Normal Varieties, Normalization, and Zariski's Main Theorem

Put $\mathcal{F}=f_{\ast }{\mathcal{O}}_X$. On $U=\mathrm{Spec}A\subseteq Y$ it is the sheaf obtained from the finite $A$-module $B=\mathrm{\Gamma }(f^{-1}(U),{\mathcal{O}}_X)$: on $D(a)$ its sections are $B_a$, since the inverse image is $D(f^{\ast }a)$. The sheaf identifications agree on restrictions, so $\mathcal{F}$ is a coherent algebra. Here coherent means locally a finitely presented module; the rings $A$ are Noetherian, so finite modules are finitely presented. Tensor products and symmetric algebras of these sheaves are defined on the affine modules and glued by localization.

For a module sheaf $\mathcal{E}$ described on affine opens by finite modules and their localizations as in step 1.1, define ${\mathbf{P}}_Y(\mathcal{E})$ by gluing the spaces of one-dimensional quotients on affine charts: a presentation $A^m\twoheadrightarrow M$ realizes this space as the closed locus in ${\mathbf{P}}_U^{m-1}$ cut out by the homogeneous linear relations of $M$. On a chart where a quotient generator has nonzero value, its remaining ratios satisfy exactly the dehomogenized relations; these charts glue by changing ratios. The resulting space is independent of the presentation, as the quotient and its ratios give inverse maps for two presentations. For $M=0$, there is no invertible quotient and this space is empty; otherwise the described presentation charts cover it. It is separated over $Y$, because locally it is closed in a projective space and the relative diagonal of projective space is cut out by the cross-product equations.

We prove the needed global generation without a cohomology theorem. Let $U_a=Y\cap D_{+}(x_a)$, omitting empty charts. Fix a section $s$ of $\mathcal{E}$ on $U_a$. On $U_b\cap U_a=D(x_a\mathrm{/}{x}_b)\subseteq U_b$, it is an element of the localized module of $\mathcal{E}{\mathrm{\mid }}_{U_b}$. Therefore multiplying by $(x_a\mathrm{/}{x}_b{)}^n$ for large enough $n$ extends it over $U_b$; using the trivialization $x_b^n$ of ${\mathcal{O}}_Y(n)$, these are extensions of $x_a^{n}s$ as sections of $\mathcal{E}(n)$. Choose one $n$ for the finite cover. On each affine overlap $U_b\cap U_c$, the two extensions agree after inverting $x_a$; their difference is consequently killed by some power of its local equation. There are finitely many overlaps, so multiplying every extension by the same further power $x_a^d$ makes them agree everywhere. They glue to a global section of $\mathcal{E}(n+d)$ which restricts to $x_a^{n+d}s$ on $U_a$.

Take $\mathcal{E}={\mathcal{O}}_Y\oplus \mathcal{F}$. The evaluation surjection $f^{\ast }\mathcal{E}\to {\mathcal{O}}_X$, $(a,b)\mapsto a+f^{\ast }b$, defines a morphism $i:X\to {\mathbf{P}}_Y(\mathcal{E})$. Its image lies in the open chart where the distinguished generator of ${\mathcal{O}}_Y$ is nonzero. Over $U=\mathrm{Spec}A$ this chart is the affine space with coordinate algebra ${\mathrm{Sym}}_{A}B$, and $i$ corresponds to the surjective algebra map ${\mathrm{Sym}}_{A}B\to B$ induced by the identity on $B$. Thus $i$ is a closed immersion into that open chart, hence an immersion into ${\mathbf{P}}_Y(\mathcal{E})$. These maps glue because evaluation does.

