme Pkn with its standard charts.

[F1]

Proj⁡k[x0,…,xn]≅Pkn canonically over Spec⁡k, with D+(xi) corresponding to the i-th standard chart Ui=Spec⁡k[xℓ(i):ℓ≠i] and transition isomorphisms xℓ(i)↦xℓ(j)/xi(j), xj(i)↦1/xi(j) on Ui∩Uj. (Projective space is Proj of a polynomial ring, Relative projective space from standard charts)

[F2]

For homogeneous f∈S of positive degree the chart map D+(f)→Spec⁡S(f) is an isomorphism of schemes. (Standard opens are affine)

[F3]

For a field k the scheme Spec⁡k has exactly one point, namely (0). (The spectrum of a field is a one-point affine scheme)

[F4]

The assumed Axiom of Choice is the choice-function principle (The Axiom of Choice); it licenses the AC-qualified Proj and associated-sheaf suppliers at step 2.1.

Verification

technique · direct: compute the degree-zero localisations of the polynomial ring at the variables and match them with the published charts and transition formulas
1.1algebra

Chart coordinates. Fix i. A degree-zero element of S[xi−1] has the form a/xid with a homogeneous of degree d, and every monomial x0a0⋯xnan of degree d gives x0a0⋯xnan/xid=∏ℓ≠i(xℓ/xi)aℓ; hence S(xi)=k[xℓ/xi:ℓ≠i], the polynomial ring in the n variables xℓ(i):=xℓ/xi.

2.1F1F2F4step 1.1

The charts. Under the assumed AC [F4], by [F2] the chart D+(xi) is Spec⁡S(xi)=Spec⁡k[xℓ/xi:ℓ≠i], which is exactly the i-th standard chart Ui=Spec⁡k[xℓ(i):ℓ≠i] of Pkn under the identification xℓ(i)=xℓ/xi of [F1].

2.2F1step 1.1algebra

The overlap. On D+(xixj) both xi and xj are invertible, so the relation xaxj=xa/xixj/xi is an identity of regular functions in the localised rings; expressed in the coordinates of step 1.1 it reads xa(j)=xa(i)/xj(i), which is exactly the transition formula of the published charts in [F1], together with xi(j)=1/xj(i) for a=i. Hence the overlapping charts are glued by the same isomorphisms.

2.3F1F3step 1.1cases: n=0

The case n=0. For n=0 we have S=k[x0] with S(x0)=k by step 1.1 with n=0 variables, so Proj⁡k[x0]=D+(x0)=Spec⁡k, which is the one-point scheme of [F3]; equivalently Pk0=Spec⁡k in the published charts.

3.1

Conclusion. Steps 2.1 and 2.2 identify the charts and gluing of Proj⁡k[x0,…,xn] with those of Pkn, in agreement with the canonical isomorphism of [F1], and step 2.3 settles n=0; the displayed coordinate change is the transition formula of the published charts. [F1, step 2.1, step 2.2, step 2.3] \qed

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-09-30Open item page →

A nilpotent irrelevant ideal gives empty Proj

Example

Let k be a field and let S=k[ε]/(ε2) be graded by deg⁡ε=1 and deg⁡k=0, so that S0=k and S+=(ε) with ε2=0. Then S+ is a nilpotent ideal, Proj⁡S=∅, and yet Spec⁡S is nonempty: the ring k[ε]/(ε2) has the single prime ideal (ε). So emptiness of Proj is not emptiness of the spectrum, and it is detected by the nilpotency of the irrelevant ideal.

Facts & Assumptions

Given: A field k, the graded ring S=k[ε]/(ε2) with deg⁡ε=1.

[F1]

Proj⁡S is the set of homogeneous prime ideals p with S+⊈p; a prime contains every nilpotent element; and S=⨁d≥0Sd with S1=kε and Sd=0 for d≥2. (Points of Proj of a graded ring)

[F2]

Proj⁡S=∅ if and only if every homogeneous element of S+ is nilpotent; if S+ is finitely generated this is equivalent to S+ being nilpotent. (Empty Proj and irrelevant torsion)

[F3]

In the ring k[ε]/(ε2) every prime ideal contains ε, so (ε) is the unique prime, and it is maximal; the localisation Sε is the zero ring. [algebra]

Verification

technique · direct: identify $S_+$, apply the emptiness criterion, and exhibit the point of $\operatorname{Spec}S$
1.1algebra

The irrelevant ideal is nilpotent. Here S+=kε consists of the multiples of ε, and S+2=kε2=0; hence every element of S+ is nilpotent and S+=(ε) is finitely generated.

1.2F1F3algebra

The standard chart is empty. For the homogeneous element ε of degree 1 we have S(ε)=(S[ε−1])0, and S[ε−1]=0 because ε is nilpotent; hence D+(ε)=Spec⁡S(ε)=Spec⁡0=∅, and since ε generates S+ this is the only standard open.

2.1F1F2step 1.1

Proj is empty. By step 1.1 every homogeneous element of S+ is nilpotent, so [F2] gives Proj⁡S=∅; equivalently, any homogeneous prime p⊆S contains the nilpotent ε, hence contains S+=(ε) and is excluded from Proj⁡S.

3.1F1F3algebra

The spectrum is nonempty. In k[ε]/(ε2) the element ε is nilpotent but nonzero, so the ideal (ε) is proper; every prime contains the nilpotent ε, so (ε) is the unique prime ideal and Spec⁡S={(ε)}≠∅, in contrast with step 2.1.

4.1

Conclusion. Steps 1.1 and 2.1 show that the nilpotent irrelevant ideal produces empty Proj, while step 3.1 shows that the underlying ring still has a point; the two conclusions are consistent because Proj discards exactly the primes containing all of S+. [F1, F2, step 2.1, step 3.1] \qed

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-09-30Open item page →

Twist transitions on the projective line

Example

Assume the Axiom of Choice (The Axiom of Choice). Let k be a field and S=k[x0,x1] with deg⁡x0=deg⁡x1=1, so that Pk1=Proj⁡S with charts U0=D+(x0) and U1=D+(x1), and let t=x1/x0, the coordinate on U0. Then for every integer n the twisting sheaf O(n)=S(n)~ has frames e0=x0 n on U0,e1=x1 n on U1, and on the overlap U0∩U1 these frames are related by e1=t ne0. Here for n<0 the symbols xin denote the corresponding units xin∈S[x0−1] or S[x1−1], and the frames are nowhere-vanishing local generators of the invertible sheaf O(n) (Invertible twists for degree-one generated rings).

Facts & Assumptions

Given: The Axiom of Choice, A field k, the graded ring S=k[x0,x1] with deg⁡xi=1, an integer n, and the charts Ui=D+(xi) of Pk1.

