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By part (a), work first with an affine base and its homogeneous equations. Put $S=k[y_0,\dots ,y_n]$, write $S_N$ for its finite-dimensional degree-$N$ vector space, and use $S_j=0$ for $j<0$. At each base point define the graded multiplication map for a projective fiberIts image is exactly $(I_x{)}_N$, since taking the degree-$N$ part of any expression in the homogeneous ideal discards all other degrees. Moreoverbecause degree-$N$ monomials generate ${\mathfrak{m}}^N$.

Choose monomial bases for the domain summands and codomain. Every entry of the resulting matrix is a coefficient of one of the $f_i$, hence a regular function of $x$. If $h_N=\dim S_N$, failure of surjectivity is the vanishing of all $h_N$-by-$h_N$ matrix minors. It is therefore a closed rank-defect locus $C_N$. If the matrix has fewer than $h_N$ columns, the failure is automatic and $C_N=X$, also closed. Constants and $N=0$ are covered by the same construction.

Part (c) gives $\pi (Y)={\bigcap }_{N\ge 0}{C}_N$, an intersection of closed subsets, hence closed. Apply part (a) to pass back to arbitrary $X$. Thus projection from $X\times {\mathbb{P}}^n$ sends every closed subset to a closed subset. This proves the closedness of projection from projective space without assuming the properness result being proved.

The field convention matters: a rational-point projection need not be Zariski closed. Over $\mathbb{R}$, the closed equation $y_0^2=t{y}_1^2$ in ${\mathbb{A}}^1\times {\mathbb{P}}^1$ has real-point image $[0,\infty )$. Indeed $y_1=0$ would force $y_0=0$, so a solution exists precisely when $t$ is a square in $\mathbb{R}$. This infinite proper subset of the affine line is not Zariski closed. The valid arbitrary-field theorem uses scheme or geometric images, for which the matrix argument remains valid at every residue field.

Use the standard convention that a projective variety is nonempty and irreducible, over an algebraically closed field. Let $f:X\to {\mathbb{A}}^1$ be a regular function, and compose with the inclusion into ${\mathbb{P}}^1$. Its graph is closed in $X\times {\mathbb{P}}^1$: on an affine open of $X$, with homogeneous target coordinates $[T_0:T_1]$, it is given by $T_1-f(x)T_0=0$. This equation has no solution at $T_0=0$, and the local closedness statements glue.

Embed $X$ as a closed subvariety of ${\mathbb{P}}^m$. The graph is then closed in ${\mathbb{P}}^m\times {\mathbb{P}}^1$. Part 1, after swapping factors, makes its projected image $f(X)$ closed in ${\mathbb{P}}^1$. Since it avoids infinity, it is a proper closed subset of the projective line and hence finite. A continuous image of an irreducible space is irreducible, and a nonempty finite closed subset over an algebraically closed field is irreducible only if it is a singleton. HenceThis proves global regular functions on an irreducible projective variety are constant, using the closed-projection result rather than assuming it. If one allows disconnected projective algebraic sets instead of irreducible varieties, different components can carry different constants; irreducibility is the relevant convention here.

The affine space $X={\mathbb{A}}_k^n$ is an integral scheme, hence its underlying topological space is irreducible. Every nonempty open subset of an irreducible topological space is irreducible and in particular connected. A section of the constant sheaf with value the function field $K=k(t_1,\dots ,t_n)$ is a locally constant function, so on any such open subset it is constant. Thus its sections are $K$ on each nonempty open subset and zero on the empty set. All restriction maps are either identities on $K$ or maps onto zero.

The constant sheaf is therefore a flasque sheaf. It has its natural ${\mathcal{O}}_X$-module structure because a regular function on a nonempty open subset embeds in the function field and acts by multiplication. Part (i) now givesThis is an instance of constant sheaves on irreducible spaces are flasque; no restriction on $n$ is needed.

Take the affine scheme $X={\mathbb{A}}_k^1$, its function field $K=k(t)$, and the constant sheaf $\mathcal{K}$ with value $K$. By Question 4(ii), $\mathcal{K}$ is a flasque sheaf. At the two distinct closed rational points $p_0=0$ and $p_1=1$, let $S_0,S_1$ be skyscraper sheaves with value $K$: on an open subset $V$, $S_i(V)=K$ if $p_i\in V$ and is zero otherwise. Both are flasque sheaves.

