Finite locally free affine equivalence relations have finite locally free scheme quotients

The rank of the outgoing-relation fibre is locally constant on $U$ and constant along every relation arrow: composing with that arrow and its inverse gives mutually inverse maps between the outgoing fibres, preserving their target point. Thus the finitely many constant-rank clopen pieces of $U$ are saturated. Their defining idempotents satisfy $s(e)=t(e)$ and belong to $C$. Split by them; it suffices to prove the claim when the two ranks have one positive constant value $r$. Positivity follows from the identity arrow. For $a\in A$, take the characteristic polynomial of multiplication by $t(a)$ on the locally free $A$-module $B$ via $s$. Its coefficients lie in $C$: after either pullback to $R$, the composition/inverse isomorphism between the two outgoing-relation fibres preserves target evaluation $t(a)$, so the corresponding multiplication operators are conjugate and have the same characteristic polynomial. Cayley–Hamilton holds here over the coordinate ring: locally write $\mathrm{adj}(ZI-T)={\sum }_{j=0}^{r-1}{B}_j{Z}^j$ for the multiplication matrix $T$. The adjugate identity in [F1] yields $-T{B}_0=c_{0}I$, $B_{j-1}-T{B}_j=c_{j}I$, and $B_{r-1}=I$; multiplying by successive powers of $T$ and adding telescopes to ${\sum }_{j=0}^r{c}_j{T}^j=0$. This identity glues for the locally free module. Follow it by the identity-arrow map $B\to A$ to obtain a monic equation for $a$ over $C$.

Hence $A$ is integral over $C$. Since $A$ is generated by finitely many elements over $k\subset C$, it is a finite $C$-module. Also $C$ is a finite-type $k$-algebra: choose finite $C$-module generators ${\alpha }_j$ of $A$, express their products, the unit, and finite $k$-algebra generators of $A$ in this generating family. Let $C_0\subset C$ be generated over $k$ by those finitely many coefficients. The $C_0$-span of the ${\alpha }_j$ contains the unit and algebra generators and is closed under multiplication, so equals $A$. The Noetherian ring $C_0$ makes its submodule $C\subset A$ finite over $C_0$. Thus $C$ is finite type and Noetherian. Injective integrality makes $\mathrm{Spec}A\to \mathrm{Spec}C$ onto by lying-over.

The map $j:R\to U\times U$ is finite: its graph into $R\times U$ is closed, and the map of that product to $U\times U$ is the base change of one finite projection. It is a monomorphism by the equivalence-relation assumption. A finite monomorphism is a closed immersion here: after any residue-field base change its affine coordinate algebra $D$ has diagonal multiplication $D\otimes D\to D$ an isomorphism; finite dimension gives $(\dim D{)}^2=\dim D$, so $D$ is zero or the residue field. Nakayama applied to the finite cokernel of the original ring map then gives surjectivity. Consequently $A{\otimes }_{C}A\to B$, $a\otimes b\mapsto s(a)t(b)$, is onto.

Localize at any prime of $C$ and faithfully flat extend this local ring, if necessary, to $C[T{]}_{\mathfrak{m}C[T]}$, whose residue field is the infinite field $\kappa (\mathfrak{m})(T)$. Formation of $C$ as the kernel of $s-t$ commutes with this flat extension. The extended $A$ is finite over the local extended $C$, so is semilocal, and $B$ is finite projective of rank $r$ over $A$ via $s$. Such a module is free: choose residue-field bases at the finitely many maximal ideals, lift them simultaneously by the Chinese remainder theorem, use Nakayama to obtain a surjection $A^r\to B$, split it by projectivity, and apply Nakayama to its finite kernel.

In this semilocal situation, the $C$-submodule $t(A)$ generates $B$ as an $s(A)$-module by step 3.1. It contains an $A$-basis of $B$. To prove this, reduce modulo the Jacobson radical of $A$ and consider the finitely many residue-field vector spaces of dimension $r$. Choose finitely many elements of $t(A)$ which generate $B$ over $A$. Because the residue field of $C$ is infinite and maps into every residue field of $A$, a linear combination with coefficients in that common field can be chosen nonzero in every factor: each forbidden condition is a proper linear subspace, and finitely many such subspaces cannot cover a vector space over an infinite field. Lift the coefficients to $C$. The resulting element generates a free direct summand of rank one, by Nakayama and the splitting argument of step 3.2. Apply the same argument to the quotient, and induct on $r$. Lifting the quotient basis elements from $t(A)$ gives a basis $t(x_1),\dots ,t(x_r)$ of $B$ over $s(A)$.

Write $t(a)={\sum }_{i}s(c_i)t(x_i)$, with $c_i\in A$. Let $c:B\to B{\otimes }_{s,A,t}B$ be the composition comorphism. The groupoid identities give $c(t(a))=t(a)\otimes 1$ and $c(s(a))=1\otimes s(a)$. Comparing composition of the displayed expansion with its first-factor pullback, and using $s(c_i)\otimes 1=1\otimes t(c_i)$, gives ${\sum }_{i}t(x_i)\otimes (s(c_i)-t(c_i))=0$. Since the $t(x_i)$ are a basis in the first tensor factor, all $c_i$ lie in $C$. Applying the identity-arrow map shows $a={\sum }_i{c}_i{x}_i$. Independence follows by applying $t$ and the basis independence in $B$. Therefore $A={⨁}_{i}C{x}_i$, and the surjection in step 3.1 is an isomorphism, since it takes the corresponding basis in the second tensor factor to $t(x_i)$.

The conclusions in step 5.1 descend to each local ring of the original $C$ by faithful flatness in [F1]. Thus $A{\otimes }_{C}A\to B$ is an isomorphism globally, and $A$ is flat over $C$. It is finite and finitely presented over the Noetherian $C$, so is finite locally free by [F1]. It is faithful because its spectrum is onto by step 2.1. The kernel pair is now exactly $R$. Every morphism into $M$ lifts after base change by the finite locally free covering $U\to M$; any two local lifts differ by the kernel pair. Since $M$ is an fppf sheaf by [F2], these statements identify it with the fppf quotient sheaf. Recombine the saturated rank pieces from step 1.1 to finish. AC is inherited from [F1] and the prime/basis selections.