[F1]

Pk1=Proj⁡S, Ui=D+(xi)=Spec⁡S(xi) with S(x0)=k[t], t=x1/x0, and S(x1)=k[t−1]; the overlap is D+(x0x1)=Spec⁡k[t,t−1]. (Projective space is Proj of a polynomial ring)

[F2]

O(n)=S(n)~ has sections Γ(Ui,O(n))=S(n)(xi)=(S(n)[xi−1])0, the degree-zero part of the homogeneous localisation, and restrictions are the canonical localisations. (Twisting sheaf on Proj)

[F3]

Since S is generated over k by S1, every O(n) is invertible, with frame xin on D+(xi): S(n)(xi)=S(xi)⋅xin is free of rank one. (Invertible twists for degree-one generated rings)

[F4]

The assumed Axiom of Choice is the choice-function principle (The Axiom of Choice); it licenses the AC-qualified Proj and associated-sheaf suppliers at step 1.1.

Verification

technique · direct: compute the degree-zero localisations of the shifted modules and compare the resulting local generators on the overlap
1.1F1F2F3F4algebra

The chart modules. The AC premise [F4] licenses the associated-sheaf and Proj charts [F1]–[F3]. For n∈Z the module S(n)(x0)=(S(n)[x0−1])0 consists of the classes a/x0k with a∈S(n)k=Sn+k homogeneous of degree n+k. Since x0 is a unit in the localisation, every such class equals (a/x0 n+k)⋅x0 n with a/x0 n+k∈S(x0)=k[t]; hence e0:=x0 n generates S(n)(x0) over k[t], and symmetrically e1:=x1 n generates S(n)(x1) over k[t−1].

2.1F1F2step 1.1algebra

The overlap. On the overlap D+(x0x1) the ring is k[t,t−1] with t=x1/x0, so x1=tx0 and therefore x1 n=t nx0 n holds in the localisation of S at x0x1 for every integer n, positive or negative; under the identifications of step 1.1 this is precisely the frame relation e1=t ne0 on U0∩U1.

3.1

Conclusion. The frame section ei is nowhere vanishing on Ui, and the transition relation e1=tne0 is exactly the change of frame of the invertible sheaf O(n) from the 0-chart to the 1-chart: for n=0 both frames are the constant function 1 and the relation is e1=e0; for n=1 it is e1=te0; for n=−1 it is e1=t−1e0 with t−1=x0/x1 the coordinate on U1. [F2, F3, step 1.1, step 2.1, cases: n=0 and negative n] \qed

ExampleConstruction: AI-generatedVerification: AI-adaptedprecheck passaudited 2026-09-30Open item page →

A nowhere-vanishing section has empty zero divisor

Example

Let X be a scheme and let 1∈Γ(X,OX) be the unit section of the structure sheaf, regarded as a section of the invertible sheaf OX (Zero scheme of a line-bundle section). Then the zero ideal of 1 is the unit ideal sheaf I1=OX, and its zero scheme is empty: Z(1)=V(I1)=∅. Moreover the empty closed subscheme is the effective Cartier divisor with unit local equation 1. This holds for every scheme X, including X=∅ and including schemes whose structure sheaf has nilpotents or zero divisors.

Facts & Assumptions

Given: A scheme X, the invertible sheaf OX, its global unit section 1, and the contraction map cs:OX−1→OX of Zero scheme of a line-bundle section.

[A1]

The Axiom of Choice states that every family of nonempty sets has a choice function. (The Axiom of Choice)

[F1]

For a global section s∈Γ(X,L) of an invertible sheaf L, the contraction map cs:L−1→OX is defined by φ↦φ(s), its image Is=Im⁡(cs) is a quasi-coherent sheaf of ideals, and the zero scheme is the closed subscheme Z(s)=V(Is)↪X; on an affine open U=Spec⁡A on which L is trivialised with local equation f, the map cs∣U becomes multiplication by f and Z(s)∩U=Spec⁡(A/(f)). (Zero scheme of a line-bundle section)

[F2]

If the section s vanishes nowhere, then each local equation f is a unit, so Is=OX and the induced closed immersion Z(s)→X is an isomorphism onto the empty subscheme; one says Z(s)=∅. Moreover Z(s) is an effective Cartier divisor precisely when each local equation is a nonzerodivisor or a unit, and if every local equation is a unit then Z(s)=∅ is the empty divisor. (Zero scheme of a line-bundle section)

Verification

technique · direct: compute the contraction map of the unit section in every trivialisation and evaluate the local chart formula $Z(s)\cap U=\operatorname{Spec}(A/(f))$ with $f=1$
1.1F1

The contraction is the identity. Take L=OX and s=1. The dual is L−1=OX and the pairing OX⊗OX→OX is multiplication, so c1:OX→OX sends a local function g to g⋅1=g; that is, c1=idOX. Consequently I1=Im⁡(c1)=OX, the unit ideal sheaf.

2.1F1step 1.1

The zero scheme is empty. Let U=Spec⁡A⊆X be any affine open; over U the structure sheaf is trivialised by the identity and the local equation of the unit section is f=1∈A. By the local chart formula of [F1], Z(1)∩U=Spec⁡(A/(1))=Spec⁡0=∅. Since the affine opens cover X, the closed subscheme Z(1) has no points; it is the empty closed subscheme.

3.1F2step 1.1step 2.1

The empty divisor. In the trivialisation of step 2.1 the local equation f=1 is a unit of A, hence in particular a nonzerodivisor, and I1=OX is an invertible sheaf of ideals; by the criterion of [F2] the zero scheme Z(1)=∅ is an effective Cartier divisor, namely the empty divisor, whose local equation is the unit 1.

4.1

Conclusion and empty scheme. Steps 1.1, 2.1 and 3.1 give I1=OX and Z(1)=∅ with unit local equation, so the empty closed subscheme of X is an effective Cartier divisor. If X=∅ then OX is the zero sheaf, Γ(X,OX)=0 and the unit section is 1=0, the unit of the zero ring; the same computation gives I1=0=OX and Z(1)=∅=X, and since OX is invertible (the zero sheaf is locally free of rank one on the empty scheme, where there is no point to test) the conclusion holds vacuously for the empty base as well. No hypothesis on X beyond the trivialisations of OX enters; in particular nilpotents or zero divisors in OX do not affect the computation, which uses only multiplication by 1. The Axiom of Choice [A1] is inherited from the affine quotient and gluing suppliers of [F1]; no choice is made here. [A1, F1, F2, step 1.1, step 2.1, step 3.1, cases: empty X and units as local equations] \qed

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-09-30Open item page →

A projective hypersurface as a homogeneous quotient

Example

Let k be a field, n≥0 and let 0≠F∈k[x0,…,xn] be homogeneous of degree e≥1. Then Proj⁡(k[x0,…,xn]/(F))=V+(F)↪Pkn is the closed subscheme of the projective space cut out by F, and on the standard chart D+(xi) its coordinate ring is k[xa/xi:a≠i]/(F/xie). The description includes the degenerate cases: for n=0 one has F=cx0e with c≠0 and the chart ring k/(c)=0, so V+(F)=∅.