There is a natural morphism of sheaveswhich sends a constant rational function to the same element in every summand present on the open subset. At $p_i$, its stalk map is the identity $K\to K$, and at any other point the target stalk is zero. Thus $\rho$ is a surjection of sheaves, although it is not surjective on global sections. Put $\mathcal{F}=\ker \rho$.

This can even be an example of sheaves of ${\mathcal{O}}_X$-modules: let regular functions act on each copy of $K$ by multiplication after embedding them in $K$. At the closed points this action is multiplication by a rational function, not evaluation at the point. With these actions, $\rho$ is ${\mathcal{O}}_X$-linear.

The exact sequence $0\to \mathcal{F}\to \mathcal{K}\to S_0\oplus S_1\to 0$, together with flasque acyclicity, givesHence $H^0(X,\mathcal{F})=0$ and $H^1(X,\mathcal{F})=K^2/\Delta K\simeq K\ne 0$. In contrast, use the one-member affine open cover $\mathcal{U}=\{X\}$. Its alternating Čech cochain complex has no positive-degree terms, so ${\check{H}}^1(\mathcal{U},\mathcal{F})=0$. We have the explicit discrepancyThe sheaf $\mathcal{F}$ is not a quasi-coherent sheaf, as also follows from its nonzero higher sheaf cohomology on an affine scheme. Thus this example does not contradict the acyclic cover theorem for quasi-coherent sheaves on affine covers.

A rational function with no poles away from $P_{\infty }$ is regular on the smooth affine curve. Its ring of regular functions is the coordinate ringso it has a unique expression $A(x)+yB(x)$ with polynomials $A,B$. This regular-function assertion uses smoothness, hence normality, of the affine curve, not Riemann-Roch theorem.

The nonconstant leading term of $A$ has even pole order $2\mathrm{deg}A$ at infinity, while $yB$ has odd pole order $3+2\mathrm{deg}B$. These different parities prevent cancellation. Thus no nonconstant such function has a pole of order at most one. Therefore

The same pole-order argument provides a basisFor $n\ge 0$ the first set contains $\lfloor n/2\rfloor +1$ elements, and the second contains $\max (0,\lfloor (n-3)/2\rfloor +1)$. Distinct pole orders imply linear independence, and the coordinate-ring expression proves spanning. HenceFor $n<0$ a function in the space would have no poles anywhere and would vanish at infinity; a regular function on a projective integral curve is constant, so it must be zero. Equivalently the coordinate-ring pole argument first makes it constant, then the vanishing condition excludes nonzero constants.

Put $A=k[X]$. There is an explicit isomorphism from the principal open subset $U_f$ to the affine algebraic setIts inverse is projection, and both maps are regular maps. The coordinate ring isby the universal property of localization. Since regular functions on an affine variety are its coordinate ring, this proves the natural $k$-algebra isomorphism

Now $W=D(x)\cup D(y)$. A regular function on $W$ determines one element of the function field $k(x,y)$, and its restrictions belong to $k[x,y{]}_x$ and $k[x,y{]}_y$. Conversely, elements of their intersection glue because they agree on $D(xy)$. ThusIf a reduced fraction has denominator dividing a power of $x$ and also a power of $y$, unique factorization makes its denominator a unit. Therefore puncturing the plane does not change its ring of global regular functions:If $W$ were an affine variety, its inclusion $j:W\hookrightarrow {\mathbb{A}}^2$ would correspond to the identity isomorphism $k[x,y]\to k[W]$. The equivalence between affine varieties and their coordinate rings would make $j$ an isomorphism, contradicting the missing origin. Hence $W$ is not affine. This illustrates regular functions on the punctured affine plane: the coordinate ring alone does not recover a nonaffine variety.

An irreducible scheme is a nonempty scheme whose underlying topological space cannot be expressed as the union of two proper closed subsets. Equivalently, any two nonempty open subsets meet. A reduced scheme is one whose local rings have no nonzero nilpotent elements. Equivalently, every affine open subscheme has a reduced ring of regular functions. We may define an integral scheme as a nonempty scheme for which the coordinate ring of every nonempty affine open subscheme is an integral domain. We shall show that this is equivalent to being reduced and irreducible. The nonempty convention matters: the empty scheme is reduced but is not irreducible or integral.

For a commutative ring $A$, the spectrum of a ring is reduced exactly when $A$ is a reduced ring. One direction follows since localization preserves reducedness. Conversely, if $a$ is a nonzero nilpotent element, choose a prime ideal containing its proper annihilator. Then $a/1$ cannot vanish at that localization, contradicting reducedness of its local ring.

