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A classical variety is affine if finitely many global regular functions generate the unit ideal and their nonvanishing opens are affine. Localization of global sections on a principal open identifies their coordinate rings with $A_{f_i}$, where $A=\Gamma (X,{\mathcal{O}}_X)$. Choose a finite-type subalgebra of $A$ containing the functions, their unit-ideal coefficients and numerators of generators of every $A_{f_i}$. Its affine variety has the same principal affine charts, whose isomorphisms glue globally. This avoids assuming finite generation of $A$ before affineness is known.

For an integral normal variety that is Noetherian, removing a closed subset of codimension at least two does not create new regular functions. Affine-locally, a normal Noetherian domain is the intersection of its localizations at height-one primes inside its fraction field. The complement still contains every height-one point, so a regular function on it lies in each such localization and therefore in the original ring. On projective space this also follows directly by excluding irreducible polynomial factor of the denominators on each polynomial affine chart.

Two nonzero elements with no common nonunit irreducible factor. In a unique factorization domain this is equivalent to their greatest common divisor being a unit. Coprimality of numerator and denominator makes a fraction reduced, so simultaneous restrictions on its denominator can force it to be a regular function.

On an integral scheme $X$, every nonempty open $U$ contains its generic point $\eta$. Taking a germ gives an injective ring homomorphism from its regular functions to the function field ${\mathcal{O}}_{X,\eta }$. Injectivity follows on each nonempty affine open subscheme from the injection of an integral domain into its field of fractions, and then from the sheaf gluing axiom. This lets regular functions on different open subsets be compared inside one field.

Over an algebraically closed field, every regular function on a nonempty irreducible projective variety is constant. Regard the function as a morphism to ${\mathbb{P}}^1$ avoiding infinity. Its graph is closed; the closedness of projection from projective space makes its image closed in ${\mathbb{P}}^1$. A proper closed subset there is finite, and an irreducible finite image is a singleton. Irreducibility is the usual variety convention; on a disconnected projective algebraic set different components may have different constants.

The only regular functions defined on all of a nonempty projective space over a field $k$ are the constants:For ${\mathbb{P}}^1$, this also follows by gluing $k[t]$ and $k[t^{-1}]$ inside $k[t,t^{-1}]$: their intersection is $k$.

For distinct closed points over an algebraically closed field, evaluation gives the short exact sequence $0\to {\mathcal{I}}_{\{P,Q\}}\to {\mathcal{O}}_X\to k_P\oplus k_Q\to 0$. If all global regular functions are constant, the long exact sequence in sheaf cohomology embeds $k^2/k(1,1)$ into $H^1(X,{\mathcal{I}}_{\{P,Q\}})$. Thus this ideal detects a concrete obstruction to affineness.

For a quasi-compact separated scheme and a global regular function $f$, global sections on $X_f$ are the localization of global sections by $f$. Choose a finite affine open cover. Its intersections are affine, so the sheaf gluing axiom describes global sections as the kernel of the difference map between two finite products of affine coordinate rings. Exactness of localization preserves this kernel, and localization commutes with those finite products. Localizing the same cover gives the claimed isomorphism. This applies before knowing whether $X$ itself is affine.

A matrix minor is the determinant of a square submatrix obtained by selecting rows and columns of a matrix. Over a field, the matrix rank is less than $r$ exactly when every $r$-by-$r$ matrix minor vanishes. In a matrix of regular functions, these equations define a closed rank-defect locus, explaining the determinantal variety construction.

Put $\mathcal{D}={\mathcal{K}}_X^{\ast }/{\mathcal{O}}_X^{\ast }$, the quotient sheaf of abelian groups. At a closed point $P$, the discrete valuation gives an isomorphismIndeed, a local parameter on a smooth algebraic curve has valuation one, and the valuation kernel is precisely the unit group of the discrete valuation ring. Thus a local section represented by $f$ determines its local principal divisor, and changing $f$ by a regular function which is a unit does not change those coefficients.

