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For an affine variety with coordinate ring $A$ acted on by a finite group $G$, the affine quotient is $\mathrm{Spec}{A}^G$. Elements of $A$ satisfy monic orbit polynomials over $A^G$, so the quotient map is a finite morphism. Over an algebraically closed field, invariant functions separate finite orbits; in characteristic not dividing $|G|$, interpolate and average a function separating two orbits. The quotient fibres are consequently the orbits.

For a finite morphism between Noetherian schemes, a quasi-coherent sheaf is coherent exactly when its direct image sheaf is coherent. Over an affine target chart, let $B$ be the finite algebra over the target ring $A$, and let $M$ represent the sheaf on $\mathrm{Spec}B$. Finite generation of $M$ over $A$ implies finite generation over $B$, using the same generators; finite generation over $B$ implies finite generation over $A$ because $B$ is finite over $A$.

For an isogeny of elliptic curves $\varphi :E_1\to E_2$, write $x_2\circ \varphi =A(x_1)/B(x_1)$ with coprime polynomials $A,B$. Each $x_i:E_i\to {\mathbb{P}}^1$ is a two-to-one finite morphism, and function field degrees multiply in a tower. Consequently $2\mathrm{deg}\varphi =2\max (\mathrm{deg}A,\mathrm{deg}B)$. This computes the total degree, including an inseparable factor. The zero homomorphism is assigned degree zero separately.

A finite étale morphism is both a finite morphism and an étale morphism. Affine-locally it is represented by a finite locally free algebra with zero module of Kähler differentials. Its fibres are finite products of finite separable field extensions. For an elliptic curve with good reduction at a prime not dividing $m$, multiplication by $m$ on its smooth proper group model is finite étale of degree $m^2$: its differential is multiplication by the unit $m$. A division torsor of a section consequently gives an unramified local extension, the good-prime input to the Weak Mordell-Weil theorem.

For a finite morphism that is flat, its direct-image algebra is finite free at every local stalk. On an irreducible scheme, every point specializes from the same generic point. Localizing each stalk further at that generic point shows that all these free ranks equal the generic rank. Hence all fibre algebras have the same dimension, even without assuming finite presentation in advance.

For a finite morphism $f:X\to Y$, the pullback of an ample line bundle is ample. If $f$ is also surjective, ampleness is equivalent in both directions. These results hold for proper schemes over a field, including nonreduced schemes. Cohomological ampleness and the exact finite pushforward prove the forward direction; coherent-sheaf devissage gives finite-surjective descent. In particular, a finite map to projective space makes the pulled-back hyperplane bundle ample.

For a finite integral inclusion $A=k[t_1,\dots ,t_r]\subset B$ of domains, a nonzero $b\in B$ has a monic minimal polynomial over the fraction field with nonzero constant term $a_0\in A$. Its equation gives $a_0\in bB\cap A$. Consequently the image of $V_B(b)$ under the finite morphism to ${\mathbb{A}}^r$ is contained in the proper Zariski-closed set $V_A(a_0)$. Finite morphisms preserve the dimension of a closed subvariety and its image, so this proves dimension drops for proper closed subsets after Noether normalization. A power of $a_0$, up to sign, is the field norm of $b$.

Let $D$ be a valuation ring, $K$ its fraction field, and let $u:\mathrm{Spec}K\to X$, $v:\mathrm{Spec}D\to S$ be a compatible pair. Since $D$ is a local ring, every open subset of $\mathrm{Spec}D$ containing its unique closed point is the whole space: a principal open subset containing the maximal ideal is defined by a unit. Choose an affine open subscheme $V=\mathrm{Spec}A\subseteq S$ containing the image under $v$ of that closed point. The entire $v$ factors through $V$.

Because $f$ is a finite morphism, $f^{-1}(V)=\mathrm{Spec}B$ with $B$ a finite $A$-module. The valuative problem is the compatible ring homomorphismsEvery $b\in B$ is an integral element over $A$. Mapping its monic polynomial equation to $K$ proves that $\psi (b)$ is an integral element over $D$. Part (c) gives $\psi (b)\in D$. Thus $\psi$ factors uniquely as a ring homomorphism $B\to D\hookrightarrow K$, producing the unique extension $\mathrm{Spec}D\to X$.

For completeness, a finite morphism is a quasi-compact morphism and of finite type: over $V$ its inverse image is an affine scheme, and finite $A$-module generators also generate it as an $A$-algebra. It is also a separated morphism. Affine-locally its diagonal morphism corresponds to the surjective multiplication ring homomorphismand hence is a closed immersion. In particular it is a quasi-separated morphism. The valuative criterion for properness now applies, giving

For genuine degrees three and two, the resultant is the Sylvester matrix determinantEquivalently, if $r_1,r_2,r_3$ are the roots of $p$, counted with multiplicity, then $R(p,q)=a^2{\prod }_{i}q(r_i)$. Thus $R(p,q)\ne 0$ means that the polynomials have no common root, or equivalently are coprime over $\mathbb{C}$. One can also see this from the matrix: its singularity gives a nonzero relation $Ap+Bq=0$ with $\mathrm{deg}A<2$, $\mathrm{deg}B<3$. If $p,q$ were coprime, $p\mid B$ would force $B=0$, then $A=0$, a contradiction. Conversely, a common factor supplies such a relation after dividing $p,q$ by that factor. When a leading coefficient is zero, use the resultant for the actual polynomial degrees rather than assuming the displayed fixed-degree criterion unchanged.