The spectrum of a ring is irreducible exactly when its nilradical $N=\sqrt{(0)}$ is a prime ideal. To see the essential implication directly, if $ab\in N$, then $D(a)\cap D(b)=D(ab)$ is empty. Irreducibility forces $D(a)$ or $D(b)$ to be empty, hence $a\in N$ or $b\in N$. Also $N$ is proper because the spectrum is nonempty. Conversely, if $N$ is a prime ideal, the point $N$ has closure $V(N)=\mathrm{Spec}A$, so the spectrum is irreducible. Combining the two criteria gives

Now suppose $X$ is reduced and irreducible. Every nonempty affine open subscheme is also reduced and irreducible: irreducibility passes to nonempty open subsets, because their nonempty open subsets are nonempty opens of $X$. The affine criterion makes each coordinate ring an integral domain, so $X$ is integral. Conversely, suppose all its nonempty affine coordinate rings are integral domains. Their localizations show that $X$ is reduced. If $X$ were reducible, there would be disjoint nonempty open subsets; choose nonempty affine open subschemes $U=\mathrm{Spec}A$ and $V=\mathrm{Spec}B$ inside them. Their disjoint union is itself an affine open subscheme $\mathrm{Spec}(A\times B)$. Since $A,B$ are nonzero, $(1,0)(0,1)=0$ contradicts the domain condition. Therefore

The generic point of an integral scheme $X$ is the unique point $\eta$ with $\overline{\{\eta \}}=X$. For existence, take a nonempty affine open subscheme $V=\mathrm{Spec}A$. Its point $(0)$ has closure containing $V$, which is dense in $X$, so its closure in $X$ is all of $X$. For uniqueness, both proposed generic points lie in every nonempty open subset, hence in $V$, where the only dense point is $(0)$. The function field isHere the stalk description shows that the field of fractions is independent of the choice of nonempty affine open subscheme.

Every nonempty open subscheme $U$ contains $\eta$, so taking a germ there defines $\Gamma (U,{\mathcal{O}}_X)\to K(X)$. If a section has zero germ, restrict it to any nonempty affine open subscheme $V\subseteq U$. Its image in $\mathrm{Frac}\Gamma (V,{\mathcal{O}}_X)$ is zero. Since this coordinate ring is an integral domain, the section is zero on $V$. Such affines cover $U$, and the sheaf gluing axiom makes the section zero on $U$. Thus the generic-point embedding of regular functions is

For a nonreduced reducible fibre between integral schemes, take both source and target to be the affine line over $\mathbb{C}$, and use the ring homomorphismBoth coordinate rings are integral domains, so both schemes are integral. At the target closed point $t=0$, the scheme-theoretic fibre has ringby the Chinese remainder theorem. Its underlying space has two distinct closed points, so it is reducible. The class of $x(x-1)$ is nonzero but has square zero, so it is not reduced. Even a morphism between integral schemes can have a fibre consisting of two nonreduced double points.

For a Young tableau $t$, let $R_t$ permute the entries within its rows and $C_t$ within its columns. We use the Young symmetrizer conventionReversing the order gives another usual realization of the same irreducible polynomial module. On the tensor power $T=V^{\otimes n}$, the actions areThey commute because applying $g$ to every factor commutes with permuting the factors.

We state the permitted combinatorial input explicitly. A Young symmetrizer satisfies $h_t^2=H_{\lambda }{h}_t$, where $H_{\lambda }$ is the nonzero hook product of a partition. Thus $e_t=H_{\lambda }^{-1}{h}_t$ is a primitive idempotent. The standard-tableau decomposition of the right regular module is $\mathbb{C}{S}_n={⨁}_{\lambda ,t\mathrm{standard}}{e}_t\mathbb{C}{S}_n$. Tensor this right-module direct sum with the left module $T$. The map $e_t\mathbb{C}{S}_n{\otimes }_{\mathbb{C}{S}_n}T\to e_{t}T$ sending $e_{t}a\otimes v$ to $e_{t}av$ is an isomorphism, with inverse $w\mapsto e_t\otimes w$. Therefore the Young-symmetrizer tensor decomposition isThis is a direct sum of $GL(V)$-modules; individual summands need not be $S_n$-invariant.

We also state the allowed Schur algebra result, namely Schur–Weyl duality: the two actions are mutual commutants, andwhere the $D_{\lambda }(V)$ are pairwise nonisomorphic irreducible homogeneous polynomial representations of degree $n$. We also use the standard Schur algebra equivalence between its modules and homogeneous degree-$n$ polynomial representations, so these exhaust the irreducibles in that category. A primitive idempotent $e_t$ has one-dimensional image on $S^{\lambda }$ and zero image on the other simple factors, so $h_{t}T\cong D_{\lambda }(V)$. This identifies the requested irreducibles. They classify the polynomial degree-$n$ representations in this tensor power, not all rational representations of every degree.

