Past exam of the mathematics course of the University of Cambridge / 2014 / iii / Paper 20 / 6

Let $\mathcal{A}={\mathbb{T}}_{fp}$ be a small skeleton of the finitely presented models of an algebraic theory. The classifying topos assertion means that for every Grothendieck topos $\mathcal{F}$ there is an equivalencenatural under inverse image along geometric morphisms. On the left, morphisms are transformations between inverse image functors, and on the right they are model homomorphisms. A model in $\mathcal{F}$ interprets the sorts by objects, the operations by arrows, and the equations by equality of the resulting arrows. Finite products suffice for these algebraic operations.

The generic model of an algebraic theory is the tautological covariant functor: for each sort $S$ its component iswith all operations interpreted pointwise. Pulling $U$ back by a geometric morphism gives its classified model. For a single-sorted theory, this is simply the underlying-set functor with its pointwise algebraic structure.

The orientation is important: $[\mathcal{A},\mathbf{Set}]$ is the presheaf topos on ${\mathcal{A}}^{\mathrm{op}}$, and the generic model is covariant on finitely presented algebras. Such an algebra is a finite-generator, finite-relation presentation. In the algebraic syntactic category it corresponds to the formula imposing its relations, with arrows reversed. The finite-presentability/filtered-colimit description of algebraic models gives the above classifying equivalence; a detailed proof is not needed for this part.

Take $\mathcal{A}$ to be the category of finitely presented commutative rings with identity, and put $\mathcal{C}={\mathcal{A}}^{\mathrm{op}}$. In its presheaf topos, the tautological ring $U(A)=A$ is generic. The additional domain axioms are coherent sequents, so they are imposed by a quotient-theory coverage on $\mathcal{C}$.

Concretely, declare the zero ring covered by the empty family. For every finitely presented $A$ and elements $a,b\in A$ with $ab=0$, declare the two opposite quotient arrows associated withto be a covering family at $A$. These quotient rings are finitely presented. Pullback and transitivity generate a Grothendieck topology $J$ from these families. The empty cover forbids $0=1$; the two quotient covers make every zero product locally have a zero factor. Conversely, any internal integral domain satisfies exactly the continuity conditions prescribed by these generating covers. Thusand its generic domain is the associated sheaf $K=a_{J}U$, with the ring operations transported through the left-exact sheaf reflector.

This coverage is not standard, meaning not all representables are sheaves; in modern terminology it is not subcanonical. For an explicit obstruction, use $A=\mathbb{Z}/4\mathbb{Z}$ and $a=b=2$. The two quotient arrows are the same map $A\to \mathbb{Z}/2\mathbb{Z}$, so their generated sieve is a singleton cover. Consider the representable on $\mathcal{C}$ corresponding to $R=\mathbb{Z}[t]$:Its distinct sections $t\mapsto 0$ and $t\mapsto 2$ become equal after restriction to $\mathbb{Z}/2\mathbb{Z}$. Hence this representable is not even separated for $J$. The nilpotent element has to disappear in the generic domain, which explains this failure of standardness.

Work in the domain-classifying site of part (ii), with generic domain $K=a_{J}U$. For finitely presented rings $A$, the object $a_{J}yA$ is an available stage. Tuples of sections of $K$ are locally represented by tuples of actual elements of $A$, because $U\to a_{J}U$ is locally surjective and sheafification preserves finite products. It is therefore enough to prove the requested implication on such representatives.

We first note the empty-cover criterion for the domain-classifying site: $a_{J}yB$ is initial if and only if the finitely presented ring $B$ is the zero ring. One direction is the generating empty cover. Conversely, any nonzero ring has a maximal ideal and hence a homomorphism to a field. That field is a set-based integral domain and defines a point of the classifying topos. The inverse image of $a_{J}yB$ at this point is ${\mathrm{Hom}}_{\mathrm{Ring}}(B,L)$, which is nonempty for the chosen field $L$. It therefore cannot be the inverse image of an initial object. This uses only the ordinary maximal-ideal existence principle externally, not excluded middle in the internal logic.

Suppose $a_1,\dots ,a_n\in A$ represent a tuple lying in the negation of the all-units subobject. Write $t=a_1\cdots a_n$ and consider the finitely presented localization of a ringEvery $a_i$ is invertible in $B$, since $({\prod }_{j\ne i}{a}_j)z$ is an inverse. But the pulled-back tuple still lies in the negation of the all-units subobject. Thus the whole stage $a_{J}yB$ maps into both that subobject and its negation, forcing this stage to be initial. The empty-cover criterion gives $B=0$.

The localization $A[1/t]$ is zero precisely when $t^N=0$ in $A$ for some integer $N\ge 1$, by the equality criterion in localization. Repeated use of the generating zero-product covers now gives a covering familyIndeed, splitting $t^N=t{t}^{N-1}=0$ and inducting first forces $t=0$ locally; splitting the product $a_1\cdots a_n=0$ then forces one $a_i=0$ locally. More formally the two inductions are coherent derivations from the zero-product axiom, so their quotient families belong to the generated topology. The sheaf semantics of disjunction consequently givesThis is the finite-tuple weak-field property of the generic integral domain. The argument is intuitionistically valid inside the topos, although its description of the site uses ordinary external set theory. There is no appeal to preservation of negation by an arbitrary geometric inverse image.

For the converse assertion, let $R$ be any internally nontrivial commutative unital ring satisfying the $n=2$ implication. Assume $xy=0$. If $x$ and $y$ were both units, multiplying by their two inverses would give $0=1$, contradicting nontriviality. ThereforeThe assumed two-variable implication then gives $x=0\vee y=0$. Together with nontriviality this is exactly the internal integral-domain axiom. Hence a nontrivial ring satisfying the two-variable case is an integral domain.

For example, the ordinary integral domain $\mathbb{Z}$ fails the one-variable implication at $2$. This does not contradict the generic result: the displayed implication uses negation and is non-coherent, so it need not survive the inverse image that classifies an arbitrary set-based domain.