The immersion in step 3.1 is closed, as follows directly from integral equations. On an affine $U$, choose finite module generators $b_1,\dots ,b_t$ of $B$; with the distinguished generator $1$ of ${\mathcal{O}}_Y$, they present $\mathcal{E}$. Write the corresponding homogeneous coordinates as $z_0,z_1,\dots ,z_t$. Define a graded algebra $T$ with $T_0=A$ and $T_n=B$ for $n\ge 1$, with multiplication induced by that of $B$. The evaluation map is the graded surjection ${\mathrm{Sym}}_A(A\oplus B)\to T$ sending $z_0$ to $1\in T_1$ and $z_i$ to $b_i\in T_1$; write $K$ for its homogeneous kernel. For each $b_i$, integrality gives $b_i^{d_i}+{\sum }_{j<d_i}{a}_{ij}{b}_i^j=0$. The homogeneous equation $z_i^{d_i}+{\sum }_{j<d_i}{a}_{ij}{z}_i^j{z}_0^{d_i-j}=0$ vanishes under evaluation and belongs to the kernel $K$. Any homogeneous prime containing $K$ and $z_0$ therefore contains every $z_i$, so it is irrelevant and gives no projective point. Consequently the closed homogeneous-kernel locus has no points outside $D_{+}(z_0)$. On that chart its coordinate algebra is precisely the quotient ${\mathrm{Sym}}_{A}B\twoheadrightarrow B$ of step 3.1; hence the locus equals $X_U$ as a scheme. These closed embeddings glue, since evaluation and its kernel commute with localization. Thus $X\hookrightarrow {\mathbf{P}}_Y(\mathcal{E})$ is a closed immersion. Also, finite algebras remain finite after arbitrary base change: the elements $1\otimes b_i$ generate $A^{\mathrm{\prime }}{\otimes }_{A}B$; quotients remain finite. By [F2] such maps are integral and closed by lying over, so this also proves universal closedness of $f$. Its affine diagonal is closed because multiplication $B{\otimes }_{A}B\twoheadrightarrow B$ is surjective, and its finite algebra is of finite type.

The finite-algebra calculations at the end of step 4.1 use no projective embedding of the target: they prove separatedness, finite type and universal closedness for any finite morphism. In the classical register a complete variety has universally closed structure morphism; composing it with a universally closed finite morphism remains universally closed after every base change, since the image of a closed subset is closed under each factor. The source is separated over $k$: its absolute diagonal factors through the relative diagonal and the inverse image of the closed absolute diagonal of the complete target. Finite type composes as well. Thus the source is complete. In particular the finite normalization over a projective curve has this property.

Apply step 2.2 to finite module generators on every $U_a$. There are finitely many such generators. Raise all resulting twists to one common $m$ by multiplying the section coming from $U_a$ by the required further power of $x_a$. On $U_a$ the factor $x_a^m$ is a trivializing unit, so the resulting global sections generate $\mathcal{E}(m)$ there. Thus there is a surjection ${\mathcal{O}}_Y^{N+1}\twoheadrightarrow \mathcal{E}(m)$. Tensoring a one-dimensional quotient with an invertible sheaf identifies ${\mathbf{P}}_Y(\mathcal{E})$ with ${\mathbf{P}}_Y(\mathcal{E}(m))$: locally the identification cancels the same unit in every homogeneous coordinate. The surjection gives a closed immersion of this projectivization into ${\mathbf{P}}_Y^N$, by the homogeneous linear relations described in step 2.1. Combining with step 4.1 gives a closed immersion $X\hookrightarrow {\mathbf{P}}_Y^N$.

Since $Y\subseteq {\mathbf{P}}_k^r$ is closed, ${\mathbf{P}}_Y^N={\mathbf{P}}_k^N\times Y$ is closed in ${\mathbf{P}}_k^N\times {\mathbf{P}}_k^r$. The Segre map sends $([u_i],[v_j])$ to $[u_i{v}_j]$ and is a closed immersion: its image is the nonzero rank-one matrices, cut out by all two-by-two minors, and on a chart with a nonzero entry the row and column ratios recover both factors regularly. Composition gives a closed immersion of $X$ into ${\mathbf{P}}_k^{(N+1)(r+1)-1}$. This proves both assertions. All chart equations and their inverse coordinate maps are over $k$; neither algebraic closure nor perfectness is used.

Cover $f^{-1}U$ by finitely many affine opens $W_i$. Cover their pairwise intersections by finitely many affine opens, using [F1]. The sheaf axiom realizes $M_U$ as the kernel of the difference map from the finite product of the coordinate rings of $W_i$ to the finite product of the overlap-chart rings. After tensoring by any flat $A$-algebra $E$, exactness and commutation with finite products identify that kernel with the sections on the base-changed affine cover. This proves the flat-base-change assertion. In particular $M_{D(a)}=(M_U{)}_a$. By [F2], $C_{D(a)}=(C_U{)}_a$. The identifications are canonical restrictions and preserve multiplication.