The length bound for a Schur module is $D_{\lambda }(V)\ne 0$ if and only if $\ell (\lambda )\le m$. A column of length greater than $m$ antisymmetrizes more than $m$ vectors and gives zero. Conversely, for $\ell (\lambda )\le m$, fill every tensor position in row $i$ with the $i$th basis vector of $V$. Row symmetrization multiplies it by ${\prod }_i{\lambda }_i!$, and column antisymmetrization is nonzero because the vectors within each column are distinct basis vectors.

A rational representation of an algebraic group is a regular morphism into the general linear group of its representation space. For $G{L}_m$, its matrix entries belong to $\mathbb{C}[g_{ij},\det (g{)}^{-1}]$; rational here permits determinant denominators but not arbitrary poles on $G{L}_m$. A one-dimensional rational character of ${\mathbb{C}}^{\times }$ is a Laurent polynomial $p(z)$ with $p(zw)=p(z)p(w)$ and $p(1)=1$. Comparing Laurent coefficients shows that just one monomial occurs and its coefficient is $1$, hence $p(z)=z^r$ for $r\in \mathbb{Z}$.

Restrict a one-dimensional rational character of $G{L}_m$ to its diagonal torus. The same argument in several variables gives ${\prod }_i{x}_i^{r_i}$. Conjugation by permutation matrices makes all $r_i$ equal. On every diagonalizable invertible matrix it consequently agrees with $(\det g{)}^r$. The allowed Zariski-density statement, and equality of regular functions on a dense subset, giveThese are the one-dimensional rational characters of the general linear group.

For complete reducibility of rational GL and SL representations, use the compact-group averaging argument. Average any positive definite Hermitian inner product over $SU(m)$ using normalized Haar measure. The orthogonal complement of an invariant subspace is then $SU(m)$-invariant. Differentiating makes it invariant under $\mathfrak{su}(m)$ and therefore under its complex span ${\mathfrak{sl}}_m(\mathbb{C})$. The elementary unipotent matrices $\exp (t{E}_{ij})$ generate $S{L}_m(\mathbb{C})$, so the complement is $S{L}_m$-invariant. This proves complete reducibility. Averaging over $U(m)$ similarly gives complete reducibility for rational $G{L}_m$ representations.

Here is an explicit rational extension from SL to GL. Decompose the representation space by the finite scalar center of $S{L}_m$ into subspaces $W_k$ on which $\zeta I$ acts as ${\zeta }^k$, with $0\le k<m$ and ${\zeta }^m=1$. These subspaces are invariant. For $g\in G{L}_m$, choose $a$ with $a^m=\det g$, set $s=a^{-1}g\in S{L}_m$, and defineChanging $a$ to $\zeta a$ changes $s$ to ${\zeta }^{-1}s$, so the two factors cancel. It is a homomorphism, since scalar roots multiply up to the same harmless factor, and it restricts to $\rho$ on $S{L}_m$.

It is rational as well. Every matrix coefficient of $\rho {|}_{W_k}$ has a polynomial representative on $S{L}_m$. Averaging that representative over the finite scalar center selects its homogeneous parts $P_d$ with $d\equiv k(\mathrm{mod}m)$. Substitution in the extension gives ${\sum }_d(\det g{)}^{(k-d)/m}{P}_d(g)$, a regular function on $G{L}_m$. Each $S{L}_m$-invariant subspace decomposes into its $W_k$ parts, and the extension acts on each part by a scalar times an $S{L}_m$ action. Thus these subspaces are also invariant under the chosen extension. Consequently $\rho$ is irreducible if and only if this ${\rho }^{\prime }$ is irreducible. For $m=1$, $S{L}_1$ is trivial and the trivial extension supplies the same conclusion.

For a decreasing integer tuple $\lambda =({\lambda }_1,\dots ,{\lambda }_m)$, let ${\mu }_i={\lambda }_i-{\lambda }_m$. Then $\mu$ is a partition and the highest-weight classification of rational GL representations definesThis is a determinant twist of a Schur module. Every irreducible rational $G{L}_m$ representation becomes polynomial after multiplication by a sufficiently large positive determinant power, which clears all matrix-entry denominators. The polynomial degree decomposition and Schur–Weyl duality then identify it with a Schur module. Undoing the twist gives exactly one decreasing integer tuple $\lambda$. Distinct tuples have distinct highest torus weights, so these are the complete pairwise nonisomorphic irreducible rational representations.

