Etale Covers and the Etale Fundamental Group

Suppose $D$ is finitely presented as an $A$-algebra. Choose finite algebra generators $d_1,\dots ,d_r$ and monic equations $P_i(d_i)=0$ by [F1]. The kernel of $A[T_1,\dots ,T_r]\to D$ is a finitely generated ideal: a finite presentation with a different generating list converts to this one by adding both finite lists, imposing the finitely many equations expressing each list in the other, and eliminating the extra variables. Put $C=A[T_1,\dots ,T_r]\mathrm{/}(P_1,\dots ,P_r)$. Reduction by the monic polynomials gives a finite free $A$-basis of $C$ consisting of the monomials $\prod T_i^{e_i}$ with $0\le e_i<\mathrm{deg}{P}_i$. The kernel $J$ of $C\to D$ is a finitely generated $C$-ideal, so multiplication of its finitely many ideal generators by this finite $A$-basis generates $J$ as an $A$-module. Thus $D$ has a finite $A$-module presentation.

Let $M$ be any finitely presented flat $A$-module. At a prime $\mathfrak{p}$, choose lifts of a basis of $M_{\mathfrak{p}}\mathrm{/}\mathfrak{p}{M}_{\mathfrak{p}}$. By [F2] they give a surjection $A_{\mathfrak{p}}^n\to M_{\mathfrak{p}}$. Its kernel $K$ is finitely generated because $M_{\mathfrak{p}}$ is finitely presented: this follows for any finite free surjection by adjoining a fixed finite presentation and eliminating its finitely many auxiliary generators. Tensoring with the residue field is exact by flatness and [F2], and the last map is the chosen basis isomorphism. Thus $K\mathrm{/}\mathfrak{p}K=0$, so $K=0$ by Nakayama. The local module is free.

Conversely, choose module generators $1=d_0,d_1,\dots ,d_r$ and finitely many generators of their $A$-linear relations. Present an algebra with variables $T_0,\dots ,T_r$, relation $T_0=1$, those finitely many linear relations, and the finitely many multiplication relations $T_i{T}_j={\sum }_k{a}_{ijk}{T}_k$ expressing products in $D$. Every polynomial in this presented algebra reduces to an $A$-linear combination of the $T_i$, by induction on its degree. The map to $D$ is surjective; a linear combination mapping to zero is a combination of the imposed module relations, so the map is injective. This is a finite algebra presentation. Applying [F1] proves the asserted étale criterion.

The local basis established in step 1.2 spreads to a principal neighbourhood of $\mathfrak{p}$. Represent the basis and its inverse over the local ring using finitely many denominators. The inverse is a map from a finitely presented module, so its values on the finite generators and the finitely many relations are defined after inverting one element outside $\mathfrak{p}$. The two composition identities are identities on finitely many generators and therefore also hold after one further localization. Hence $M$ is free on a neighbourhood of every prime. The resulting rank function is locally constant. Applied to $D$, its geometric fibre is a finite-dimensional algebra with zero differentials over an algebraically closed field; étaleness, or equivalently the zero-dimensional geometrically regular fibre in its definition, makes it a product of copies of that field. The product description is Finite and finite type etale schemes over an algebraically closed field. Its vector-space dimension is the local rank of $D$, proving the fibre-count assertion.

A finite locally free, finitely presented module is projective. To see this explicitly, take finitely many principal opens $D(a_i)$ on which it is free. Local basis vectors and dual coordinate maps have finitely many denominators, since the module is finitely presented. Clearing these denominators expresses $a_i^{N_i}{\mathrm{id}}_M$ as a finite sum of maps $m\mapsto \lambda (m)v$, with $v\in M$ and $\lambda \in {\mathrm{Hom}}_A(M,A)$. The powers $a_i^{N_i}$ generate the unit ideal because the principal opens cover the spectrum. A linear combination equal to $1$ therefore expresses ${\mathrm{id}}_M$ as a finite dual-basis sum. The associated maps $M\to A^r\to M$ compose to the identity, so $M$ is a summand of a finite free module and hence projective. This completes the strengthened assertion.