Facts & Assumptions

Given: A field k, an integer n≥0, the graded polynomial ring B=k[x0,…,xn] with deg⁡xi=1, a nonzero homogeneous F∈B of degree e≥1, and the homogeneous principal ideal I=(F)⊆B.

[A1]

The Axiom of Choice states that every family of nonempty sets has a choice function. (The Axiom of Choice)

[F1]

For a commutative ring A, a homogeneous ideal I⊆A[x0,…,xn] determines a closed subscheme V+(I)↪PAn whose intersection with the chart D+(xi) is Spec⁡(B(xi)/I(xi)), where I(xi)=(I[xi−1])0; equivalently V+(I)=Proj⁡(B/I) under the canonical closed immersion Proj⁡(B/I)→Proj⁡B=PAn. (Closed subschemes of projective space and saturated ideals)

[F2]

There is a canonical isomorphism Proj⁡A[x0,…,xn]≅PAn over Spec⁡A, and over the field k the standard chart D+(xi) has coordinate ring k[xa(i):a≠i] with xa(i)=xa/xi. (Projective space is Proj of a polynomial ring, Closed subschemes of projective space and saturated ideals)

Verification

technique · direct: read the hypersurface off the homogeneous quotient theorem, compute the degree-zero localisation of the principal ideal at each chart variable, and identify the resulting equation with $F/x_i^{e}$
1.1A1F1F2

The quotient and the closed subscheme. Since F is homogeneous, I=(F) is a homogeneous ideal, so by [F1] the subscheme V+(I)↪Pkn exists and equals Proj⁡(B/I)=Proj⁡(k[x0,…,xn]/(F)); by [F2] the ambient space Proj⁡B is Pkn with charts D+(xi).

2.1F1F2step 1.1algebra

The chart ring. Fix i. By [F2] the chart ring of Pkn on D+(xi) is B(xi)=k[xa(i):a≠i] with xa(i)=xa/xi. In the localisation B[xi−1] the element xi is a unit and F=xie⋅(F/xie), so the ideal generated by F is generated by the degree-zero element F/xie; hence I(xi)=(I[xi−1])0=(F/xie), and the chart ring of V+(I) on D+(xi) is B(xi)/I(xi)=k[xa(i):a≠i]/(F/xie), that is, k[xa/xi:a≠i]/(F/xideg⁡F).

3.1

Conclusion and degenerate cases. Steps 1.1 and 2.1 identify Proj⁡(k[x0,…,xn]/(F)) with V+(F)⊆Pkn and compute its chart rings. If F=cxie is a nonzero multiple of a single coordinate power, then on the i-th chart the equation is the unit c and the chart ring is k[xa(i)]/(c)=0, so that chart meets V+(F) in the empty scheme; in particular for n=0 one has F=cx0e and V+(F)=Spec⁡0=∅, the empty hypersurface. If F is not such a monomial then for every i the element F/xie is the dehomogenisation of F, a nonzero element of the polynomial ring k[xa(i):a≠i] that is not a unit because some monomial of F has a positive exponent at a variable other than xi; each chart is then the genuine affine hypersurface k[xa(i)]/(F/xie), which is a nonzero ring. The zero polynomial is excluded by hypothesis. The Axiom of Choice [A1] is inherited from the affine quotient and gluing suppliers of [F1]; no choice is made here. [A1, F1, F2, step 1.1, step 2.1, cases: empty chart and n=0] \qed

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-09-30Open item page →

O(-1) has no global generator

Counterexample

Assume the Axiom of Choice (The Axiom of Choice). Let k be a field and let O(−1) be the twisting sheaf on Pk1=Proj⁡k[x0,x1] (Invertible twists for degree-one generated rings, Projective space is Proj of a polynomial ring). Then Γ(Pk1,O(−1))=0, so the evaluation morphism Γ(Pk1,O(−1))⊗ZOPk1⟶O(−1) is the zero morphism, which is not surjective because O(−1) is a nonzero sheaf. Hence O(−1) is not globally generated (Global generation by the evaluation map), even though it is invertible.

Facts & Assumptions

Given: The Axiom of Choice, A field k, the graded ring S=k[x0,x1] with deg⁡xi=1, the scheme Pk1=Proj⁡S with charts U0=D+(x0), U1=D+(x1), and the sheaf O(−1)=S(−1)~.

[F1]

U0=Spec⁡k[t] with t=x1/x0 and U1=Spec⁡k[t−1]; the overlap is D+(x0x1)=Spec⁡k[t,t−1], and restriction of sections is the canonical localisation. (Projective space is Proj of a polynomial ring, Twisting sheaf on Proj)

[F2]

On U0 one has Γ(U0,O(−1))=S(−1)(x0)=k[t]⋅x0−1 and on U1 one has Γ(U1,O(−1))=S(−1)(x1)=k[t−1]⋅x1−1: in the localisation, x0−1 and x1−1 are units of degree (−1), with x1−1=t−1x0−1. (Twisting sheaf on Proj)

[F3]

O(−1) is invertible, in particular nonzero on the nonempty scheme Pk1: its restriction to Ui is free of rank one with frame xi−1. (Invertible twists for degree-one generated rings)

[F4]

A global section of a sheaf on Pk1 is exactly a pair of chartwise sections on U0 and U1 whose restrictions to U0∩U1 agree; the global section is zero exactly when both chartwise sections are zero. (Twisting sheaf on Proj)

[F5]

A sheaf M is globally generated if the evaluation morphism Γ(X,M)⊗ZOX→M is surjective; the zero morphism out of a zero module is not surjective onto a sheaf with a nonzero stalk. (Global generation by the evaluation map)

[F6]

The Axiom of Choice is the choice-function principle (The Axiom of Choice). It licenses the AC-qualified supplier used at step 1.1.

Refutation

technique · direct: compute both chartwise modules of global sections of $\mathcal O(-1)$ and show that agreement on the overlap forces both to vanish
1.1F1F2F6

A global section has two chart expressions. Let s∈Γ(Pk1,O(−1)). Under the AC premise [F6], by [F2] its restriction to U0 has the form s0=a(t) x0−1 with a∈k[t], and its restriction to U1 has the form s1=b(t−1) x1−1 with b∈k[t−1]; these are finite polynomials a(t)=∑m≥0αmtm and b(u)=∑m≥0βmum with u=t−1.