These local valuations identify $\Gamma (X,\mathcal{D})$ with $\mathrm{Div}(X)$. To see that they give finite support, represent a section locally by finitely many rational functions on an algebraic variety, using Noetherian Zariski topology to take a finite cover. Each representative has only finitely many zeros and poles, so their combined support is finite. Conversely, for $D=\sum n_{P}P$, choose at each support point a local parameter $t_P$ and shrink its neighbourhood to exclude every other zero or pole of $t_P$ and every other support point. The local rational equation $t_P^{n_P}$ represents the desired valuations there; on the complement of the support use $1$. On overlaps these equations differ by units, so their classes glue in the quotient sheaf. Finally, a section with every valuation zero has zero germ everywhere and is the identity section. Hence the identification is both surjective and injective.

The map on global sections sends $f\in k(X{)}^{\ast }$ to $\mathrm{div}(f)$. Therefore its cokernel is

For a sheaf of abelian groups $\mathcal{F}$ on a topological space $X$, putThese arbitrary families of germs form a sheaf, and restrictions are projections. Extending a family by zero proves that $C(\mathcal{F})$ is flasque. Sending a section to all its germs gives an injective sheaf map $\mathcal{F}\to C(\mathcal{F})$, since a section with every germ zero vanishes locally and hence globally.

Set $Q^{-1}=\mathcal{F}$, $I^j=C(Q^{j-1})$ and $Q^j=\mathrm{coker}(Q^{j-1}\to I^j)$, using the sheaf cokernel. Compose $I^j\to Q^j\to I^{j+1}$ to obtain the Godement resolutionIt is exact by construction, with flasque terms. Define sheaf cohomology by the cohomology of its global-section complex:The resolution principle for sheaf cohomology identifies this with the computation using any flasque resolution.

The elementary input for acyclicity is the flasque-kernel section-lifting lemma: in a short exact sequence with flasque kernel, sections of the quotient lift over every open subset. Consequently a quotient of two flasque sheaves is flasque, because a quotient section lifts and that lift extends. If $\mathcal{F}$ itself is flasque, induction through $0\to Q^{j-1}\to I^j\to Q^j\to 0$ makes every $Q^j$ flasque. The same lifting lemma makes these sequences exact on global sections. Hence the global-section complex has no positive-degree cohomology, provingThis deduction uses the lifting lemma, not the desired cohomological vanishing as an assumption.

For the rational-section construction, add two representative sections after restricting to the intersection of their dense domains. Multiply by a representative rational function in the same way. Finite intersections of dense open sets are dense, and further restricting both representatives does not change their resulting class. The module axioms hold on their common domain. Thus rational sections form a module over $\mathrm{Rat}(X)$; irreducibility is not needed for this module statement. When $X$ is irreducible, $\mathrm{Rat}(X)$ is the function field $K=k(X)$.

Suppose now that $\mathcal{F}$ is locally free and $X$ irreducible. A germ at $P$ is represented on a nonempty open neighbourhood, which is dense, so it defines a rational section. Two representatives of the germ agree on a neighbourhood of $P$ and therefore give the same class. If the class is zero, trivialize $\mathcal{F}$ on a neighbourhood of $P$. Its coefficient functions vanish on a dense open subset; regular functions on a reduced irreducible variety with this property vanish identically. The germ is therefore zero. This proves the stalk inclusion into rational sections of a locally free sheaf:Local freeness matters: a nonzero torsion germ can vanish on a dense open set.

On an irreducible algebraic curve, every proper closed subset is finite, and every open subset is quasi-compact because the Zariski space is Noetherian. Let $R=\mathrm{Rat}(\mathcal{F})$. The constant sheaf $\mathcal{R}(\mathcal{F})$ has sections $R$ on every nonempty open set: such opens are irreducible and hence connected. DefineRestriction drops the components outside the smaller open set. For a compatible family over an open cover of $U$, take a finite subcover by quasi-compactness. The union of the finite supports of these finitely many sections is finite. Their agreeing components therefore glue to a unique finite-support tuple on $U$. This verifies the sheaf gluing axiom, including arbitrary covers, rather than merely asserting that a presheaf direct sum is always a sheaf.