Let $S_i$ be the coordinate ring of $W_i$ and $\mathfrak{q}$ a minimal prime of $S_i$; set $D=S_i\mathrm{/}\mathfrak{q}$ and $A_0=A\mathrm{/}\ker (A\to D)$. The corresponding irreducible component maps dominantly onto its image closure, with finite fibres because it is a closed subvariety of $W_i$. By [F3] its function field $L=\mathrm{Frac}D$ is a finite extension of $\mathrm{Frac}{A}_0$: the general fibre is zero-dimensional, so the relative transcendence degree is zero, and both fields are finitely generated. Choose a polynomial subring $R\subseteq A_0$ for which $A_0$ is module-finite by [F4]. Then $L\mathrm{/}\mathrm{Frac}R$ is finite. The integral closure $B$ of $R$ in $L$ is finite over $R$. Every element integral over $A_0$ is integral over $R$ by [F2]; hence the integral closure of $A_0$ in $D$ embeds as an $R$-submodule of $B$, and is finite over $R$ by [F4]. Its $R$-generators are also $A_0$-generators, since it is an $A_0$-module. It is therefore finite over $A$.

The reduced ring $S_i$ injects into the finite product of its minimal-prime domains $D$. Restricting an element integral over $A$ into each of these domains gives an element of the finite module found in step 1.2. Thus ${\mathrm{Int}}_A(S_i)$ is a submodule of a finite $A$-module and is finite by [F4]. The restriction map $M_U\hookrightarrow {\prod }_i{S}_i$ is injective by the sheaf axiom. Consequently $C_U$ is a submodule of the finite product ${\prod }_i{\mathrm{Int}}_A(S_i)$ and is finite. It is reduced because it is a subring of the ring of actual functions on $f^{-1}U$. If the inverse image is empty, $M_U=C_U=0$ and all assertions hold.

On every principal open of an affine target chart, step 1.1 identifies the finite algebra restrictions. On an overlap of target charts use a finite principal-open refinement; the identifications agree because each comes from restriction of actual functions. They satisfy the cocycle condition and glue their finite affine spaces and sheaves to $N$. Finiteness is affine local by its defining finite-module charts, so $N\to Y$ is finite. On $f^{-1}U$, every element of $C_U$ is a global regular function and evaluation is a regular map to the affine space of that algebra: choose finitely many algebra generators, evaluate them, and note that their defining relations vanish. The maps agree on the same refinements and give $j:X\to N$. Thus neither the charts nor this map depend on the covers chosen in the proof.

By [F1], each displayed zero-dimensional chart is flat and every geometric fibre is regular of dimension zero. A finite-type zero-dimensional algebra over a field is Artinian, and a regular zero-dimensional local ring is a field. Thus the fibres are finite products of finite separable field extensions; at a classical point over $k$ the selected factor is $k$. Base change and composition preserve the presentation and relative dimension zero by [F1]. Open restrictions localize these presentations. At a selected local ring the flat map is local, hence faithfully flat by [F4]. If the base algebra $A$ is reduced, its injection into the finite product of the fraction fields of its minimal-prime quotients remains injective after tensoring with the flat chart. The resulting factors are reduced by the geometric-fibre assertion. The chart is therefore reduced, as a subring of a product of reduced rings.

We justify openness on prime spectra before restricting to classical points. For an injective map of finite-type $k$-domains $R\hookrightarrow S$, generic affine normalization [F3] gives $0\mathrm{\ne }a\in R$ such that $S_a$ is finite and integral over an injected polynomial ring $R_a[z_1,\dots ,z_r]$. Each prime $\mathfrak{p}$ of $R_a$ extends to the prime $\mathfrak{p}{R}_a[z]$ and lifts to $S_a$ by lying over. Thus the spectral image contains $D(a)$ in its image closure. For a general affine source, pass to its finitely many minimal-prime quotients and the corresponding image closures. On each irreducible component remove the inverse image of such a principal image open and induct on the proper closed source complement. Noetherian induction terminates this argument and proves constructibility of the whole image. An open source has a finite principal-open cover, so the same argument applies to images of opens. Hence the image of any open under a finite-type map of the present Noetherian affine spaces is constructible. By going down, the image under a flat map is stable under generization: apply [F3] to the localization at a source point in that open. A constructible generization-stable subset of a Noetherian spectrum is open. Indeed its complement is constructible and specialization-stable; write that complement as a finite union of locally closed subsets, include the generic points of their irreducible closures, and specialization stability then includes their entire closures. The complement is closed. Thus the chart map is open. This remains true after base change between the present finite-type algebras. Restricting to closed $k$-points gives openness of the classical map: a nonempty fibre over a closed point has a closed point because it is a finite-type $k$-algebra.