1.2F1F2algebra

Agreement on the overlap. By [F4] the two expressions agree on U0∩U1, where x1=tx0 is invertible. Substituting x1−1=t−1x0−1 turns the agreement into the identity b(t−1)=t a(t) in k[t,t−1]. The right-hand side is a finite sum of monomials tm+1 with m≥0, so it involves only strictly positive powers of t; the left-hand side ∑m≥0βmt−m involves only nonpositive powers of t. Comparing coefficients in the basis {tj:j∈Z} of k[t,t−1] gives αm=0 for all m and βm=0 for all m.

2.1F2F4step 1.1step 1.2

Vanishing of all global sections. By step 1.1 every global section is given by its two chart expressions, and by step 1.2 those expressions have a=0 and b=0; hence s0=0 and s1=0, so s=0 by [F4]. Therefore Γ(Pk1,O(−1))=0.

3.1

Failure of global generation. With Γ(Pk1,O(−1))=0 the evaluation morphism of [F5] is the zero morphism; since O(−1) is invertible and Pk1≠∅, it has a nonzero stalk at every point and the zero morphism is not surjective. Hence O(−1) is not globally generated, although it is invertible by [F3]. This is the standard contrast with the positive twists: O(1) is generated by its two coordinate sections x0,x1. [F3, F5, step 2.1] \qed

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-09-30Open item page →

Two graded rings with the same Proj

Counterexample

Assume the Axiom of Choice (The Axiom of Choice). Let k be a field and let S=k[x,y] be graded by total degree, so deg⁡x=deg⁡y=1, with second Veronese regrading T=S(2)=⨁n≥0S2n=k[x2,xy,y2] (graded so that S2n sits in degree n; Nonnegatively graded rings and modules, homogeneous elements, and twists, Proj is invariant under Veronese regrading). Then:

  1. Proj⁡S≅Proj⁡T canonically, and under this isomorphism the twist OT(1) corresponds to OS(2), not to OS(1);
  2. nevertheless S and T are not isomorphic as graded k-algebras: dim⁡kS1=2 while dim⁡kT1=3.

So Proj⁡S does not determine the graded ring up to graded isomorphism, and the twist data must be carried separately .

Facts & Assumptions

Given: The Axiom of Choice, A field k, the graded polynomial ring S=k[x,y] with deg⁡x=deg⁡y=1, its second Veronese regrading T=S(2)=⨁n≥0S2n, and the scheme Proj⁡S=Pk1.

[F1]

For every commutative nonnegatively graded ring R and every d≥1 there is a canonical isomorphism of schemes Proj⁡R≅Proj⁡R(d) mapping the chart D+(f), for homogeneous f∈R+ of positive degree, to D+(fd) with the same coordinate ring, and under it the twist OProj⁡R(d)(1) corresponds to OProj⁡R(d). (Proj is invariant under Veronese regrading)

[F2]

A nonnegatively graded ring is a commutative ring R=⨁n≥0Rn with RiRj⊆Ri+j; a homomorphism of graded rings is a ring homomorphism carrying Rn into Rn′ for every n, so an isomorphism of graded k-algebras restricts to a k-linear isomorphism of degree-one parts. (Nonnegatively graded rings and modules, homogeneous elements, and twists)

[F3]

Proj⁡k[x0,x1]=Pk1 with the standard charts D+(xi), so the two constructions of the counterexample take place on the same scheme. (Projective space is Proj of a polynomial ring)

[F4]

On Pk1 the twist O(n) has frames xn on D+(x) and yn on D+(y), related by yn=tnxn on the overlap, where t=y/x; the chart rings are k[t], k[t−1] and k[t,t−1] respectively. (Twist transitions on the projective line)

[F5]

The Axiom of Choice is the choice-function principle (The Axiom of Choice). It licenses the AC-qualified supplier used at step 2.1.

Verification

technique · direct: compute the degree-one parts of the two rings, apply Veronese invariance with $d=2$ to identify the Proj's and the twists, and rule out a graded isomorphism by comparing dimensions
1.1F2algebra

The rings and their degree-one parts. By [F2] the ring S=k[x,y] with deg⁡x=deg⁡y=1 is nonnegatively graded with S1=kx⊕ky, of dimension 2, and the Veronese T=S(2)=⨁nS2n is the graded k-subalgebra k[x2,xy,y2]⊆S generated by the three degree-two monomials: every S2n is spanned by the monomials xay2n−a=(x2)a′(xy)b(y2)c with 2a′+b=a and b+2c=2n−a. Hence T1=S2=kx2⊕kxy⊕ky2, of dimension 3, so dim⁡kS1=2 and dim⁡kT1=3.

1.2F3F4algebra

The twists are different. By [F4], a global section of OS(n) for n=1 or 2 is a pair a(t)xn on D+(x) and b(t−1)yn on D+(y), where a∈k[t], b∈k[t−1], and the overlap condition is a(t)=tnb(t−1). This holds exactly when a(t) has degree at most n, with b(t−1)=t−na(t). Hence dim⁡kΓ(Pk1,OS(1))=2 and dim⁡kΓ(Pk1,OS(2))=3. An isomorphism of these sheaves would give an isomorphism of their global-section k-vector spaces, which is impossible. Thus the two twists are not isomorphic.

2.1F1F3F5step 1.1

Same Proj. Under the AC premise [F5], since T=S(2), [F1] with R=S and d=2 gives a canonical isomorphism Proj⁡S≅Proj⁡T which maps the chart D+(f) of Proj⁡S to the chart D+(f2) of Proj⁡T, and under which OT(1)=OProj⁡T(1) corresponds to OProj⁡S(2); by [F3] the scheme Proj⁡S is the projective line Pk1.

2.2F2step 1.1

No graded isomorphism. Suppose φ:S→T is an isomorphism of graded k-algebras, that is, a k-algebra isomorphism with φ(Sn)=Tn for all n; by [F2] it restricts to a k-linear isomorphism S1→T1, so dim⁡kS1=dim⁡kT1. By step 1.1 this would require 2=3, which is impossible; hence S and T are not isomorphic as graded k-algebras.