A rational section is regular on some dense open set, whose complement on a curve is finite. Its classes $[s{]}_P\in R/{\mathcal{F}}_P$ therefore have finite support, defining a sheaf map $\mathcal{R}(\mathcal{F})\to \mathcal{P}(\mathcal{F})$. At $P$, the first sheaf has stalk $R$. The second has stalk $R/{\mathcal{F}}_P$: any finite tuple can be restricted to a neighbourhood omitting all its other support points, and its $P$-component is unaffected by such restriction. The resulting sequence on stalks isExactness of sheaves can be tested on stalks, so this proves the rational principal-parts resolution on an algebraic curveBoth right-hand sheaves are flasque: restrictions for the constant rational-section sheaf are identities between nonempty opens, and restrictions for the rational principal-parts sheaf are projections, with extension by zero. ConsequentlySurjectivity on stalks has not been confused with surjectivity on global sections.

For an explicit failure of global surjectivity take $X={\mathbb{P}}^1$ and $\mathcal{F}=\mathcal{O}(-2)$. Use coordinate $z=X_1/X_0$ and frame $e_0=X_0^{-2}$ on the finite chart. At infinity use $w=1/z$ and $e_{\infty }=X_1^{-2}=z^{-2}{e}_0$. Prescribe the rational principal part of $z^{-1}{e}_0$ at zero and zero at every other point. If a rational section $r(z)e_0$ realized this tuple, then $r-z^{-1}$ would be regular at zero and $r$ would be regular at every other finite point. Thus $r=z^{-1}+p(z)$ with $p\in k[z]$. Regularity at infinity would requireto be regular at $w=0$. Its $w^{-1}$ term cannot cancel with any term of the polynomial contribution, whose powers are at most $-2$. This is impossible. Hence the global rational-principal-part map need not be surjective.

For an affine curve it is always surjective, and here is a direct construction. Write $X=\mathrm{Spec}A$, $K=\mathrm{Frac}A$ and $\mathcal{F}=\tilde{M}$. By the affine module sheaf construction, $R=M{\otimes }_{A}K=:M_K$ and ${\mathcal{F}}_P=M_{{\mathfrak{m}}_P}$. Choose representatives $v_P\in M_K$ for a prescribed finite tuple, and choose one nonzero $a\in A$ with $a{v}_P\in M$ for all of them. The zero set $T=V(a)$ is finite. The ring $A/(a)$ is Artinian, since it is Noetherian of dimension zero, and its Artinian decomposition into local factors givesPrescribe the class of $a{v}_P$ in the indicated component for each requested point in $T$, and zero in the other components. A requested point outside $T$ already has zero principal part because $a$ is a unit there. Lift the tuple to $w\in M$. Then $s=w/a\in M_K$ satisfies $s-v_P\in M_{{\mathfrak{m}}_P}$ at every requested point in $T$, and is regular at every other point of $T$. Outside $T$ it is regular because $a$ is invertible. It therefore realizes exactly the prescribed tuple. This proves affine principal-parts interpolation for a locally free sheaf and givesNo nonsingularity assumption was used; the construction applies to singular irreducible affine curves as well.

On the standard affine charts $U_i=\{X_i\ne 0\}$, use the formal frame $e_i=X_i^m$ and glue rank-one free sheaves byThe transition functions are regular units on overlaps and satisfy the cocycle identity. This constructs the twisting sheaf on projective space $\mathcal{O}(m)$ as an invertible sheaf for every integer $m$, including negative $m$.

On a nonempty $U$, a section has local expressions $h_i{e}_i$, where $h_i\in \mathcal{O}(U\cap U_i)$. Pulling to the punctured affine cone gives $h_i(X/X_i)X_i^m$. Their agreement on overlaps makes them one homogeneous rational function $r$ of degree $m$, regular on ${\pi }^{-1}U$. To make its polynomial representation explicit, write one nonzero local coefficient as $p(y)/q(y)$ in the affine coordinates. Homogenize $p,q$ to degrees $a,b$; the resulting rational expression isIf the exponent of $X_i$ is negative, move its power to the denominator. Cancel common factors; the greatest common divisor of homogeneous polynomials can be chosen homogeneous. Thus $r=F/G$ with $F,G$ coprime homogeneous polynomials, $G\ne 0$, and $\mathrm{deg}F-\mathrm{deg}G=m$.