We prove the monic local form at a classical point $q$ of a chart $A\to S$. By step 1.1 this is quasi-finite everywhere. Apply [F2] at $q$ to obtain $g$ in the relative integral closure $C\subseteq S$, nonzero at $q$, with $C_g=S_g$. Choose finite $A$-algebra generators of $S_g$, express them with numerators in $C$ and powers of $g$ as denominators, and let $T\subseteq C$ be generated by those numerators and $g$. Then $T$ is finite over $A$ and $T_g=S_g$. Thus an affine neighbourhood of $q$ is open in $\mathrm{Spec}T$. The fibre $T{\otimes }_{A}k$ is Artinian. Its selected local factor is $k$, because its localization agrees with the selected fibre local ring of $S$. Choose $t\in T$ whose image is $1$ in that factor and $0$ in every other factor; lifting is possible because the residue field of the base point is $k$. Put $D=A[t]\subseteq T$. The selected prime $q_D$ has exactly one prime of $T$ above it, because the selected nonzero value separates it from all other fibre factors. Consequently $T_{q_D}$ is local, finite over $D_{q_D}$, and its closed fibre is generated by $t$ and equals $k$. Nakayama applied to the finite cokernel of $D_{q_D}\hookrightarrow T_{q_D}$ makes that injection an isomorphism. Vanishing of its finite cokernel after one further localization identifies the original chart near $q$ with a localization of the monogenic finite algebra $D$.

Let $I=\ker (A[T]\to D)$ and let $Q$ be the selected prime of $A[T]$. The localized relative differentials of $D$ vanish by [F5] and the preceding identification. Thus a finite linear combination of derivatives of elements of $I$ is a unit at $Q$. Lift the coefficients to fractions of polynomials, clear a denominator outside $Q$, and sum the coefficient multiples of these relations. The derivatives of the coefficients multiply relations vanishing in $D$, so this produces $P_1\in I$ with $P_1^{\mathrm{\prime }}(t)$ a unit at the selected point. Since $t$ is integral, choose monic $P_0\in I$. For $N\ge 2$ and $N\mathrm{deg}{P}_0>\mathrm{deg}{P}_1$, put $P=P_0^N+P_1$. It is monic, belongs to $I$, and $P^{\mathrm{\prime }}(t)=P_1^{\mathrm{\prime }}(t)$ is a unit. The selected local fibre of $D^{\mathrm{\prime }}=A[T]\mathrm{/}(P)$ is a field: the selected root is simple, so localizing the finite polynomial fibre removes every other factor and leaves $k$. The fibre surjection $D_{Q\mathrm{/}(P)}^{\mathrm{\prime }}\to D_{q_D}$ is consequently an isomorphism.

Both local rings in step 3.1 are flat over the local base: the source is a localization of the finite free monic algebra $D^{\mathrm{\prime }}$, and the target agrees with the original flat chart. Tensor the kernel sequence with the base residue field; flatness of the target gives an injection on the kernel, and the identical fibres give $K\mathrm{/}\mathfrak{m}K=0$. The kernel is finite because the rings are Noetherian. Nakayama yields $K=0$. Finite generation permits a principal shrinking on which $D^{\mathrm{\prime }}\to D$ is an isomorphism; shrink inside the original chart and invert also $P^{\mathrm{\prime }}$. This is the desired monic presentation. Such neighbourhoods at all closed points cover the chart, since a nonempty closed subset of a finite-type $k$-space has a closed point. Conversely a monic presentation with invertible derivative is one of the square-Jacobian presentations of [F1]. Steps 1.1 and 2.1 give all the remaining assertions.

First lift a monic coprime factorization $\bar{P}=\bar{I}\bar{H}$ over $k$, of positive degrees $r,s$, for a monic $P\in A[T]$. Introduce the $r+s$ coefficients of monic universal factors $I,H$ and impose the $r+s$ coefficient equations $IH=P$. Their square Jacobian acts by $(\delta I,\delta H)\mapsto H\delta I+I\delta H$, with $\mathrm{deg}\delta I<r$, $\mathrm{deg}\delta H<s$. At the chosen coprime factors its kernel is zero: the equation implies $\bar{H}$ divides $\delta H$, hence $\delta H=0$, and then $\delta I=0$. Invert the Jacobian determinant $d$. The resulting algebra $E$ is a square-Jacobian algebra, giving an elementary change by [F1] and [F4]. Substitution of the chosen coefficients gives a point $t$ with residue field $k$. The same invertible matrix solves $aI+bH=1$, so the factors are coprime over $E$. This construction works over any finite-type $k$-base and is compatible with restriction.