3.1

Conclusion. Steps 2.1 and 2.2 exhibit the two graded k-algebras S=k[x,y] and T=k[x2,xy,y2] with canonically isomorphic Proj but no graded isomorphism between them; steps 2.1 and 1.2 show that the isomorphism matches OT(1) with OS(2) and not with OS(1), so not even the degree-one twists correspond. Since k[x,y] and its Veronese are nonnegatively graded with nonzero degree-one parts, neither Proj is empty and the invariant dim⁡k(−)1 is defined; the case d=1 of [F1] is excluded here because the two rings are then equal, while d=2≥1 is the smallest regrading for which S(d)≠S in this example. No choice principle is used beyond the inherited Proj construction. [F1, F2, step 1.1, step 2.1, step 2.2, step 1.2, cases: d=1 excluded and d=2 smallest] \qed

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-09-30Open item page →

The conic map from O(2)

Example

Let k be a field and let Pk1 have homogeneous coordinates x0,x1, with twisting sheaf O(2)=O(1)⊗2 (Relative very ampleness in the finite projective-space convention). Then:

  1. the three global sections x02,  x0x1,  x12 generate O(2) (Global generation by the evaluation map) and define a closed immersion ν2:Pk1⟶Pk2,[x0:x1]⟼[x02:x0x1:x12], the degree-two Veronese, with ν2∗O(1)≅O(2) (Veronese embedding pulls O(1) back to O(d));
  2. the image of ν2 is the plane conic V+(Z0Z2−Z12)⊆Pk2, where Z0,Z1,Z2 are the target coordinates; on the chart Z0≠0 the map is [1:x1/x0]↦[1:x1/x0:(x1/x0)2], so ν2 identifies Pk1 with that conic;
  3. since ν2 is a closed immersion, the scheme-theoretic image of ν2 is exactly the conic V+(Z0Z2−Z12).

The computation is valid over an arbitrary field, with no restriction on the characteristic: the conic equation Z0Z2−Z12 and the kernel computations below are polynomial identities with integer coefficients.

Facts & Assumptions

Given: A field k, the projective line Pk1 with coordinates x0,x1 and twisting sheaf O(2), the projective plane Pk2 with coordinates Z0,Z1,Z2 and twisting sheaf O(1), and the Axiom of Choice as inherited from the projective-space and sheaf constructions.

[A1]

The Axiom of Choice states that every family of nonempty sets has a choice function. (The Axiom of Choice)

[F1]

(Veronese in degrees n=1, d=2.) The monomial sections s(2,0)=x02, s(1,1)=x0x1, s(0,2)=x12 generate O(2), and the associated morphism ν2:Pk1→Pk2 is a closed immersion with ν2∗O(1)≅O(2) carrying the target coordinate ym to sm; here N=(1+22)−1=2. (Veronese embedding pulls O(1) back to O(d))

[F2]

For a morphism φ attached to generating sections tm of an invertible sheaf: φ−1(D+(ym))=Xtm and, on the chart where tm0 is a trivialising section, the chart coordinates satisfy (ym/ym0)∘φ=tm/tm0. In particular, if φ∗ym=tm for all m then the ratios of the sections are the ratios of their pullbacks. (Generating line-bundle sections define a morphism to projective space, Veronese embedding pulls O(1) back to O(d))

[F3]

On the standard chart Ui=D+(xi) of Pk1 the sheaf O(1) has frame xi, so O(2) has frame xi2, the section xi is a unit on Ui, and Xxi=Ui; the charts U0,U1 are Spec⁡k[x1(0)] and Spec⁡k[x0(1)] with x1(0)=x1/x0 and x0(1)=x0/x1 on the overlap. The standard charts of Pk2 are the three affine planes Spec⁡k[Z1/Z0,Z2/Z0], Spec⁡k[Z0/Z1,Z2/Z1], Spec⁡k[Z0/Z2,Z1/Z2]. (Relative projective space from standard charts, Relative very ampleness in the finite projective-space convention)

[F4]

For a commutative ring A, an integer n≥0, B=A[x0,…,xn] and a homogeneous ideal I⊆B, the closed subscheme V+(I)↪PAn has V+(I)∩D+(xi)=Spec⁡(B(xi)/I(xi)); every closed subscheme of PAn is recovered from its chart ideals, and V+(I)=V+(J) as closed subschemes exactly when I and J have the same saturation. (Closed subschemes of projective space and saturated ideals)

[F5]

Kernel computations over a field k: the k-algebra homomorphism k[u,v]→k[t] with u↦t, v↦t2 has kernel (v−u2), because k[u,v]/(v−u2)≅k[u] via elimination of v and k[u]→k[t], u↦t, is injective; the homomorphism k[u,v]→k[t,t−1] with u↦t−1, v↦t has kernel (uv−1) by the same elimination, using k[u,v]/(uv−1)≅k[u,u−1]; and the homomorphism k[a,b]→k[t−1] with a↦t−2, b↦t−1 has kernel (a−b2). [algebra]

[F6]

A morphism of affine schemes whose associated ring map is surjective with kernel K has image the closed subscheme Spec⁡(k[u,v]/K), and a closed immersion is in particular injective, so its image is the closed subscheme it defines. (Closed subschemes of projective space and saturated ideals, Immersion of schemes)

Verification

technique · direct: invoke the Veronese theorem in degree two on the projective line, read the chart formulas for the ratios of the monomial sections, compute the chart rings of the conic $Z_0Z_2-Z_1^2$, and compare them with the chartwise images of the morphism
1.1F1F3

The monomials and the morphism. Put M={(2,0),(1,1),(0,2)}, so ∣M∣=3=(1+22) and N=2, and set sm=xm. By [F1] the sections s(2,0)=x02, s(1,1)=x0x1, s(0,2)=x12 generate the invertible sheaf O(2), and ν2:Pk1→Pk2 is a closed immersion with ν2∗O(1)≅O(2), carrying the target coordinate ym to sm; write Z0=y(2,0), Z1=y(1,1), Z2=y(0,2).

1.2F1F2F3algebra

The chart formulas for ν2. On U0=D+(x0) the section s(2,0)=x02 is a frame of O(2) by [F3], and by [F2] ν2−1(D+(Z0))=Xs(2,0)=U0, with (Z1/Z0)∘ν2=s(1,1)/s(2,0)=x1/x0 and (Z2/Z0)∘ν2=s(0,2)/s(2,0)=(x1/x0)2 on U0. Symmetrically on U1=D+(x1) one has ν2−1(D+(Z2))=U1, (Z0/Z2)∘ν2=(x0/x1)2 and (Z1/Z2)∘ν2=x0/x1, while on the overlap U0∩U1=Xs(1,1) one has (Z0/Z1)∘ν2=x0/x1 and (Z2/Z1)∘ν2=x1/x0. In particular, on the chart Z0≠0 the morphism sends a point with coordinate u=x1/x0 to [1:u:u2], which is the displayed formula [x0:x1]↦[x02:x0x1:x12].