Conversely, for such a homogeneous rational function regular on ${\pi }^{-1}U$, its coefficient $r/X_i^m$ is regular on $U\cap U_i$: restrict to the slice $X_i=1$. These coefficients obey the displayed transition rule and hence define a section. This proves the homogeneous rational sections of a twisting sheaf description in both directions, for arbitrary open $U$ rather than only a standard chart.

Now lift a regular degree-zero rational function $f=P/Q$ to the punctured cone. The quotient rule givesThis is a rational function homogeneous of degree $-1$. It is independent of the chosen representation because differentiation is a derivation of the rational function field. It is regular wherever $f$ is regular: a derivation of a polynomial ring extends to each localization by the quotient rule, and regular functions are locally such fractions. The preceding homogeneous-section description therefore provesZeros of a particular displayed denominator do not invalidate this argument; regularity is a property of the rational function, which can have another local representation.

The map $f\mapsto ({\partial }_{0}f,\dots ,{\partial }_{n}f)$ is a $k$-linear sheaf derivation with values in $\mathcal{O}(-1{)}^{\oplus (n+1)}$. The universal property of Kähler differentials consequently definesMultiplication by the coordinate section $X_i\in \Gamma ({\mathbb{P}}^n,\mathcal{O}(1))$ defines the other mapFor a degree-$d$ homogeneous polynomial, differentiating each monomial gives ${\sum }_i{X}_i{\partial }_{i}P=dP$. Applying this to the equal-degree numerator and denominator gives ${\sum }_i{X}_i{\partial }_{i}f=0$ for degree-zero $f$, so $\sigma D=0$. This identity is valid in every characteristic.

To prove all the exactness assertions, fix $U_i$ and put $y_j=X_j/X_i$ for $j\ne i$. The Kähler differential sheaf is free there on $d{y}_j$, because a derivation of the polynomial ring is determined freely by its values on the coordinates. Trivialize each $\mathcal{O}(-1)$ with frame $X_i^{-1}$. In this frame,where $e_j$ now denotes the $j$th coordinate vector of the direct sum. The map $\sigma$ is surjective, since its $i$th coefficient is one. Its kernel consists exactly of tuples with $b_i=-{\sum }_{j\ne i}{y}_j{b}_j$, so the $n$ displayed vectors $e_j-y_j{e}_i$ form a free basis of that kernel. The map $D$ sends the differential basis bijectively to this kernel basis and is therefore injective. Exactness on these charts proves the cotangent Euler sequence in homogeneous coordinatesThe proof uses no division by an integer and hence no characteristic-zero assumption.

For $n\ge 1$, the cohomology of twisting sheaves on projective space givesIn particular, $H^0(\mathcal{O})=k$, $H^{q>0}(\mathcal{O})=0$, and $H^q(\mathcal{O}(-1))=0$ for every $q$. The long exact sequence in sheaf cohomology of the displayed Euler sequence beginsand gives zero in the remaining degrees. ThusThe nonzero group is generated by the connecting image of the section $1$ of $\mathcal{O}$. If $n=0$ is admitted, ${\mathbb{P}}^0$ is a point and its cotangent sheaf is zero, so the sole requested group $H^0$ is zero.

Let $u,v,w$ be the consecutive primitive toric ray vectors around the compact invariant curve $F=F_v$. Smoothness gives oriented determinants $\det (u,v)=\det (v,w)=1$, hence $u+w=bv$ for an integer $b$. Choose an algebraic torus character $m$ with $⟨m,u⟩=0$ and $⟨m,v⟩=1$. The principal divisor on a toric variety has coefficient one at $F$, coefficient $b$ at $F_w$, and zero at $F_u$. The other invariant curves do not meet $F$, while each adjacent one meets it transversely once. Intersecting this principal divisor with the proper curve $F$ gives $0=F^2+b$. This proves the self-intersection formula for a toric surface divisor even when the ambient surface is not complete.