The Weyl character formula specializes toIt is a symmetric Laurent polynomial in the eigenvalues. Equality extends from the dense set of diagonalizable matrices to all invertible matrices: both the character and the expression in the characteristic-polynomial coefficients are regular functions on $G{L}_m$. For a polynomial representation this also extends to every endomorphism of $V$. For a general rational representation, the printed claim at singular endomorphisms needs this qualification: for example ${\det }^{-1}$ is undefined at a singular matrix. The displayed formula is valid on $G{L}_m$, and on all of $\mathrm{End}(V)$ when ${\lambda }_m\ge 0$.

To compute the degree, set $x_j=e^{t{c}_j}$ with distinct $c_j$ and take $t\to 0$. For ${\ell }_i={\lambda }_i+m-i$ and $d=\left(\genfrac{}{}{0ex}{}{m}{2}\right)$, the leading coefficient of an exponential alternant isIndeed, expand every exponential in powers of $t$. The first nonzero determinant uses the distinct powers $0,1,\dots ,m-1$; its coefficient is the product of the two Vandermonde determinants divided by $\prod k!$. Taking the same expansion in the denominator cancels the powers and the $c$ factors, giving the Weyl dimension formulaThis proof works for negative ${\lambda }_m$ as well, since determinant twists have dimension one.

Every finite-dimensional rational module is completely reducible. The characters of its irreducible constituents are linearly independent: each Schur Laurent character has its highest dominant monomial $x^{\lambda }$ with coefficient $1$, and only lower weights besides it. In a finite relation, choose a lexicographically highest remaining weight; its coefficient must vanish, and iterate. Therefore equal characters give equal multiplicities of every irreducible constituent, proving rational modules with the same character are isomorphic.

Finally the symmetric algebra of $V\oplus {\Lambda }^{2}V$ has the formal torus characterEach factor sums the symmetric powers of a one-dimensional weight space; the exterior square has weights $x_i{x}_j$ for $i<j$. The permitted Schur identity makes this ${\sum }_{\lambda }{s}_{\lambda }(x)$. In each fixed scalar degree there are only finitely many terms, so complete reducibility and character independence apply degree by degree without a convergence assumption. Thus the multiplicity-free symmetric-algebra model for polynomial GL representations contains each irreducible polynomial representation exactly once. The word irreducible is necessary: arbitrary reducible polynomial modules, such as two copies of the trivial module, do not each occur once in a multiplicity-free sum.

For an affine variety $V\subseteq {\mathbb{A}}_k^N$, one construction defines ${\mathcal{O}}_V(U)$ as the functions $U\to k$ which locally have the form $p/q$, where $p,q$ are elements of the coordinate ring $k[V]$ and $q$ does not vanish on the neighbourhood in question. Restrictions are restrictions of functions, and locality gives the sheaf axioms. In particularThe second formula identifies its local rings.

In the classical convention, an algebraic variety over $k$ is an irreducible ringed space with a sheaf of $k$-valued functions, admitting a finite open cover by spaces isomorphic to affine varieties, and satisfying the separated variety condition that its diagonal is closed in $X\times X$. Its topology is Noetherian. One can instead allow reducible reduced varieties; the arguments below still work with the denominator-clearing version of the localization argument. A morphism of varieties is a continuous map whose pullback takes local regular functions to regular functions, equivalently a morphism of locally ringed $k$-spaces. The maps on local rings are local because a function nonzero at the image point stays nonzero at the source point.

An affine variety supplies its own finite affine cover, is Noetherian, and is separated: its diagonal in $V\times V$ is cut out by the differences of corresponding coordinates. A regular map between affine varieties pulls coordinate functions back to regular functions. It is continuous because the inverse image of any polynomial zero set is the zero set of the pulled-back regular functions; on affine charts such zero sets are closed. Pulling back a locally represented fraction $p/q$ gives a regular fraction wherever its denominator is nonzero. Hence a regular map is a morphism of varieties in this definition.

For a morphism $\varphi :X\to Y$ which is an isomorphism over each member $U_i$ of an open cover of $Y$, every fibre contains exactly one point. Thus $\varphi$ is bijective. Its inverse is continuous and regular on every $U_i$, since there it is the given inverse of the local isomorphism. These inverses agree on overlaps, being inverses of the same map. They glue to a global inverse morphism of varieties. Therefore being an isomorphism is local on the target.