1.3F4algebra

The conic and its chart rings. Let F=Z0Z2−Z12∈k[Z0,Z1,Z2], homogeneous of degree 2. By [F4] the closed subscheme V+(F)⊆Pk2 has chart ideals generated by the dehomogenisations: (F)(Z0)=(Z2/Z0−(Z1/Z0)2), (F)(Z1)=((Z0/Z1)(Z2/Z1)−1) and (F)(Z2)=(Z0/Z2−(Z1/Z2)2), so its chart rings are k[u,v]/(v−u2) with u=Z1/Z0, v=Z2/Z0; k[u,v]/(uv−1) with u=Z0/Z1, v=Z2/Z1; and k[a,b]/(a−b2) with a=Z0/Z2, b=Z1/Z2.

2.1F5F6step 1.2step 1.3algebra

The image on each target chart. On D+(Z0) the morphism ν2 restricts on U0 to the morphism corresponding to the k-algebra map k[u,v]→k[t], u↦x1/x0=t, v↦t2 by step 1.2, whose kernel is (v−u2) by [F5]; hence the image of U0 is the closed subscheme cut out by v−u2, which is exactly V+(F)∩D+(Z0) by step 1.3, and U0→V+(F)∩D+(Z0) is an isomorphism. On D+(Z1) the restriction corresponds on U0∩U1 to k[u,v]→k[t,t−1], u↦t−1, v↦t, with kernel (uv−1); on D+(Z2) the restriction corresponds on U1 to k[a,b]→k[t−1], a↦t−2, b↦t−1, with kernel (a−b2). In each case the image chart is the corresponding chart of V+(F) from step 1.3 and the restriction is an isomorphism onto it.

3.1F1F4step 1.3step 2.1

The image is the conic. The morphism ν2 is a closed immersion by [F1], so its image is a closed subscheme Z⊆Pk2; by [F4] such a closed subscheme is recovered from its chart ideals. Step 2.1 computes the chart of Z over each of D+(Z0), D+(Z1), D+(Z2) to be the corresponding chart of V+(F) computed in step 1.3, so Z=V+(Z0Z2−Z12): the image of the Veronese ν2 is exactly the conic Z0Z2=Z12, and ν2 identifies Pk1 with it.

4.1

Conclusion. Steps 1.1 and 1.2 show that the global sections x02,x0x1,x12 generate O(2) and define the degree-two Veronese closed immersion [x0:x1]↦[x02:x0x1:x12] with ν2∗O(1)≅O(2), and steps 1.3 to 3.1 identify its image, hence its scheme-theoretic image, with the conic Z0Z2−Z12=0. No division by 2 or by any other nonzero scalar occurs: the quadratic equation is integral and the kernels (v−u2), (uv−1), (a−b2) of [F5] are computed by elimination of a variable in every characteristic, so the verification is uniform, including characteristic two. The Axiom of Choice [A1] is inherited from the Veronese and projective-space suppliers; the only objects chosen are the three monomials and the three target charts, so no choice is made here. [A1, F1, F5, step 1.2, step 3.1, cases: characteristic two and general characteristic] \qed

CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-09-30Open item page →

Global generation does not imply very ampleness

Counterexample

Let k be a field and let X=Pk1=Proj⁡k[x0,x1] with its structure sheaf OX (Projective space is Proj of a polynomial ring). Then:

  1. OX is globally generated by its unit section 1∈Γ(X,OX) (Global generation by the evaluation map);
  2. OX is not H-very ample over Spec⁡k (Relative very ampleness in the finite projective-space convention): the datum (OX;1) with its single generating section 1 has n=0 and corresponds to the constant structure morphism X→Pk0=Spec⁡k, and for no n≥0 is there a quasi-compact Spec⁡k-immersion i:X→Pkn with i∗O(1)≅OX.

So global generation of an invertible sheaf does not imply relative very ampleness, even over a field. This distinguishes global generation from relative very ampleness.

Facts & Assumptions

Given: A field k, the scheme X=Pk1=Proj⁡S with S=k[x0,x1] graded by total degree, the standard opens D+(x0), D+(x1) and the coordinate t=x1/x0 on D+(x0), the structure sheaf OX=S~, and the Axiom of Choice as inherited from the projective-space and sheaf constructions.

[A1]

The Axiom of Choice states that every family of nonempty sets has a choice function. (The Axiom of Choice)

[F1]

For homogeneous f∈S+ of positive degree the standard open D+(f) is the affine chart Spec⁡S(f) of X; the charts D+(x0) and D+(x1) cover X with overlap D+(x0x1); and Γ(D+(f),OX(n))=S(n)(f), with restrictions induced by homogeneous localisation, so that Γ(D+(f),OX)=S(f) for the structure sheaf. (Standard opens of Proj, Proj carries a scheme structure, Twisting sheaf on Proj, Sections of a graded-module sheaf on a standard open)

[F2]

A section of a sheaf on X is the same as a compatible family of sections on the members of an open cover, and compatible local sections glue uniquely; the charts U0=D+(x0) and U1=D+(x1) are affine, hence quasi-compact, so their union X is quasi-compact, and the structure morphism X→Spec⁡k is quasi-compact. (A sheaf on a topological space, Every affine scheme is quasi-compact, A closed subspace of a compact space is compact, and a finite union of compact subspaces is compact, Quasi-compact and quasi-separated schemes, Quasi-compact and quasi-separated morphisms)

[F3]

The structure sheaf OX is invertible; for a section s of an invertible sheaf the nonvanishing locus Xs is the set of points at which the image of s in the fibre is nonzero; and the evaluation morphism of OX with its unit section is the canonical identification. (Invertible sheaves, Absolute ampleness by affine section opens, Global generation by the evaluation map)

[F4]

Pkn has standard charts Ui=Spec⁡k[xℓ(i):ℓ≠i], and O(1) is glued from frames ei on Ui by ej=xj(i)ei, with coordinate sections xj∈Γ(Pkn,O(1)) restricting to xj(i)ei on Ui and to ej on Uj; also Pk0=Spec⁡k. A line bundle L on X is H-very ample relative to Spec⁡k exactly when there are an integer n≥0 and a quasi-compact Spec⁡k-immersion i:X→Pkn with i∗O(1)≅L. (Relative projective space from standard charts, Relative very ampleness in the finite projective-space convention)

[F5]

For every S-scheme Y the assignment φ↦(φ∗O(1);φ∗x0,…,φ∗xn) is a natural bijection from S-morphisms Y→PSn to isomorphism classes of pairs (L;s0,…,sn) with L invertible and s0,…,sn generating L; the S-morphism φ attached to such data satisfies φ−1(D+(xi))=Xsi and xj(i)∘φ=sj/si on Xsi, the quotient being a regular function because si trivialises L there. (Maps to projective space equal generating line-bundle data, Generating line-bundle sections define a morphism to projective space)