If $F^2=-1$, then $b=1$ and $v=u+w$. The vectors $u,w$ form a lattice basis since $\det (u,w)=1$, and their toric cone is the union of the two toric cones adjacent to $v$. Deleting $v$ is consequently a smooth toric fan coarsening. Its inverse is precisely the toric blowup at a torus-fixed point, obtained by inserting the sum of the two primitive basis vectors. Conversely, that insertion gives $u+w=v$, and hence algebraic self-intersection $-1$. Thus the toric contraction of a minus-one curve satisfiesFor the cyclic quotient description, first take a strongly convex toric cone of dimension two. Write its primitive boundary vectors as $a,b$ and put $N_0=\mathbb{Z}a+\mathbb{Z}b$, $r=[N:N_0]$. Since $a$ is primitive, extend it to a basis of $N$ and write $b=ua+rc$ with $\gcd (u,r)=1$. Thus $N/N_0$ is cyclic of order $r$. In the lattice $N_0$, the toric cone is smooth and its toric variety is ${\mathbb{A}}^2$. Passing to the larger lattice $N$ takes the finite quotient whose invariant monomials are precisely those in the smaller dual lattice $M=N^{\ast }\subset N_0^{\ast }$. Relative to the two boundary coordinates, $c=(b-ua)/r$, so a generator acts with weights $(-u,1)$ modulo $r$. Replacing that generator by a suitable power makes the first weight one and the second some $q$ with $0<q<r$ and $\gcd (q,r)=1$. Therefore, for the singular case,The cyclic quotient surface singularity can equivalently be described by the first-quadrant toric cone inIndeed its coordinate ring is $k[x^i{y}^j:i,j\ge 0,i+qj\equiv 0(\mathrm{mod}r)]=k[x,y{]}^{{\mu }_r}$. Characteristic zero ensures the stated ordinary finite-group quotient interpretation.

The restriction to singular toric cones of dimension two is necessary for $r>1$. A smooth toric cone has $r=1$ and gives ${\mathbb{A}}^2$ with the trivial quotient. A toric cone of dimension one gives ${\mathbb{A}}^1\times k^{\ast }$, and the zero toric cone gives $(k^{\ast }{)}^2$. These latter surfaces cannot be a quotient of the displayed kind, since they have nonconstant invertible regular functions, whereas $k[x,y{]}^{{\mu }_r}$ has only constant units. Thus the blanket affine-surface wording requires these separate cases.

For the singular case, compute the negative continued fraction by the ceiling form of the Euclidean algorithm. Start with $a_0=r$, $a_1=q$, and, while $a_i>0$, setThen $0\le a_{i+1}<a_i$, so the process terminates at $a_{s+1}=0$. The gcd is unchanged at each step, hence $a_s=1$. Since $a_{i-1}>a_i>0$, every $b_i\ge 2$. Rearranging the recurrence givesNow define lattice vectorsTo verify this Hirzebruch–Jung resolution algorithm, write $v_i=(p_i,a_i)/r$ with $p_0=0$, $p_1=1$ and $p_{i+1}=b_i{p}_i-p_{i-1}$. Induction gives $a_i\equiv q{p}_i(\mathrm{mod}r)$, so $v_i\in L$. It also gives $p_i{a}_{i+1}-a_i{p}_{i+1}=-r$. Thus every adjacent pair has absolute determinant $1/r$, the covolume of $L$, and is a basis of $L$. The $p_i$ increase strictly, while the $a_i$ decrease to zero, so these toric rays occur in order inside the first quadrant. At termination, $a_s=1$ and the determinant identity gives $p_{s+1}=r$, hence $v_{s+1}=(1,0)$. All boundary and inserted vectors are primitive because they occur in lattice bases.

Subdivide the quadrant by $v_1,\dots ,v_s$. Every resulting toric cone is smooth by the smoothness criterion for a toric variety, and the fan subdivision gives a proper birational toric morphism, hence a resolution. Each interior toric ray has a complete one-dimensional star, so its exceptional curve of a surface resolution is ${\mathbb{P}}^1$. The recurrence yieldsThis proves minimality: no exceptional curve of a surface resolution can be contracted to a smooth point by the Castelnuovo contraction criterion. There is also a direct toric fan check. For fixed $i$, the determinants of $v_i$ with $v_{i+h}$, measured in units of the lattice covolume, satisfy the same recurrence with coefficients at least two and initial values zero and one. They are at least $h$. Omitting any intervening toric ray therefore leaves a toric cone of determinant at least two, so no nontrivial coarsening is smooth. Thus this is the minimal resolution of a cyclic quotient surface singularity.

If all exceptional algebraic self-intersections are $-2$, all $b_i=2$ and induction gives $[2,\dots ,2{]}^{-}=(s+1)/s$. Since $r,q$ are coprime, this forces $r=s+1$, $q=s$. Conversely $r/(r-1)$ has exactly $r-1$ coefficients equal to two. ThereforeFor the final toric fan, the minimal smooth fan subdivision of the toric cone from $(0,1)$ to $(r,1)$ inserts the toric rays $(j,1)$ for $1\le j<r$. Adjacent determinants have absolute value one, and $(j-1,1)+(j+1,1)=2(j,1)$, so its inserted curves have algebraic self-intersection $-2$ and it is the minimal fan subdivision just described. The second original toric cone, from $(r,1)$ to $(1,0)$, is already smooth.

Write $v_j=(j,1)$ and $e=(1,0)$. In the resolved toric fan, the rightmost interior toric ray obeys $v_{r-1}+e=v_r$, so its curve has algebraic self-intersection $-1$ and can be blown down to a smooth point. After deleting it, the same relation $v_{r-2}+e=v_{r-1}$ applies to the new rightmost interior toric ray. Continue deleting $v_r,v_{r-1},\dots ,v_1$ in that order. Each deletion is the inverse of a toric blowup at a torus-fixed point. The last toric fan has only the quadrant toric cone and defines ${\mathbb{A}}^2$. ConsequentlyAll maps induce the identity on the dense algebraic torus, so this composite is the same morphism as $X_{{\Sigma }^{\prime }}\to X_{\Sigma }\to {\mathbb{A}}^2$. The intermediate smooth toric fans need not include the singular toric fan $\Sigma$.

Use the usual irreducible algebraic variety convention here, and write $N=\dim X$, $r=\dim Y$. Fix a closed point $y$ and an affine neighborhood $Y_0$ of it. Noether normalization gives a finite surjective map $\pi :Y_0\to {\mathbb{A}}^r$. Put $a=\pi (y)$. Its fibre is a finite set $\{y_1,\dots ,y_s\}$, one of whose members is $y$, because a finite morphism has finite fibres over closed points.

Choose an affine open $U\subseteq {\varphi }^{-1}(Y_0)$ meeting $X_y$. Since $X$ is irreducible, $U$ still has dimension $N$. In $U$, the fibre of $\pi \varphi$ over $a$ is cut out by the $r$ regular functions ${\pi }_j\varphi -a_j$. Apply the supplied principal hypersurface dimension lemma successively on each irreducible component: if a function vanishes identically the dimension does not drop, if its zero set is empty the component disappears, and otherwise each new component loses exactly one dimension. Hence every surviving component after the $r$ cuts has dimension at least $N-r$.

But that composite fibre is the disjoint union of $U\cap {\varphi }^{-1}(y_i)$. Each member is closed, and since there are finitely many members it is also open in the composite fibre. Thus $U\cap X_y$ is a nonempty union of its irreducible components, givingThe same argument around a point on any fibre component gives the componentwise bound. This is the closed-fibre lower bound by finite normalization; it avoids assuming that $Y$ is smooth or that its point ideal has only $r$ generators.

Irreducibility matters if “variety” is generalized to arbitrary reduced reducible spaces. For example $X={\mathbb{A}}^2\bigsqcup {\mathbb{A}}^1\to {\mathbb{A}}^1$ can map the first component constantly to zero and the second by the identity. It is surjective, but nonzero fibres have dimension zero while $\dim X-\dim Y=1$. For reducible sources one uses the individual component dimensions instead; the preceding proof establishes the stated theorem under its standard irreducible-variety convention.