First consider a cover $g:T\to S$ with a section $s:S\to T$ and a module $N$ on $T$ with the given transport. Put $M=s^{\ast }N$. Pull back $\theta$ along $T\to T{\times }_{S}T$, $t\mapsto (s(g(t)),t)$. This gives an isomorphism $g^{\ast }M\to N$. Its compatibility with the original datum follows by pulling the cocycle identity back along $(s(g(t_1)),t_1,t_2)$ on $T{\times }_{S}T$. Pulling the cocycle back to the diagonal shows that diagonal transport is an invertible idempotent and hence the identity. Pulling back along $(t,s(g(t)),t)$ then shows that reverse transport is the inverse. These identities prove both effectiveness and that a map between descended objects is determined uniquely by its pullback: its descent-compatible upstairs map is recovered by pulling back along the section. This argument applies to affine modules, and to algebras when the transports preserve multiplication and unit.

Define $M$ to be the equalizer inside $N$ displayed in the Statement, considered as an $A$-module. Flat tensor product preserves this equalizer. For any flat base extension $A\to A^{\mathrm{\prime }}$, the invariant module for the base-changed datum is therefore $A^{\mathrm{\prime }}{\otimes }_{A}M$, since it is the kernel of the base-changed difference of the two transport maps. Take $A^{\mathrm{\prime }}=B$. The new covering ring is $B\to B{\otimes }_{A}B$ and has a retraction given by multiplication; equivalently its affine covering has a section. By step 1.1 the base-changed datum is effective. For an effective datum coming from a module $Q$, the invariant equalizer is $Q$: for a split cover, apply the section to an invariant element to recover its unique downstairs element. Thus the base change of $B{\otimes }_{A}M\to N$ is an isomorphism. Faithful flatness in [F1] proves that the original map is an isomorphism.

For completeness, the invariant equalizer of the canonical datum on $B{\otimes }_{A}Q$ is $Q$ also before this split base change. Indeed $Q\to B{\otimes }_{A}Q\rightrightarrows B{\otimes }_{A}B{\otimes }_{A}Q$ becomes a split exact equalizer after tensoring with $B$: multiplication supplies the section and step 1.1 applies. Its kernel and cokernel are detected by [F1]. Consequently a descent-compatible map $N\to N^{\mathrm{\prime }}$ takes invariants to invariants, and the isomorphisms of step 2.1 identify it uniquely with the base change of that restricted $A$-linear map. This proves full faithfulness as well as effectiveness.

If $N$ is an algebra, its invariant subset is closed under sums, products, scalar multiplication and the unit because $\theta$ is an algebra isomorphism. It is therefore an $A$-algebra. The map of step 2.1 is an algebra homomorphism and an isomorphism of modules, hence an algebra isomorphism. Restriction to invariants preserves algebra maps and step 3.1 proves their full faithfulness. This proves both equivalences and the displayed construction.

For an affine faithfully flat map $A\to B$, let $N$ be the upstairs finite étale $B$-algebra and let $D$ be its invariant algebra. By [F1], $D{\otimes }_{A}B\cong N$. To descend finite generation, express finitely many $B$-module generators of $N$ as finite sums of tensors $d\otimes b$. The finitely many occurring $d$ generate a submodule $D_0$ whose base change surjects onto $N$. Thus $(D\mathrm{/}{D}_0){\otimes }_{A}B=0$, and faithful flatness gives $D=D_0$.

Choose a finite free surjection $A^r\to D$ with kernel $K$. Flatness of $B\mathrm{/}A$ identifies $K{\otimes }_{A}B$ with the kernel of $B^r\to N$. This kernel is finitely generated because $N$ is finitely presented as a $B$-module by [F2]; the assertion for an arbitrary finite free surjection follows by comparing it with a fixed finite presentation and eliminating finitely many auxiliary generators. Applying the same finite-tensor-generator argument as step 1.1 to $K$ proves that $K$ is finitely generated. Hence $D$ is finitely presented as an $A$-module. It is flat by [F2]. Since ${\mathrm{\Omega }}_{D\mathrm{/}A}{\otimes }_{A}B\cong {\mathrm{\Omega }}_{N\mathrm{/}B}=0$, faithful flatness gives ${\mathrm{\Omega }}_{D\mathrm{/}A}=0$. Therefore $D$ is finite étale by [F2]. Maps descend with their algebra structures by [F1], proving the entire assertion for affine faithfully flat maps.

Now restrict the target to an affine open $U=\mathrm{Spec}A\subseteq S$. By [F4], $S_U^{\mathrm{\prime }}$ is quasi-compact, so choose finitely many affine opens $V_j=\mathrm{Spec}{B}_j$ covering it. Their disjoint union is affine, with ring $B={\prod }_j{B}_j$. The map $A\to B$ is flat and surjective on spectra, hence faithfully flat by [F3]. Pull the upstairs cover and cocycle back to this disjoint union, and use steps 1.1 and 2.1 to construct a finite étale cover of $U$. Its pullback is canonically the given cover on each $V_j$; the isomorphisms agree on intersections because their restrictions to $V_j{\times }_U{V}_k$ are precisely the original descent datum. Thus they glue to identify its pullback over all of $S_U^{\mathrm{\prime }}$ with the original cover and datum.

On the intersections of two affine opens of $S$, cover the intersection by affine opens and repeat step 3.1. Full faithfulness in step 2.1 makes the resulting comparison isomorphisms unique after requiring compatibility upstairs; uniqueness makes them agree on further overlaps and satisfy the cocycle identity. The algebras and then their relative spectra glue by [F3]. Finiteness and étaleness are affine local, so the glued cover has both properties. A compatible upstairs morphism descends on these affine opens by [F1] and its unique local descents agree, hence glue. This proves the claimed equivalence, including effectiveness and uniqueness, over arbitrary base schemes. AC is used through the suppliers recorded in [F4].

Given $u:Y\to Z$, its graph in $Y{\times }_{X}Z$ is the pullback of the diagonal of $Z\mathrm{/}X$, hence is an open and closed immersion by [F1]. The projection $Y{\times }_{X}Z\to Z$ is finite étale, so the graph factorization shows that $u$ is finite étale. The equalizer of two maps is likewise the pullback of an open and closed diagonal. If two maps agree on $F(Y)$, their equalizer contains the entire fibre; its complementary open and closed cover has rank zero at $\bar{x}$ and hence everywhere on connected $X$ by [F2], so it is empty. This proves faithfulness.

A nonempty open and closed piece of a cover is itself finite étale. Its image is open and closed by [F1] and nonempty, hence all of $X$. Each such piece therefore contributes at least one point to every geometric fibre. A cover of rank $n$ cannot have more than $n$ disjoint nonempty open and closed pieces. Repeatedly split a disconnected piece: every split increases their number, so after at most $n-1$ splits all pieces are connected. This proves the finite connected decomposition without a Noetherian assumption. A map to a connected cover is open and closed by step 1.1 and [F1], and thus is surjective if its source is nonempty.

If $F(u)$ is a bijection, its source-to-target map has degree one on each component of $Z$, by [F2], step 2.1 and connectedness. A finite locally free map of degree one is an isomorphism: locally its algebra is free of rank one, and the unit generates its residue fibre at every prime; Nakayama as used in [F2] makes the unit a local basis, so the ring map is an isomorphism. Hence $F$ reflects isomorphisms. For maps from a connected cover, agreement at one fibre point makes their open and closed equalizer nonempty and therefore the whole source.

Let $Y$ have rank $n>0$. In $Y^n=Y{\times }_X\cdots {\times }_{X}Y$, remove the finitely many open and closed loci on which two coordinates agree. The resulting open and closed cover $P$ has as its fibre the ordered lists of all $n$ distinct points of $F(Y)$. Choose one such list $p$, and let $T$ be the connected component of $P$ containing it. By step 2.1, $T\to X$ is nonempty finite étale and surjective. The coordinate sections of $T{\times }_{X}Y\to T$ have disjoint open and closed graphs by [F1]. On every geometric fibre they exhaust the $n$ points, since the list in $P$ has no repeated coordinate. Their disjoint union is thus the whole $T{\times }_{X}Y$, and $T$ trivializes $Y$. For $n=0$ take $T=X$; the empty cover is already trivialized.

Permuting the $n$ coordinates gives an action of $S_n$ on $P$. Any other point $p^{\mathrm{\prime }}\in F(T)$ is obtained from $p$ by a unique permutation $\sigma$. The image $\sigma (T)$ is a connected component of $P$ meeting $T$ at $p^{\mathrm{\prime }}$, and therefore equals $T$. Thus the subgroup preserving $T$ acts transitively on $F(T)$. An automorphism of $T$ fixing one geometric point is the identity by step 3.1, so $H={\mathrm{Aut}}_X(T)$ acts freely as well. This proves simple transitivity. Any map $T\to Y$ is a section of the trivial cover $T{\times }_{X}Y$; connectedness of $T$ makes it one of the coordinate sections. These are in bijection with $F(Y)$ by evaluation at a fixed point of $T$, proving the uniqueness and existence assertion.

The map ${\coprod }_{h\in H}T\to T{\times }_{X}T$ given on the $h$-summand by the graph of $h$ is finite étale by step 1.1. It is bijective on the geometric fibre because $H$ acts simply transitively; step 3.1 makes it an isomorphism. Consequently $T\mathrm{/}X$ is an $H$-torsor, and $T\to X$ is an fpqc cover. Any finite $H^{\mathrm{o}\mathrm{p}}$-set $E$ determines a descent datum on ${\coprod }_{E}T$ over that fpqc cover: on the overlap identified with ${\coprod }_{H}T$, use the permutation of $E$ associated to $h$, with the opposite-group convention matching composition of changes of trivialization. The group-action law is precisely the cocycle identity. By [F3] this datum descends to a finite étale $X$-cover with fibre $E$.

For $K\le H$, use the coset action in step 5.1. The quotient map of the finite trivial fibre is compatible with the descent datum, so [F3] descends it to $q:T\to T\mathrm{/}K$. Its fibre is $F(T)\mathrm{/}K$, and $q$ is finite étale and surjective by step 1.1 and fibrewise surjectivity; thus it is fpqc. The relation $T{\times }_{T\mathrm{/}K}T$ is the disjoint union of the graphs of $k\in K$, as seen after the faithfully flat cover $T\to X$ and then descended by [F3]. A $K$-invariant map $a:T\to W$ to any $X$-scheme has equal pullbacks on that relation. For each affine open of $W$, its preimage in $T$ is saturated, hence descends to an open of $T\mathrm{/}K$ by submersiveness in [F3]. Cover that open by affine opens $U$: since $q$ is finite, $q^{-1}(U)$ is affine, and the ring equalizer argument of [F3] descends $a$ on $U$. The unique descended morphisms agree on overlaps and glue. Thus $T\mathrm{/}K$ satisfies the universal property of the scheme quotient. This completes the assertions with the AC use recorded in [F4].

Choose a set of pointed representatives $(T_i,t_i)$ for the connected Galois covers. Write $i\ge j$ when there is a pointed map $u_{ij}:T_i\to T_j$; it is unique by [F2]. The relation is a directed partial order up to pointed isomorphism: antisymmetry follows because surjections in both directions give equal finite degrees, hence degree-one maps and isomorphisms; directedness follows by taking the component of $T_i{\times }_X{T}_j$ through $(t_i,t_j)$ and then a pointed Galois refinement of it. A Galois refinement of a disconnected cover trivializes all its components at once by the ordered-fibre construction in [F2]. Thus for every finite cover $Y$, evaluation yields a natural bijection ${<mrow><mi\; mathvariant="normal">c</mi><mi\; mathvariant="normal">o</mi><mi\; mathvariant="normal">l</mi><mi\; mathvariant="normal">i</mi><mi\; mathvariant="normal">m</mi></mrow>}_i{\mathrm{Hom}}_X(T_i,Y)⟶F(Y),a⟼a(t_i)\mathrm{.}$ It is onto by such a trivializing refinement. If two maps evaluate to the same point, take a common refinement and apply the connected-source uniqueness in [F2]; they are equal in the colimit.

Put $H_i={\mathrm{Aut}}_X(T_i)$. For $i\ge j$ and $h\in H_i$, simple transitivity gives a unique $h_j\in H_j$ with $h_j(t_j)=u_{ij}(h(t_i))$. Both maps $h_j{u}_{ij}$ and $u_{ij}h$ agree at $t_i$, hence agree everywhere by [F2]. This defines a homomorphism $H_i\to H_j$. It is surjective because $u_{ij}$ is surjective on the geometric fibre. The maps compose compatibly. Let and $G=H^{\mathrm{o}\mathrm{p}}$. The inverse limit is a closed subgroup of a product of finite discrete groups, hence compact, Hausdorff and totally disconnected by [F3]. Its projections onto $H_i$ are surjective: imposing any prescribed coordinate together with finitely many compatibility constraints can be solved at a common upper index, and compactness makes the resulting family of closed conditions simultaneously satisfiable.

A compatible family $h=(h_i)$ acts on the colimit in step 1.1 by precomposition $a\mapsto a{h}_i$. Precomposition reverses multiplication, so this gives a homomorphism $G\to \mathrm{Aut}(F)$. Its restriction on $F(T_i)$, identified with $H_i$ by evaluation at $t_i$, is right multiplication by $h_i$. Conversely, any natural automorphism $\gamma$ determines $h_i\in H_i$ by $h_i(t_i)={\gamma }_{T_i}(t_i)$. Naturality for $u_{ij}$ makes these compatible. Naturality for every map $a:T_i\to Y$ forces ${\gamma }_Y(a(t_i))=a(h_i(t_i))$, which determines $\gamma$ on every fibre by step 1.1. Thus $G\cong \mathrm{Aut}(F)$. This is a topological isomorphism: each finite fibre is represented at a single trivializing refinement, so its action factors through $H_i^{\mathrm{o}\mathrm{p}}$; conversely the action on $F(T_i)$ detects the entire $i$th coordinate. Hence both topologies have the same finite-coordinate neighbourhood basis. In particular the group in [F1] is profinite.

The group acts transitively on the fibre of every nonempty connected cover. Indeed choose a pointed Galois refinement surjecting onto that cover; the projection $H\to H_i$ is onto by step 2.1, and its regular action on $F(T_i)$ is transitive, so the induced action on the target fibre is transitive. For a general cover its connected decomposition is carried to its orbit decomposition: the group preserves every component by naturality for its inclusion, and acts transitively within it.

A continuous action on a finite set $E$ has an open kernel: intersect its finitely many open point stabilizers. By the inverse-limit topology of step 3.1, that kernel contains the kernel of $G\to H_i^{\mathrm{o}\mathrm{p}}$ for some $i$; finitely many coordinate constraints can be combined at one upper index. The projection is onto by step 2.1, so the action factors through $H_i^{\mathrm{o}\mathrm{p}}$. The contracted-cover construction in [F2] gives a finite étale cover with precisely that action on its fibre. This proves essential surjectivity.

Let $q:F(Y)\to F(Z)$ be equivariant. Its graph is an invariant subset of $F(Y{\times }_{X}Z)$, and hence, by step 4.1, a union of fibres of connected components of $Y{\times }_{X}Z$. Take the corresponding open and closed union of components $W$. Its projection to $Y$ is bijective on the fibre and hence an isomorphism by [F2]. The composite $Y\cong W\to Z$ has fibre map $q$. Faithfulness follows from [F2]; therefore the functor is fully faithful.

Steps 4.2 and 5.1 prove the equivalence, step 3.1 proves profiniteness, and step 4.1 identifies the connected covers. AC enters in choosing the pointed set of representatives and through [F2], and its additional compactness use is exactly step 2.1 via [F3]. This proof supplies reconstruction explicitly and does not invoke an unproved Galois-category theorem or a universal cover as an actual finite scheme.

The open and closed diagonal in $\mathrm{Spec}(D{\otimes }_{A}D)$ corresponds to an idempotent $e$ on whose summand the multiplication map $\mu :D{\otimes }_{A}D\to D$ is an isomorphism, while the complementary summand is its kernel. Hence $\mu (e)=1$. For $a\in D$, $a\otimes 1-1\otimes a$ is in that kernel and therefore annihilates $e$. Expressing the tensor $e$ as a finite sum gives the asserted identities.

Agreement of the two $A$-actions makes $\delta$ preserve $A$-multilinearity and makes the tensor formula for $h$ well-defined. Expand $\delta (hc)+h(\delta c)$ using the displayed differential. The initial term in $h\delta c$ is ${\sum }_i{x}_i{y}_{i}c(a_1,\dots ,a_n)=c(a_1,\dots ,a_n)$. Every term combining two adjacent inputs cancels with its counterpart of opposite sign in $\delta hc$, and the last right-action terms cancel as well. The remaining pair is $a_1{\sum }_i{x}_{i}c(y_i,a_2,\dots )-{\sum }_i{x}_{i}c(y_i{a}_1,a_2,\dots )$, which is zero by $(a_1\otimes 1)e=(1\otimes a_1)e$ and multilinearity. Thus $\delta h+h\delta =\mathrm{id}$ for $n\ge 1$. For a cocycle $c$, this gives $c=\delta (hc)$. An input equal to $1$ after the inserted first input still makes each summand zero, proving the normalization assertion.

First assume $I^2=0$. Lift the finite projective module $D_0$ to a finite projective $A$-module $P$. Explicitly, write $D_0$ as the image of an idempotent matrix over $A\mathrm{/}I$, lift that matrix to $p$, and put $ϵ=p^2-p$. Then $p$ commutes with $ϵ$, ${ϵ}^2=0$, and $p^{\mathrm{\prime }}=p+(1-2p)ϵ$ has $(p^{\mathrm{\prime }}{)}^2=p^{\mathrm{\prime }}$ and the same reduction; take $P=\mathrm{im}{p}^{\mathrm{\prime }}$. The locus where $D_0$ has rank zero is open and closed by [F1]; its idempotent lifts by the same scalar formula. On that factor lift by zero. On the complementary factor the unit of $D_0$ is unimodular: at every prime it is a nonzero unit in a nonzero fibre algebra, so the ideal of its values under the dual module is the unit ideal. Hence it admits a functional ${\lambda }_0$ with ${\lambda }_0(1)=1$. Lift the unit to $u\in P$ and the functional to $\lambda :P\to A$ by projectivity. Since $\lambda (u)$ reduces to $1$ and $I$ is nilpotent it is invertible; rescale $\lambda$ so that $\lambda (u)=1$. Thus $P=Au\oplus \ker \lambda$.

Lift the multiplication on $D_0$ to an $A$-bilinear multiplication $\mu$ on $P$ with unit $u$. Set the products involving $u$ as required by the unit, and lift the product on $\ker \lambda {\otimes }_A\ker \lambda$ using its projectivity. Its associator $a(v,w,z)=\mu (\mu (v,w),z)-\mu (v,\mu (w,z))$ takes values in $IP$. It vanishes if any input is $u$ and factors through $D_0$ in each input, because $I^2=0$. The left and right $D_0$-actions on $IP$ are well-defined from the reduced multiplication. Expanding the five terms of $\delta a$ shows that they cancel: each nested triple product occurs twice with opposite signs, while changing brackets inside a term involving $a$ is harmless modulo $I^2$. Thus $a$ is a normalized Hochschild $3$-cocycle. By [F1], $b=ha$ is a normalized $2$-cochain with $\delta b=a$. Replace $\mu$ by $\mu +b$, interpreting $b$ as a map $P{\otimes }_{A}P\to IP$. The new associator is $a-\delta b=0$; terms involving two values of $b$ vanish because $I^2=0$. The unit is preserved by normalization.

The associative lifted algebra is commutative. For each $v\in P$, the inner derivation $w\mapsto vw-wv$ has values in $IP$, vanishes on $IP$, and hence defines a derivation $D_0\to IP$. The bimodule $IP$ is symmetric, since its actions come from the commutative reduced algebra. By [F1], every $1$-cocycle is the coboundary of an element of this bimodule, and such a coboundary is $w\mapsto wm-mw=0$. Thus the inner derivation vanishes. The lifted algebra is finite projective as a module and therefore finitely presented and flat. Its finite module of differentials reduces to ${\mathrm{\Omega }}_{D_0\mathrm{/}(A\mathrm{/}I)}=0$ by [F2]; nilpotence, or Nakayama, makes it zero. By [F1] the lifted algebra is finite étale.

If $I^N=0$, lift successively through $A\mathrm{/}{I}^{j+1}\to A\mathrm{/}{I}^j$ for $1\le j<N$; each kernel is square-zero because $2j\ge j+1$. Steps 1.1, 2.1 and 3.1 give existence at each stage. Given two finite étale lifts and a map between their reductions, apply the unique infinitesimal lifting in [F2] to the source algebra and to the nilpotent quotient of the target algebra. This yields one and only one algebra map upstairs. Identities and compositions are preserved by uniqueness, proving full faithfulness. Together with existence this proves the equivalence. The AC use is exactly the inherited use in [F3].

By [F1] construct compatible finite étale algebras $D_n$ over $A\mathrm{/}{I}^n$. Each is free of the same finite rank $r$: its residue-field dimension is constant under reduction. Choose a basis of $D_1$ and lift it successively to $D_{n+1}$. Nakayama makes each lifted list a basis, since the source and target are free of the same rank and its determinant reduces to a unit. Thus these identifications respect the transition maps, and is $A^r$ as a module. The compatible multiplication tables and units define a commutative associative unital algebra structure on it. Its reduction modulo $I^n$ is $D_n$.

The algebra $D$ is finite free, hence flat and finitely presented as an algebra by [F1]. Its finite module of differentials has reduction modulo $I$ equal to zero by [F2], and Nakayama makes it zero. Thus $D$ is finite étale. Given two finite étale $A$-algebras and a map between their reductions, [F1] gives compatible maps modulo all $I^n$. Their matrices have a unique limit because finite free modules are complete and separated. The limit preserves multiplication and unit, since these identities hold modulo every $I^n$; conversely any map is determined by all those reductions. This proves the equivalence.

If $A$ is initially complete for its maximal ideal, it is complete for every ideal $I\subseteq \mathfrak{m}$. Choose representatives for a compatible system modulo $I^n$. They form a maximal-adic Cauchy sequence since $I^n\subseteq {\mathfrak{m}}^n$, and hence have a limit in $A$. Each $I^n$ is closed in the maximal-adic topology, because $A\mathrm{/}{I}^n$ is a complete separated finite module by [F3]; consequently the limit has every prescribed residue modulo $I^n$. Injectivity follows from $\bigcap I^n\subseteq \bigcap {\mathfrak{m}}^n=0$. This proves the parameter-ideal application in the Statement without an additional completeness assumption.

By [F1], both $x$ and $y$ act injectively on $M$. Moreover $y$ acts injectively on $M\mathrm{/}{x}^{r}M$ for every $r\ge 1$: its filtration by the powers of $x$ has successive quotients isomorphic to $M\mathrm{/}xM$, on which $y$ is injective. Thus, inside $M_{xy}$, the intersection $M_x\cap M_y$ is $M$. Indeed if $a\mathrm{/}{x}^r=b\mathrm{/}{y}^t$, injectivity of $xy$ gives $y^{t}a=x^{r}b$ after clearing denominators. Injectivity of $y^t$ on $M\mathrm{/}{x}^{r}M$ gives $a=x^{r}c$, so the common element is $c\in M$.

A section of $\tilde{M}$ on $U$ restricts to elements of $M_x$ and $M_y$ agreeing in $M_{xy}$, hence comes from some $c\in M$ by step 1.1. The difference from the section defined by $c$ vanishes on $D(x)$. On any affine principal open contained in $U$, an element whose localization at $x$ is zero is killed by a power of $x$; injectivity of $x$ on the localized module forces it to vanish. Thus the difference vanishes on all of $U$. Conversely $M\to M_x$ is injective, so restriction is injective. This proves the Hartogs assertion.

Assume $\mathrm{depth}A\ge 2$ for these remaining assertions. Generate $\mathfrak{m}$ by $a_1,\dots ,a_s$, so that $U$ is covered by the finitely many affine opens $D(a_i)$. Sections of $\tilde{M}$ are the kernel of the difference of the two maps ${\prod }_i{M}_{a_i}\rightrightarrows {\prod }_{i,j}{M}_{a_i{a}_j}$. A flat base change commutes with this finite kernel and these localizations. For $M=A$, step 2.1 therefore gives $\mathrm{\Gamma }(U_C,\mathcal{O})=C$. A finite projective $P$ is an idempotent summand of $C^r$, so applying that idempotent to the equality of sections gives $\mathrm{\Gamma }(U_C,\tilde{P})=P$. The module ${\mathrm{Hom}}_C(P,Q)$ for two finite projectives is again finite projective (it is $P^{\ast }{\otimes }_{C}Q$); its sections on $U_C$ are therefore exactly its module elements. This proves full faithfulness.

Extend $f$ to a regular system of parameters $f,x_2,\dots ,x_d$. By [F1], $A$ has depth at least three, while $A\mathrm{/}{f}^{n}A$ has the regular sequence $x_2,x_3$: this follows by filtering it with the powers of $f$ and using regularity modulo $f$. Thus [F2] gives $\mathrm{\Gamma }(U,\mathcal{O})=A$ and $\mathrm{\Gamma }(U_n,\mathcal{O})=A\mathrm{/}{f}^{n}A$. Also $A$ is complete for the $f$-adic topology. An $f$-adic Cauchy sequence is $\mathfrak{m}$-adically Cauchy; its limit remains in every prescribed residue class modulo $f^n$, since $A\mathrm{/}{f}^{n}A$ is a separated complete finite module by [F2]. Injectivity follows from $\bigcap f^{n}A\subseteq \bigcap {\mathfrak{m}}^n=0$. Therefore .

Every finite locally free sheaf $H$ on the quasi-compact open $U$ admits an exact sequence $0\to H\to {\mathcal{O}}_U^r\to {\mathcal{O}}_U^s$ whose first cokernel is finite locally free. Here are the globalization details. For any quasi-coherent sheaf on $U$, its sections are a finite equalizer over principal affine opens $D(a_i)$ covering $U$ and their affine intersections. Localizing that equalizer shows $\mathrm{\Gamma }(U,H{)}_a=\mathrm{\Gamma }(D(a),H)$ for $D(a)\subseteq U$, since localization is flat. Take local bases of $H^{\ast }$ on a finite principal cover. Multiply each basis section by a sufficient power of its defining element to make it a global section; the resulting finitely many global sections still span $H^{\ast }$ on the corresponding opens, hence give a surjection ${\mathcal{O}}_U^r\to H^{\ast }$. Its kernel is locally free, since the quotient is locally free and the sequence splits locally. Dualizing gives an injection $H\to {\mathcal{O}}_U^r$ with locally free cokernel $K$. Apply the same construction to $K^{\ast }$ and dualize to embed $K\to {\mathcal{O}}_U^s$. This is the desired sequence, and its restriction to every $U_n$ remains exact because the local splitting makes it universally exact.

Taking sections expresses $\mathrm{\Gamma }(U,H)$ as the kernel of $A^r\to A^s$ using step 1.1. Similarly $\mathrm{\Gamma }(U_n,H{\mathrm{\mid }}_{U_n})$ is the kernel of $(A\mathrm{/}{f}^{n}A{)}^r\to (A\mathrm{/}{f}^{n}A{)}^s$. Inverse limits preserve kernels, and the compatible matrices on the right are reductions of the original matrix. Step 1.1 therefore gives . Apply this to the locally free sheaf $\mathcal{H}om(E,G)$ to obtain the claimed bijection. Apply it also to the Hom sheaf in the opposite direction: compatible inverse maps extend, and their compositions are identities because this bijection is injective. Multiplication and unit identities are maps between tensor products of locally free sheaves, so their equality is likewise detected on all $U_n$.

Trace and its determinant commute with every scalar extension because they are the trace and determinant of matrices of multiplication on the specified free module. Over an algebraically closed field, each local Artinian factor $C$ has residue field that field. If its nilradical is nonzero, each nonzero nilpotent $z$ is orthogonal under trace to every $w\in C$, since multiplication by $zw$ is nilpotent and has trace zero; hence the pairing is degenerate. If the nilradical is zero, the algebra is a product of copies of the field, with trace pairing the ordinary diagonal nondegenerate pairing. Therefore the determinant is nonzero exactly when the geometric fibre is reduced, or equivalently regular of dimension zero. By [F1] this is exactly étaleness.

Under the local hypotheses, generic étaleness says $\mathrm{\Delta }\mathrm{\ne }0$. If $\mathrm{\Delta }$ were not a unit, a prime minimal over $(\mathrm{\Delta })$ would have height at most one by [F2]; because $A$ is a domain and $\mathrm{\Delta }\mathrm{\ne }0$, its height is one. It is therefore in the punctured spectrum of the local ring of dimension at least two, contradicting step 1.1 and the assumed étaleness there. Thus $\mathrm{\Delta }$ is a unit and $D$ is étale everywhere.

The normalization construction provides a finite normal $A$-algebra $B$ whose restriction is $V$. Indeed $U$ is integral and normal by [F1]. The finite étale algebra on each principal open is normal by [F2]: applying integral-closure compatibility to the inclusion of its normal base ring into its fraction field identifies that algebra with the integral closure in its generic, separable algebra, a product of fields. Thus $V$ is a finite disjoint union of integral normal schemes, and its generic fibre is a finite product of finite separable extensions of $\mathrm{Frac}A$. Normalize $A$ in these fields; [F2] makes their product $B$ finite. On every principal open of $U$ it agrees with the original finite normal algebra by uniqueness of integral closure, so it restricts to $V$. Empty covers use $B=0$.

Suppose $d=2$ and $B\mathrm{\ne }0$. Choose $0\mathrm{\ne }x\in \mathfrak{m}$; it acts injectively on each normal domain factor of $B$. If $\mathfrak{q}$ is associated to $B\mathrm{/}xB$, then [F3] makes $(B\mathrm{/}xB{)}_{\mathfrak{q}}$ have depth zero, so $B_{\mathfrak{q}}$ has depth one. By $S_2$ in [F2] and $x\mathrm{\ne }0$, this implies $\mathrm{ht}\mathfrak{q}=1$. Its contraction to $A$ also has height one by [F2]. By prime avoidance in [F3] choose $y\in \mathfrak{m}$ outside these finitely many contractions. Then $x,y$ is a $B$-regular sequence from $A$, giving ${\mathrm{depth}}_{A}B\ge 2$. Its projective dimension is finite by the regular-ring global-dimension assertion in [F1], so Auslander–Buchsbaum makes $B$ free over $A$. It is generically étale and étale on $U$, so [F4] makes it étale everywhere. This proves existence when $d=2$.