For one selected point choose affine charts $\mathrm{Spec}S\subseteq X$ and $\mathrm{Spec}A\subseteq Y$ around it and its image. Put $C={\mathrm{Int}}_A(S)$, finite by [F2]. At the selected maximal ideal CA-20 gives $g\in C$ with $C_g=S_g$. The selected point of the finite fibre of $C$ is isolated, and is the only source fibre point above it, because $g$ is nonzero there and identifies the two open fibre neighbourhoods. In the Artinian ring $C{\otimes }_{A}k$, take the idempotent equal to $1$ on that local factor and $0$ on every other factor, and lift it to $c\in C$; the residue field is $k$, so the quotient map onto this fibre is surjective. Choose monic $P$ with $P(c)=0$, multiply it by $T$ if necessary, and factor $\bar{P}=T^e\bar{H}$ with $e\ge 1$ and $\bar{H}(0)\mathrm{\ne }0$. The selected value of $c$ is $1$, so $\bar{H}(1)=0$ and $\mathrm{deg}\bar{H}\ge 1$.

Apply step 1.1 to $T^e\bar{H}$, obtaining $E$ and coprime monic factors $I,H$. In $E{\otimes }_{A}C$ and $E{\otimes }_{A}S$, the equations $I(c)H(c)=0$ and $a(c)I(c)+b(c)H(c)=1$ give compatible product decompositions. The factor where $I(c)$ is invertible, equivalently the quotient by $H(c)$, has just the selected point over $t$: on the fibre $c$ equals $1$ on the selected local factor and $0$ elsewhere, so $I(c)$ equals $1$ there and is nilpotent on the other factors. Let $C_1$ and $S_1$ be these factors. The ring $C_1$ is finite over $E$. Its closed locus $V(g)$ has closed image in $\mathrm{Spec}E$ by [F3], and that image omits $t$. Shrink to a principal neighbourhood of $t$ avoiding this image. Then $g$ is a unit in $C_1$, and also in $S_1$. The equality $C_g=S_g$ base changes and takes factors to give $(C_1{)}_g=(S_1{)}_g$, hence $C_1=S_1$. Thus $\mathrm{Spec}{S}_1$ is an open neighbourhood in the changed affine source, finite over the new base, with the required singleton chosen fibre.

This finite open neighbourhood $V$ is also closed in the entire changed source $X_T$. The finite map $V\to T$ is universally closed: tensoring its finite algebras by any base algebra stays finite and lying over on quotients proves closedness. The graph of $V\hookrightarrow X_T$ is closed in $V{\times }_T{X}_T$ by separatedness of $X_T\to T$, and projection of the graph to $X_T$ is closed by universal closedness of $V\to T$. Hence $V$ is open and closed. This argument uses separatedness exactly here.

Induct on the selected list. The first construction gives a clopen finite piece for $x_1$. In its clopen complement select the lift of $x_2$, apply the constructions of steps 1.2, 2.1 and 3.1 again, and base change the previous pieces. The lift of each $x_i$ is unique with unchanged residue field, because $k{\otimes }_{k}k=k$. By [F1] the composite base change is still elementary and the previous pieces remain clopen and finite. At each step the new piece has only its selected point in the chosen fibre, so it does not remove a later selected point. After finitely many steps the clopen complement $W$ has exactly the stated fibre exclusion. The empty list uses the identity change and $W=X$. If every fibre point was selected, the complement fibre is empty. No perfectness or source normality is used.

Flatness of $E$ preserves the injection $C\hookrightarrow B$. The image of $E{\otimes }_{A}C$ is integral over $E$, since each of its elements uses finitely many integral elements of $C$ and sums and products of integral elements remain integral by [F2]. For the opposite inclusion it suffices to work on the monic principal-open charts of [F1]; an element belongs to a submodule if its class in the quotient vanishes locally on a covering family.

