Alphabeta Math
ExampleConstruction: AI-adaptedVerification: AI-adaptedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07
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.

The ideal of a polynomial graph

Example

Let k be a field, n,m0, and let u:AknAkm be given by polynomials f1,,fmk[x1,,xn]. Its graph is the closed subscheme of Akn+m with ideal (y1f1(x),,ymfm(x)).

Facts & Assumptions

Given: The objects, hypotheses and conventions in the statement above.

[F1]

For an S-morphism u:XY (as in def-scheme-over-base), the graph morphism is Γu=(idX,u):XX×SY, supplied by thm-fibre-products-of-schemes-exist. Its first projection is the identity and its second projection is u. The definition alone does not assert that its image is closed. (The graph morphism over a base)

[F2]

For an S-morphism u:XY, put H=(uprX,prY):X×SYY×SY. The square with top arrow Γu:XX×SY, bottom arrow ΔY/S:YY×SY, left arrow u, and right arrow H is Cartesian. (The graph is a pullback of the diagonal)

[F3]

For a ring A, closed immersions ZSpecA are, up to unique isomorphism over SpecA, precisely the morphisms Spec(A/I)SpecA for ideals IA. (Closed immersions into affine schemes are quotient spectra)

[F4]

Let AC be a unital ring map. For any set of variables (ti) and any ideal IA[ti], (A[ti]/I)ACC[ti]/IC[ti]. Here the extended ideal is generated by the coefficient images of all elements of I. For a multiplicative subset MA, (M1A)ACM1C. These are ring isomorphisms; no flatness, finite-generation or nonzero-ring hypothesis is required. (Presentations and localization under base extension)

Verification

1.1

By F1 the graph has coordinate map sending xi to xi and yj to fj(x). This is a surjection to k[x1,,xn]. In the quotient by the displayed ideal, successive substitution replaces every yj by fj(x), giving inverse ring maps with k[x]. Thus the displayed ideal is exactly the kernel; F3 makes the morphism a closed immersion.

givenF1F3algebra
2.1

The diagonal in Akm×kAkm has ideal (zjyj) by the same substitution calculation. Pulling it back along (uprX,prY) sends those generators to fj(x)yj. By F2 this pullback is the graph; F4 computes its quotient ideal and gives the identical presentation. If m=0 the list is empty and the graph is the identity; if n=0 it is the point given by the constants fj.

F2F4step 1.1

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources