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CounterexampleConstruction: Literature-sourcedVerification: AI-adaptedPipeline-generatedprecheck passaudited 2026-09-30
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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

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