Let $P\in A[T]$ be monic of degree $d>0$, put $D=A[T]\mathrm{/}(P)$, and take $c\in B[T]\mathrm{/}(P)$ integral over $D$. Since $D$ is finite free over $A$, transitivity makes $c$ integral over $A$. Write $c={\sum }_{i<d}{b}_i{T}^i$. Form a splitting algebra over $B$ by adjoining a root of $P$, dividing by its monic linear factor, then adjoining a root of the quotient and continuing. Every extension is finite free with a basis containing $1$, so $B$ injects into the splitting algebra, and its roots ${\alpha }_1,\dots ,{\alpha }_d$ are integral over $A$. The polynomial identity $P^{\mathrm{\prime }}(T)c(T)\equiv {\sum }_{i=1}^{d}c({\alpha }_i){\prod }_{j\mathrm{\ne }i}(T-{\alpha }_j)(\mathrm{m}\mathrm{o}\mathrm{d}P(T))$ holds universally: over independent formal roots, interpolation proves it after inverting the Vandermonde determinant, and injectivity of that localization in the universal polynomial domain proves the original polynomial identity. Specialization therefore permits repeated roots. Each $c({\alpha }_i)$ is integral over $A$, because evaluation carries its monic equation to a monic equation. Every coefficient on the right is integral over $A$. The degree-less-than-$d$ remainder on the left has coefficients in the injected ring $B$, so each of those coefficients belongs to $C$. Thus $P^{\mathrm{\prime }}c\in C[T]\mathrm{/}(P)$.

On the chart $E=D_h$ of [F1], let $b\in (B[T]\mathrm{/}(P){)}_h$ be integral over $E$. Localization compatibility in [F2] gives some $N$ such that $c=h^{N}b$ lies in $B[T]\mathrm{/}(P)$ and is integral over $D$. By step 1.2, $P^{\mathrm{\prime }}c$ lies in $C[T]\mathrm{/}(P)$. Both $h$ and $P^{\mathrm{\prime }}$ are invertible in $E$, whence $b=h^{-N}(P^{\mathrm{\prime }}{)}^{-1}(P^{\mathrm{\prime }}c)$ belongs to $E{\otimes }_{A}C$. Covering by these charts proves the opposite inclusion. Degree-zero monic charts are empty and cause no exception; the zero algebras are likewise harmless. Together with step 1.1 this proves the equality. AC is inherited through [F1]; the finite splitting-algebra construction adds no further choice.

The affine case. Let $U\subseteq X$ be a nonempty affine open with $A=k[U]$, and put $V=f^{-1}(U)$, $B=k[V]$, so that $B$ is a finite $A$-module and the pullback $A\to B$ exhibits $f{\mathrm{\mid }}_V$ [F1]. By [F2], $A$ is an integrally closed domain, and $B$ is a domain. By [F3], $\mathrm{Frac}(B)=\mathrm{Frac}(A)$ under the pullback, so every $b\in B$ lies in the fraction field of $A$ and is integral over $A$ because $B$ is a finite, hence integral, $A$-module [F4]. Since $A$ is integrally closed, $b\in A$; hence $B\subseteq A$, and with $A\subseteq B$ we get $A=B$. By the anti-equivalence [F4] the map $f{\mathrm{\mid }}_V$ is an isomorphism.

The general case. Cover $X$ by finitely many nonempty affine opens $U_1,\dots ,U_n$ as in [F1] and [F5]; by step 1.1 each restriction $f_i\text{\hspace{0.17em}⁣}:f^{-1}(U_i)\to U_i$ is an isomorphism, with inverse $g_i\text{\hspace{0.17em}⁣}:U_i\to f^{-1}(U_i)$. On an overlap $U_i\cap U_j$, the maps $g_i$ and $g_j$ both invert the same map $f$ on that overlap, so they agree there. Since being a morphism is a local condition on the source and the $U_i$ cover $X$, the $g_i$ glue to a morphism $g\text{\hspace{0.17em}⁣}:X\to Y$; the identities $f\circ g={\mathrm{i}\mathrm{d}}_X$ and $g\circ f={\mathrm{i}\mathrm{d}}_Y$ hold because they hold locally on the cover. Thus $f$ is an isomorphism.

If every $A_i$ is normal, each maximal localization $(A_i{)}_{{\mathfrak{m}}_x}$ is an integrally closed domain by [F2]. These are the local rings of $X$ by [F1], so $X$ is normal.