Choose affine neighborhoods $U_0=\mathrm{Spec}A$ of $x$ and $V_0=\mathrm{Spec}B$ of $y$, and orient the given local ring isomorphism asBoth algebras are finitely presented over $k$. Choose finitely many generators of $B$. Their images under $\alpha$ are fractions in $A$ whose denominators do not vanish at $x$. After inverting the product of those denominators, they define regular functions on an affine open neighborhood of $x$. The finitely many defining relations of $B$ vanish as germs at $x$; each vanishes on a further open neighborhood. Shrinking finitely many times therefore turns those fractions into an actual algebra homomorphism $B\to \mathcal{O}(U_1)$, and hence a morphism of algebraic varieties $F:U_1\to V_0$ inducing $\alpha$ on the local rings.

A local-ring isomorphism preserves the maximal ideal. Its induced residue-field map is the identity of $k$, because it is a $k$-algebra map. Evaluating every coordinate generator consequently shows $F(x)=y$. Apply the identical finite-generator and finite-relation construction to ${\alpha }^{-1}$ to obtain $G:V_1\to U_0$ with $G(y)=x$.

On $U_2=U_1\cap F^{-1}(V_1)$ the composition $GF$ is defined, and on $V_2=V_1\cap G^{-1}(U_1)$ the composition $FG$ is defined. Their maps on local rings are the identities. Thus the coordinate differences between $GF$ and the identity vanish as germs at $x$; finitely many coordinate generators allow us to shrink to $U_3\subseteq U_2$ on which $GF$ actually is the identity. Likewise shrink to $V_3\subseteq V_2$ on which $FG$ is the identity. Finally takeThese are open neighborhoods of the specified points. If $u\in U$, then $F(u)\in V_3$ and $G(F(u))=u\in U_3$, so $F(u)\in V$. Conversely, if $v\in V$, then $G(v)\in U_3$ and $F(G(v))=v\in V_3$, so $G(v)\in U$. Therefore the restrictions are mutual inverse regular maps, provingThis proves isomorphic local rings determine isomorphic neighborhoods using actual regular functions and finite presentations. The conclusion concerns neighborhood isomorphisms, rather than just isomorphisms of formal completions.

Choose $P\in X(k)$ and put $D=(g+1)P$. The Riemann-Roch theorem givesAs constants supply only a one-dimensional subspace, the Riemann-Roch space $L(D)$ contains a nonconstant rational function $f$. It has no poles except possibly at $P$. It must have a pole, since a global regular function on an irreducible projective variety is constant. Let its pole order at $P$ be $m$; then $1\le m\le g+1$.

On the open set where $f$ is regular, define $\varphi =[f:1]$. At its pole use the other affine chart, $\varphi =[1:1/f]$. The local ring of a smooth algebraic curve is a discrete valuation ring, and the positive valuation of $1/f$ at $P$ makes the latter expression regular there. The two expressions agree on their overlap, so they give a regular map $X\to {\mathbb{P}}^1$.

This map is nonconstant. It is proper because its source is projective and its target is separated, so its image is closed; a nonconstant image in the projective line must be the whole line. Each finite-value fibre consists of zeros of a nonzero rational function $f-a$, and the infinity fibre is the finite pole set. Hence every fibre is finite, giving a quasi-finite morphism. The proper quasi-finite criterion therefore makes $\varphi$ a finite morphism.

Finally its pullback of the infinity divisor is exactly the pole divisor $mP$. The degree of a morphism of curves equals the degree of this pullback, or equivalently the function-field extension degree $[k(X):k(f)]$. ThusThe argument includes inseparable maps: the degree here is the full field-extension degree, not just the separable degree. This proves the gonality bound from Riemann-Roch with an explicitly constructed finite regular map.

An affine algebraic group is a linearly reductive algebraic group if every finite-dimensional rational representation is completely reducible; equivalently the functor of invariants is exact. It is a geometrically reductive algebraic group if, whenever $0\ne v\in V^G$, there is a homogeneous invariant polynomial $F\in k[V{]}^G$ of positive degree with $F(v)\ne 0$. Linear reductivity permits degree one; geometric reductivity allows higher degree.