Let $A=\Gamma (X,{\mathcal{O}}_X)$, and let $Y$ be affine with coordinate ring $B$. Given a $k$-algebra homomorphism $\alpha :B\to A$, choose a presentation $B=k[y_1,\dots ,y_N]/I$. The maplands in $Y$ because all the equations in $I$ become zero functions. Its coordinates are global regular functions, so it is a morphism on every affine chart of $X$, and therefore globally. On a target neighbourhood where a fraction $b/c$ is defined its pullback is $\alpha (b)/\alpha (c)$, proving that the induced map on global sections is exactly $\alpha$. This also proves independence of the chosen generators and uniqueness. This is the affine-target adjunction for varieties.

For $f\in A$, the natural map $A_f\to \Gamma (X_f,{\mathcal{O}}_X)$ sends $a/f^N$ to that regular function. In the irreducible convention, if $f\ne 0$, $X_f$ is dense, so this map is injective: a global regular function vanishing there vanishes everywhere. Cover $X$ by finitely many affine charts $U_j$. A section $s$ on $X_f$ restricts on $U_j\cap X_f=D_{U_j}(f{|}_{U_j})$ to an element of $\Gamma (U_j,{\mathcal{O}}_X{)}_{f{|}_{U_j}}$. A common power $f^N$ clears all these finitely many denominators. The resulting regular sections on $U_j$ agree on the dense principal open in each overlap, hence agree there altogether and glue to a global section $a$. Thus $s=a/f^N$. If $f=0$, both sides are the zero ring of sections on the empty open. ConsequentlyThis is localization of global sections on a principal open. For reduced reducible varieties, equality on a principal open instead means that a sufficiently large power of $f$ annihilates the difference; finitely many charts and overlap refinements allow one common extra power. The same denominator-clearing proof then gives the stated localization isomorphism without a density assumption.

Now suppose ${\sum }_i{g}_i{f}_i=1$ in $A$ and every $X_{f_i}$ is affine. Each $A_{f_i}=\Gamma (X_{f_i},{\mathcal{O}}_X)$ is a finitely generated $k$-algebra. Select finitely many generators, writing them as $a_{ij}/f_i^{N_{ij}}$. Let $B\subseteq A$ be the $k$-subalgebra generated by all $f_i$, all $g_i$, and all $a_{ij}$. It is finitely generated, andThe inclusion from left to right is immediate, while the chosen generators give the reverse inclusion. This construction does not assume that $A$ was finitely generated in advance.

The reduced algebra $B$ is the coordinate ring of an affine variety $Y$ (irreducible when $X$ is). The map $B\hookrightarrow A$ produces $X\to Y$ by the preceding construction. The sets $D_Y(f_i)$ cover $Y$, since their functions generate the unit ideal already in $B$. Their inverse images are $X_{f_i}$, and on them the map is the isomorphism corresponding to $B_{f_i}=A_{f_i}$. The target-local argument above now provesThis is affineness from a unit-ideal principal affine cover. The printed sets are $X_{f_i}$; the missing index in the TeX aid is not a different hypothesis.

The localization of global sections on a principal open gives $A_{f_i}=\Gamma (X_{f_i},{\mathcal{O}}_X)$. Each of these rings is a finitely generated algebra because $X_{f_i}$ is an affine variety. Choose finite generators of every $A_{f_i}$ and write them as $a_{ij}/f_i^{e_{ij}}$ with $a_{ij}\in A$. Let $B$ be the $k$-subalgebra of $A$ generated by all $f_i,g_i,a_{ij}$. It is a finitely generated algebra, and $B_{f_i}=A_{f_i}$ since these localizations contain the chosen generators and $f_i^{-1}$.

For any integer $N\ge 1$, raise ${\sum }_i{g}_i{f}_i=1$ to the power $r(N-1)+1$. Every resulting monomial contains some $f_i^N$, so it gives a unit-ideal identity for powersFor an arbitrary $a\in A$, equality $A_{f_i}=B_{f_i}$ implies $f_i^{N_i}a\in B$ for some $N_i$: multiply by an additional power if equality of localized fractions requires it. Choose a common $N$ and the identity above. Then $a={\sum }_i{c}_i(f_i^{N}a)\in B$. Consequently $A=B$, so $A$ is a finitely generated algebra. It is a reduced ring, since a nilpotent element global regular function has zero germ everywhere on the reduced variety $X$. This is finite generation from finitely many principal localizations.