Conversely suppose $X$ is normal. For a prime $\mathfrak{p}$ of $A_i$, choose a maximal ideal $\mathfrak{m}$ containing it, using AC. By [F1] and [F2], $(A_i{)}_{\mathfrak{m}}$ is an integrally closed domain. Its further localization at $\mathfrak{p}(A_i{)}_{\mathfrak{m}}$ is $(A_i{)}_{\mathfrak{p}}$ and is integrally closed by [F3]. Thus $A_i$ is normal. For an irreducible chart, [F3] identifies this with integral closedness of its coordinate domain.

For $x\in U_i$, [F1] identifies its germ ring with $(A_i{)}_{{\mathfrak{m}}_x}$, so this one ring decides normality of $x$ independently of the chart. The empty variety and empty charts have no points or primes, and all conditions are vacuous. This proves the chartwise and pointwise assertions.

First the affine case: let $Y$ have coordinate ring $A=k[Y]$, let $X$ have coordinate ring $B=k[X]$, let $A\to B$ be the pullback of $f$, and assume $B$ is a finite $A$-module; let $Z\subseteq X$ be closed with radical ideal $J=I(Z)\subseteq B$. Then $f(Z)=V(J\cap A)$, which is closed. Indeed, if $z\in Z$ and $g\in J\cap A$, then $f^{\ast }(g)=g\circ f\in J$ vanishes at $z$, so $g(f(z))=0$, giving $f(Z)\subseteq V(J\cap A)$. Conversely let $y\in V(J\cap A)$ and let $\mathfrak{m}={\mathfrak{m}}_y\subseteq A$ be the corresponding maximal ideal, which contains $J\cap A$. The induced map $A\mathrm{/}(J\cap A)\to B\mathrm{/}J$ is integral, because a monic integrality equation for $b\in B$ over $A$ reduces modulo $J\cap A$ to a monic equation for the class of $b$; and $\bar{\mathfrak{m}}=\mathfrak{m}\mathrm{/}(J\cap A)$ is maximal. By lying over [F3] there is a prime $\bar{\mathfrak{q}}\subseteq B\mathrm{/}J$ contracting to $\bar{\mathfrak{m}}$, and $\bar{\mathfrak{q}}$ is maximal by [F4]; it is the image of a maximal ideal $\mathfrak{q}\subseteq B$ containing $J$. By [F5] the maximal ideal $\mathfrak{q}$ is ${\mathfrak{m}}_z$ for a point $z\in X$, and $J\subseteq \mathfrak{q}$ with $Z=V(J)$ gives $z\in Z$. Finally $\mathfrak{q}\cap A=\mathfrak{m}$, since both ideals contain $J\cap A$ and have the same image in $A\mathrm{/}(J\cap A)$; as contraction along the pullback is precomposition, $\mathfrak{q}\cap A=\{g\in A:g\circ f\in {\mathfrak{m}}_z\}=\{g\in A:g(f(z))=0\}={\mathfrak{m}}_{f(z)}$. Hence ${\mathfrak{m}}_{f(z)}={\mathfrak{m}}_y$, so $f(z)=y$ by the bijectivity in [F5], and $y\in f(Z)$.

Finite fibres: let $y\in Y$ and choose an index $i$ with $y\in U_i$; then $f^{-1}(y)\subseteq f^{-1}(U_i)$, and the restriction $f_i\text{\hspace{0.17em}⁣}:f^{-1}(U_i)\to U_i$ is a module-finite morphism of affine classical algebraic sets, because $B_i$ is a finite $A_i$-module by the choice of the cover [F1, F2]. By [F6] the map $f_i$ is quasi-finite, so its fibre $f_i^{-1}(y)=f^{-1}(y)$ is a finite set.

Now the general closedness statement. Let $Z\subseteq X$ be closed. A subset of $Y$ is closed exactly when its traces in the members of the open cover $U_1,\dots ,U_n$ are closed, and $f(Z)\cap U_i=f_i(Z\cap f^{-1}(U_i))$, where $Z\cap f^{-1}(U_i)$ is closed in the affine variety $f^{-1}(U_i)$. By step 1.1 applied to the affine map $f_i$, each $f(Z)\cap U_i$ is closed in $U_i$. Hence $f(Z)$ is closed in $Y$, so $f$ is closed.