Alphabeta Math
ExampleConstruction: Literature-sourcedVerification: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-13
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.

Z[x] represents the underlying-set functor on unital rings

Example

Let U:RingSet send a unital ring to its underlying set and a unit-preserving ring homomorphism to its underlying function. The ring Z[x] represents U. Explicitly, for every unital ring R, not assumed commutative, evaluation at rR gives the natural bijection

Ring(Z[x],R)U(R),ϕϕ(x),

whose inverse is

revr,evr(iaixi)=i(ai1R)ri.

Facts & Assumptions

Given: An arbitrary unital ring R, an element rR, and finitely supported integer coefficient sequences p=(ai) and q=(bj).

[F1]

Unital rings and unit-preserving ring homomorphisms form the large locally small category Ring (Unital rings and unit-preserving ring homomorphisms form the large locally small category Ring).

[F2]

For a commutative ring A, A[x] consists of finitely supported coefficient sequences, with coefficientwise addition and convolution (pq)n=i+j=naibj (The polynomial ring over a commutative ring as finitely supported coefficient sequences with convolution).

[L1]

The integers form a commutative unital ring, and the operations of [F2] make Z[x] a commutative unital ring whose constant-polynomial map is an injective unital homomorphism (The integers form a commutative ring, Polynomial convolution makes R[x] a commutative ring containing R as its constant subring).

[L2]

In any ring, a1R is central and (a1R)(b1R)=(ab)1R for integers a,b; more generally integer multiples distribute over addition and multiplication (Integer multiples in a ring: (m+n)a=ma+na, m(a+b)=ma+mb, (ma)b=m(ab)=a(mb) and (ma)(nb)=(mn)(ab) for all m,nZ and a,bR).

[F3]

A ring homomorphism preserves addition, multiplication, and one (Ring homomorphism: additive, multiplicative, and required to send 1 to 1).

[L3]

Finite sums in a commutative monoid may be reindexed, split, and summed in either order over a finite product (Finite commutative-monoid sums are invariant under bijective reindexing, split over disjoint unions, and satisfy the finite Fubini rule).

[F5]

A covariant functor is represented by A when it is naturally isomorphic to the hom-functor Ring(A,) (Presheaves, covariantly and contravariantly representable functors, and representations).

Verification

technique · constructive
1.1

The assignment U preserves identities and composition because it leaves their underlying functions unchanged, so [F1] makes it a functor RingSet.

F1
1.2

Because p has finite support, the displayed sum defining evr(p) is finite and is independent of any larger finite support bound by adjoining zero terms. It sends 1 to (11R)r0=1R.

F2L1L2F4construct
1.3

Coefficientwise addition, distributivity of integer multiples, and finite-sum splitting give evr(p+q)=evr(p)+evr(q).

F2L2L3
1.4

Conversely, let ϕ:Z[x]R be a unital ring homomorphism and put r=ϕ(x). From ϕ(1)=1R and additivity, including additive inverses, ϕ sends the constant a to a1R for every integer a; multiplicativity gives ϕ(xi)=ri. Additivity over the finite expression p=iaixi then gives ϕ(p)=i(ai1R)ri=evr(p).

L1L2F3F4
2.1

Expanding a product of the two finite evaluation sums and using [L3] gives evr(p)evr(q)=i,j(ai1R)ri(bj1R)rj. Since bj1R is central by [L2], each summand is ((aibj)1R)rirj=((aibj)1R)ri+j by [F4].

step 1.2L2F4L3
2.2

The polynomial x has only coefficient 1 at index 1, so evr(x)=r. Hence evaluation at x after revr recovers r.

step 1.2F2F4
3.1

Reindex the last finite sum by n=i+j and regroup its fibres. By [F2] and [L3] the inner coefficient sum is (pq)n, so the result is evr(pq). Together with steps 1.2 and 1.3, [F3] makes evr:Z[x]R a unital ring homomorphism, without any commutativity hypothesis on R.

step 1.2step 1.3step 2.1F2F3L3
4.1

Steps 2.2 and 1.4 make ϕϕ(x) and revr inverse bijections. If h:RS is a unital ring homomorphism, then hevr and evh(r) have the same value h(r) at x, so step 1.4 makes them equal; the bijection is natural in R.

step 3.1step 2.2step 1.4F3
5.1

By [F5], Z[x] represents U. The proof includes the zero ring: when 0R=1R, its underlying set and the hom-set from Z[x] are both singletons, and the same formulas apply.

step 1.1step 4.1F5discharge-construct

Depends on

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Dependency tree · next 3 levels

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Sources