[F6]

For a scheme Y and a ring A, taking global sections is a bijection Hom⁡(Y,Spec⁡A)→Hom⁡CRing(A,Γ(Y,OY)), so a morphism into an affine scheme is determined by its ring map on global sections; a morphism into the affine space ASn=Spec⁡OSOS[t1,…,tn] is given by its n coordinate functions. (Morphisms to an affine scheme and global sections, Affine n-space over an arbitrary base, Relative projective space from standard charts)

[F7]

A closed immersion is a homeomorphism onto a closed subset and an open immersion identifies its source with an open subscheme, so both are injective on points; an immersion is a composite of a closed immersion into an open subscheme followed by the inclusion of that open subscheme, hence is injective on points. A morphism that factors through a one-point scheme is constant on points, so it is not injective whenever its source has at least two points. (Closed immersions of schemes, Open immersions of schemes, Immersion of schemes)

[F8]

The polynomial ring k[t] over the field k is a domain, and (0) and (t) are two distinct prime ideals of k[t]; since U0≅Spec⁡k[t] is an open subscheme of X, the space X has at least two points. (A polynomial ring over an integral domain is an integral domain, The underlying space of an affine spectrum, Projective space is Proj of a polynomial ring)

Refutation

technique · direct: compute the global sections of the structure sheaf by gluing its sections over the two standard charts and find that all of them are constants; then show that any morphism to projective space whose generating data consist of these constant sections factors through the structure morphism to the one-point scheme $\operatorname{Spec}k$, hence is constant, while an immersion is injective
1.1F1algebra

The chart rings. Every element of S(x0)=(S[x0−1])0 is a class a/x0m with a∈Sm homogeneous of degree m, and division by x0 rewrites it as a polynomial in t=x1/x0; conversely every polynomial in t arises this way. Hence S(x0)=k[t], and symmetrically S(x1)=k[t−1], so by [F1] one has Γ(U0,OX)=k[t] and Γ(U1,OX)=k[t−1].

2.1F1step 1.1algebra

The overlap. The intersection U0∩U1=D+(x0x1) has S(x0x1)=k[t,t−1], and by [F1] the restriction maps are the homogeneous localisations Γ(U0,OX)→Γ(U0∩U1,OX) and Γ(U1,OX)→Γ(U0∩U1,OX), that is, the inclusions k[t]↪k[t,t−1] and k[t−1]↪k[t,t−1].

3.1F1F2step 1.1step 2.1algebra

Global sections of the structure sheaf. By [F2] a global section of OX is exactly a pair (a,b) with a∈k[t], b∈k[t−1] whose images in k[t,t−1] agree, that is, a(t)=b(t−1). Every element of k[t,t−1] is a finite sum ∑jαjtj with αj∈k; the left side involves only powers j≥0 and the right side only powers j≤0, so all αj with j≠0 vanish and a=b=α0∈k. Hence Γ(X,OX)=k, consisting of the constant global functions.

4.1F3step 3.1algebra

The unit section generates. The unit section 1∈Γ(X,OX)=k is nonzero, and the evaluation morphism Γ(X,OX)⊗ZOX→OX sends 1⊗g to g⋅1=g on every open U⊆X, so it is surjective and OX is globally generated by 1. With [F3] it follows that X1=X, and more generally Xc⋅1=X for c∈k× while X0⋅1=∅, since a constant section has the same nonzero or zero value in every fibre.

5.1F2F4F5step 3.1step 4.1

A hypothetical immersion and its data. Suppose OX were H-very ample over Spec⁡k. By [F4] there are n≥0 and a quasi-compact Spec⁡k-immersion i:X→Pkn with i∗O(1)≅OX; the structure morphism X→Spec⁡k is quasi-compact by [F2], so the definition applies. By [F5] the morphism i is the one attached to the data (i∗O(1);s0,…,sn) with sj=i∗xj, and these sections generate i∗O(1). Under the isomorphism i∗O(1)≅OX and the identification Γ(X,OX)=k of step 3.1 the sections sj correspond to constants cj⋅1 with cj∈k, and not all cj vanish, since the cj⋅1 generate the nonzero sheaf OX on the nonempty scheme X; fix j with cj≠0.

6.1F4F5step 4.1step 5.1

The image lies in one affine chart. By [F5] the morphism i satisfies i−1(D+(xj))=Xsj, and under the isomorphism i∗O(1)≅OX the nonvanishing locus Xsj is the nonvanishing locus of cj⋅1, which is all of X by step 4.1 because cj≠0. Hence i−1(D+(xj))=X: the image of i is contained in the single standard chart Uj=D+(xj)=Spec⁡k[xℓ(j):ℓ≠j].

7.1F5step 5.1step 6.1algebra

The chart coordinates are constant. Again by [F5], on Xsj=X one has xℓ(j)∘i=sℓ/sj for every ℓ≠j; under the identifications of step 5.1 this quotient is (cℓ⋅1)/(cj⋅1)=cℓ/cj∈k, a constant regular function on X.

8.1F6step 6.1step 7.1

The morphism is constant. Write Rj=k[xℓ(j):ℓ≠j], so that Uj=Spec⁡Rj; since i maps into Uj, it factors as ι∘i′ with i′:X→Uj and ι:Uj↪Pkn the open immersion. By [F6] the morphism i′ is determined by the ring map (i′)#:Rj→Γ(X,OX)=k, which sends xℓ(j) to the global function xℓ(j)∘i′=cℓ/cj of step 7.1. This ring map is the composite of the ring homomorphism Rj→k, xℓ(j)↦cℓ/cj, with the structure map k→Γ(X,OX)=k; let d:Spec⁡k→Uj be the morphism corresponding to Rj→k, xℓ(j)↦cℓ/cj under the bijection of [F6], and let p:X→Spec⁡k be the structure morphism. Then (d∘p)#=(i′)#, so i′=d∘p and i=ι∘d∘p factors through the one-point scheme Spec⁡k.

9.1F7F8step 5.1step 8.1

Contradiction. By step 8.1 the morphism i is constant on points, its image being the single point d(Spec⁡k); but by [F7] the immersion i is injective on points, and by [F8] the source X has at least two points. A constant map from a set with at least two points into any set is not injective, so no such i exists for any n≥0, and OX is not H-very ample over Spec⁡k.