An algebraic addition theorem for a meromorphic function $f$ means a nonzero polynomial $P(U,V,W)$ withidentically where the values are finite. A local relation extends meromorphically to the other regular arguments.

For a polynomial $f$ of positive degree $d$, let $z_i$ be the $d$ roots of $f(z)-U$ and $w_j$ the $d$ roots of $f(w)-V$. FormIts coefficients are separately symmetric polynomials in the two sets of roots. The Fundamental theorem of symmetric polynomials and the fixed nonzero leading coefficient of $f$ show they are polynomials in $U,V$. The expression is monic of degree $d^2$ in $W$, so it is nonzero. If $U=f(z)$, $V=f(w)$, one factor vanishes at $W=f(z+w)$. This proves the addition relation, including at exceptional multiple-root values by polynomial identity. A constant polynomial instead has the relation $W-f(0)=0$.

For the elliptic case, let $E=\mathbb{C}/\Lambda$. A nonconstant elliptic function $f$ defines a finite morphism $E\to {\mathbb{P}}^1$, so the function field of $E$ is a finite algebraic extension of $\mathbb{C}(f)$. The elliptic function-field decomposition expresses it as $\mathbb{C}(\wp ,{\wp }^{\prime })$, withThe Weierstrass addition formula makes translation algebraic:Differentiating and using the cubic differential equation also makes ${\wp }^{\prime }(z+w)$ rational in the four separate values. Therefore $f(z+w)$ lies in a finite algebraic extension of $\mathbb{C}(f(z),f(w))$. Its algebraic equation, after clearing denominators, is the required polynomial relation. This sketches why every elliptic function, not only $\wp$, has an algebraic addition theorem. Constants are already covered.

For the final assertion, consider $f(z)=e^{e^z}$, an entire function. If $c$ is a period, then $e^{e^z(e^c-1)}=1$ for every $z$. The continuous exponent must be a constant integer multiple of $2\pi i$; its derivative forces $e^c=1$. Hence its periods are exactly $2\pi i\mathbb{Z}$, making it a simply periodic function.

Suppose it had an algebraic addition theorem $P$. There are only finitely many values $V_0$ for which $P(U,V_0,W)$ is identically zero: a nonzero coefficient polynomial in $V$ already bounds this exceptional set. Choose a positive irrational $\alpha$ with $e^{\alpha }$ outside this set, and set $c=\log \alpha$. Specializing $w=c$ gives a nonzero polynomial $Q(U,W)=P(U,e^{\alpha },W)$ and, with $t=e^z$,All exponents with nonzero coefficients are distinct, by irrationality of $\alpha$. Divide by the exponential with largest real exponent and let $t\to +\infty$; its nonzero coefficient would have to tend to zero. This contradiction proves that simple periodicity does not imply an algebraic addition theorem. It is the simply periodic entire function without an algebraic addition theorem example.

The usual field norm $N_{L/K}$ is defined for a finite field extension. The printed hypothesis “algebraic” alone allows an infinite extension, for which this determinant norm is not defined. We therefore assume $[L:K]<\infty$. For an arbitrary algebraic extension the meaningful elementwise statement is instead $N_{K(b)/K}(b)\in A$, and we prove this as well.

First a unique factorization domain is an integrally closed domain. Indeed, write an integral fraction as $u/v$ with relatively prime numerator and denominator. A monic integral equation, after multiplication by $v^r$, shows that $v$ divides $u^r$. Unique factorization forces $v$ to be a unit.

Let $m_b(T)=T^r+a_{r-1}{T}^{r-1}+\cdots +a_0\in K[T]$ be the monic minimal polynomial of $b$. Because $b$ is an integral element over $A$, it satisfies a monic polynomial $f\in A[T]$, and $m_b$ divides $f$ in $K[T]$. Every root of $m_b$ in an algebraic closure is therefore integral over $A$. This assertion includes repeated roots when the extension is inseparable. The coefficients of $m_b$ are elementary symmetric expressions in these roots. The integral elements form a ring: finitely many integral elements generate a finite $A$-module algebra, and the determinant trick shows every element of that algebra is integral. Thus all $a_i$ are integral over $A$. Since $a_i\in K$ and $A$ is integrally closed, $a_i\in A$.