Over the algebraically closed field, an algebraic torus is $T\cong ({\mathbb{G}}_m{)}^r$. Its coordinate ring is the Laurent polynomial ring. A rational action writes the coaction of a vector as a finite sum $\delta (v)={\sum }_{\chi }{v}_{\chi }\otimes t^{\chi }$. The coaction identity and linear independence of Laurent monomials give $\delta (v_{\chi })=v_{\chi }\otimes t^{\chi }$ and $v={\sum }_{\chi }{v}_{\chi }$. ThusEach weight space has scalar action, so any vector-space complement within each weight space is invariant. This proves complete reducibility, and hence tori are linearly reductive in every characteristic. The argument uses characters themselves, not their Lie algebra differentials, which could coincide modulo $p$.

We now prove geometric reductivity of $G=S{L}_2(k)$ by determinant separation for positive-characteristic SL2. Fix a rational $G$-module $V$ and a nonzero fixed vector $v$. For the diagonal torus, choose a weight-zero functional $l\in V^{\ast }$ with $l(v)=1$, using the torus decomposition just proved. Define matrix coefficientsThese are regular functions on $G$, and $\Phi (hw)(g)=\Phi (w)(h^{-1}g)$ makes $\Phi$ equivariant for left translations. Moreover $\Phi (w)(gt)=\Phi (w)(g)$ because $l$ is torus-invariant. Thus $\Phi$ maps into $k[G{]}^T$ for the right torus action, and $\Phi (v)=1$.

Write a matrix's two columns as $a=(a_0,a_1)$ and $b=(b_0,b_1)$, with $\Delta =a_0{b}_1-a_1{b}_0=1$. Right torus multiplication scales these columns by $t$ and $t^{-1}$. Every right-torus-invariant regular function has a weight-zero polynomial lift, hence a sum of polynomials of bidegrees $(j,j)$ in $(a,b)$. To see that a weight-zero lift exists, take any lift and retain its zero-weight component; the ideal $(\Delta -1)$ is torus-stable and the quotient map preserves weights. Multiply each piece by ${\Delta }^{n-j}$ to put the entire sum in bidegree $(n,n)$ without changing its restriction to $G$.

Let $P_n$ be the space of degree-$n$ polynomials in one column. Restriction gives an injective equivariant map $P_n\otimes P_n\to k[G{]}^T$. Indeed if a bidegree-$(n,n)$ polynomial vanishes on determinant-one matrices, scale the first column of any invertible matrix to make its determinant one. Homogeneity then makes the polynomial vanish on every invertible matrix, a dense open subset, so it is zero. Denote the image by $R_n$. These spaces increase because multiplication by $\Delta$ realizes $R_{n-1}\subset R_n$, and their union is $k[G{]}^T$. The finite-dimensional image $\Phi (V)$ therefore lies in one $R_n$, and in any sufficiently larger filtration piece.

Choose $q=p^s$ with $q-1$ at least that bound, and set $n=q-1$. The identity $(1+z{)}^q=1+z^q$ in characteristic $p$ givesThe invariant tensortherefore defines an invertible equivariant linear map $D:P_n^{\ast }\to P_n$: its matrix is anti-diagonal with all entries nonzero. Interpret any tensor $B\in P_n\otimes P_n$ as a map $P_n^{\ast }\to P_n$ and send it to $B{D}^{-1}\in \mathrm{End}(P_n)$. Equivariance of $D$ makes this identification intertwine the diagonal tensor action with conjugation on endomorphisms. Crucially, $D$ restricts to the constant function $1$ on $G$, and is sent to the identity endomorphism.

Under the resulting equivariant linear map $V\to \mathrm{End}(P_n)$, the vector $v$ is sent to $I$. Since $\dim P_n=n+1=q$, the endomorphism determinant is a homogeneous polynomial of positive degree $q$, invariant under conjugation. Pull it back to $V$. The resulting polynomial satisfiesThis constructs the required invariant for every fixed nonzero vector and proves $S{L}_2(k)$ is geometrically reductive, including characteristic two. No averaging by a group order or division by $q$ is used; those operations would fail in the field's characteristic.