10.1

Conclusion. The structure sheaf OX is globally generated by its unit section by step 4.1, while steps 5.1 to 9.1 show that it is not H-very ample over Spec⁡k. For n=0 the same computation reads: the data (OX;1) consist of a single generating section and correspond by [F5] to the morphism X→Pk0=Spec⁡k, which is the structure morphism and is constant, so the associated map to Pk0 is constant and is not an immersion. Thus global generation does not imply relative very ampleness; OX is generated by one global section but is not H-very ample. The Axiom of Choice [A1] is inherited through the projective-space and data-equivalence suppliers; the only further data used are the two charts and the finite list of constants, so no choice is made here. [A1, F5, step 4.1, step 9.1, cases: n=0 and n at least 1] \qed

ExampleConstruction: Literature-sourcedVerification: AI-adaptedprecheck passaudited 2026-09-30Open item page →

Projective bundle of a trivial module

Example

Let S be a scheme and let OS r be the free OS-module of rank r≥0. Then, with the projective bundle PS(E)=Proj⁡SSym⁡(E) in the quotient convention (Projective bundle in the quotient convention):

  1. for r≥1 there is a canonical isomorphism PS(OS r)≅PSr−1 of S-schemes, carrying the tautological quotient π∗OS r→O(1) to the standard quotient OS r→OPSr−1(1) whose components are the coordinate sections;
  2. for r=1 one has PS(OS)=S, and under the identification OPS(OS)(1)≅OS given by the coordinate frame the tautological quotient is the identity morphism OS→OS;
  3. for r=0 one has PS(0)=∅.

Facts & Assumptions

Given: A scheme S, an integer r≥0, the free OS-module OS r, and the Axiom of Choice as inherited from the relative Proj construction.

[A1]

The Axiom of Choice states that every family of nonempty sets has a choice function. (The Axiom of Choice)

[F1]

Symmetric algebra: for a commutative ring A and the free module Ar one has Sym⁡A(Ar)=A[T1,…,Tr] with deg⁡Ti=1, generated by its degree-one part; for a quasi-coherent OS-module F the symmetric algebra Sym⁡(F) is the quasi-coherent graded OS-algebra glued from these affine models, with Sym⁡0(F)=OS and Sym⁡1(F)=F, and it is generated as an OS-algebra by F. (Symmetric algebra of a quasi-coherent module, Relative Proj of a graded quasi-coherent algebra)

[F2]

Projective bundle: PS(E)=Proj⁡SSym⁡(E) with structural morphism π and tautological quotient π∗E→OPS(E)(1); if E∣U≅OU r with r≥1 over an open U⊆S, then PS(E)∣U≅PUr−1 by the absolute case of relative Proj, the twist O(1)∣U corresponds to the standard twist, and the tautological quotient restricts to the standard quotient OU r→OPUr−1(1) whose components are the coordinate sections; if E=0 then PS(0)=∅. The standard charts of PSr−1 are the affine spaces Spec⁡OS[xℓ(i):ℓ≠i], with twisting sheaf glued from frames ei and coordinate sections xj satisfying xj∣Ui=xj(i)ei for j≠i and xi∣Ui=ei; the case r−1=0 gives PS0≅S with a single frame. (Projective bundle in the quotient convention, Relative Proj of a graded quasi-coherent algebra, Relative projective space from standard charts, Relative very ampleness in the finite projective-space convention, Projective space is Proj of a polynomial ring)

[F3]

Representing property: for every S-scheme g:T→S, S-morphisms T→PS(E) correspond naturally to isomorphism classes of surjections g∗E→L with L invertible on T, the universal element being the tautological quotient. (Projective bundle represents line quotients)

[F4]

A surjective morphism between invertible sheaves is an isomorphism: locally on an affine chart both sides are free of rank one, so the morphism is multiplication by a section which must be a unit at each point of the source, and invertibility of the map is local. (Invertible sheaves, Locally free sheaves of finite rank, Pullback of a module along a morphism of ringed spaces)

Verification

technique · direct: identify the symmetric algebra of a free module with a polynomial algebra, apply the frame description of the projective bundle on the whole of $S$, and compute the cases $r=1$ and $r=0$ from the chart and rank-zero descriptions
1.1F1F2

The symmetric algebra. Since OS r is free of rank r, [F1] identifies Sym⁡(OS r) with the graded OS-algebra OS[T1,…,Tr] with deg⁡Ti=1, generated by its degree-one part; hence PS(OS r)=Proj⁡SOS[T1,…,Tr].

1.2F2F3F4

The case r=0. The zero module E=0 has Sym⁡(0)=OS concentrated in degree 0 and PS(0)=∅ by [F2]; consistently, for every S-scheme T a surjection 0→L onto an invertible sheaf exists only when T=∅, since an invertible sheaf on a nonempty scheme is nonzero, and morphisms T→∅ likewise exist only for T=∅.

2.1F1F2step 1.1

The isomorphism with projective space. Applying [F2] with U=S, where OS r∣S=OS r is free of rank r≥1, gives PS(OS r)≅PSr−1; under this isomorphism the twist O(1) corresponds to the standard twist and the tautological quotient restricts to the standard quotient OS r→OPSr−1(1) whose components are the coordinate sections x0,…,xr−1. This is the isomorphism of (1), and it is canonical because it is the chart-gluing identification of the two constructions.

3.1F2F3F4step 2.1

The case r=1. Here PS(OS)≅PS0=S by step 2.1 and [F2], and PS0 has the single chart U0=S with frame e0, so O(1)≅OS with frame the coordinate section x0; the tautological quotient OS=Sym⁡1(OS)→O(1) sends the generator to the coordinate section, which is the frame e0, hence is an isomorphism OS→O(1)≅OS: under the identification by the frame it is the identity. Equivalently, by [F3] the right side for E=OS consists of isomorphism classes of surjections g∗OS=OT→L with L invertible, and every such surjection is an isomorphism by [F4], so there is exactly one class; correspondingly T→PS(OS) is the single structure morphism T→S, and the two descriptions agree.

4.1

Conclusion. Steps 1.1 and 1.2 give the isomorphism PS(OS r)≅PSr−1 for r≥1 together with the identification of the universal quotients, step 3.1 computes the case r=1 as PS(OS)=S with tautological quotient the identity OS→OS, and step 1.2 records PS(0)=∅. The identification of universal quotients is what makes the isomorphism an isomorphism "in the quotient convention" of [F3]: for E=OS r the functor T↦{S-morphisms T→PSr−1} is the functor of surjections OT r→L in both models. The Axiom of Choice [A1] is inherited from the relative Proj construction; no further choice is made. [A1, F3, step 2.1, step 3.1, cases: r=0 and r=1 and r at least 2] \qed

5 · Examples, counterexamples and false statements

None yet.

Sources