Multiplication by $b$ on $K(b)$, in the basis $1,b,\dots ,b^{r-1}$, has the companion matrix of $m_b$, whose determinant is $(-1{)}^r{a}_0$. If $e=[L:K(b)]$, take a $K(b)$-basis of $L$. Multiplication by $b$ preserves each summand in the resulting direct sum of $e$ copies of $K(b)$, so its determinant over $K$ isIn particular $N_{K(b)/K}(b)=(-1{)}^r{a}_0\in A$. The argument requires no separability; in fact it holds for every integrally closed domain, not just a unique factorization domain.

For the connection with dimension, let $B$ be the coordinate ring of an irreducible affine variety. By Noether normalization there is a finite integral inclusion $A=k[t_1,\dots ,t_d]\subset B$, with $d=\dim B$. Their fraction fields form a finite extension. For any $0\ne b\in B$, its minimal-polynomial constant term $a_0$ is nonzero and lies in $A$. Its equation givesThus a nonzero function on the normalized variety produces a nonzero function on the base that vanishes wherever $b$ vanishes. Up to sign, the field norm is a power of this $a_0$, so it too belongs to $bB\cap A$.

Under the finite morphism to ${\mathbb{A}}^d$, the image of $V_B(b)$ is therefore contained in the proper Zariski-closed set $V_A(a_0)$ of dimension at most $d-1$. Finite morphisms preserve the dimension of a closed subvariety and its image. Applying this to a nonzero element of the ideal of any proper closed subvariety proves proper closed subvarieties have strictly smaller dimension. This is the useful role of the norm argument: nonzero equations upstairs give nonzero equations downstairs under a finite normalization.

The determinant-one cyclic quotient of the affine plane has the closed embeddingHere ${\mu }_n$ denotes the cyclic group generated by the given diagonal matrix. We establish the presentation of the polynomial invariant ring and verify that its fibres really are the group orbits.

On the coordinate ring $A=\mathbb{C}[x,y]$, the inverse-pullback action multiplies $x^a{y}^b$ by ${\zeta }^{b-a}$. Since these monomials are a basis, a polynomial is invariant precisely when each of its nonzero monomials satisfies $a-b\equiv 0(\mathrm{mod}n)$. If $a\ge b$, such a monomial is $(x^n{)}^{(a-b)/n}(xy{)}^b$; if $b\ge a$, it is $(y^n{)}^{(b-a)/n}(xy{)}^a$. ThereforeThese generators give a surjective homomorphism $\mathbb{C}[u,v,w]\to A^{{\mu }_n}$, and $uv-w^n$ lies in its kernel of a ring homomorphism. To prove there are no further relations, reduce any monomial using $uv=w^n$. The result is a linear combination ofTheir images are respectively $x^j{y}^j$, $x^{ni+j}{y}^j$ and $x^j{y}^{ni+j}$. All have distinct exponent pairs, so they are linearly independent in $A$. Any polynomial in the kernel of a ring homomorphism has zero reduced expression and hence belongs to $(uv-w^n)$. ConsequentlyThis proves the claimed closed subvariety embedding of the affine quotient.

For surjectivity of $\pi$, if $u\ne 0$ choose $x^n=u$ and set $y=w/x$. Then $y^n=w^n/u=v$. If $u=0$, the relation forces $w=0$; choose $x=0$ and any $n$th root of $v$ for $y$. If two points have the same image and $x\ne 0$, equality of their $n$th powers gives $x^{\prime }={\zeta }^{k}x$ and equality of $xy$ gives $y^{\prime }={\zeta }^{-k}y$. If $x=0$, both first coordinates vanish, and the same argument using $y^n$ gives the orbit relation; the case $x=y=0$ has only that one point in its fibre. Thus every fibre is exactly one group orbit.

The map is finite: $x$ and $y$ satisfy the monic equations $T^n-u=0$ and $T^n-v=0$, and the finitely many monomials $x^a{y}^b$, $0\le a,b<n$, span $A$ as a module over its polynomial invariant ring. A finite morphism is closed when it is surjective as here and therefore gives the quotient topology. For a principal open subset defined by an invariant $h$, its regular functions pull back to $A_h$ andIndeed, because $h$ is invariant, an invariant fraction with denominator $h^k$ has invariant numerator. These principal open subsets cover the affine hypersurface, proving the required local equality of invariant regular functions. Together with the orbit fibres, this verifies the geometric quotient, not merely a bijection of underlying sets.

At a point $P$ of the smooth algebraic curve $X$, the local ring ${\mathcal{O}}_{X,P}$ is a discrete valuation ring. Embed the projective variety $Y$ in ${\mathbb{P}}^N$. A rational map of algebraic varieties to $Y$ has homogeneous coordinates $[f_0:\cdots :f_N]$ in the function field $k(X)$, not all zero. Let $v_P$ be the local valuation, choose a uniformizer $u$, and set $m={\min }_{f_i\ne 0}{v}_P(f_i)$. The coordinates $u^{-m}{f}_i$ are regular at $P$, and at least one is a unit. On a sufficiently small open subset containing $P$, all these functions are regular and that coordinate stays nonzero. Their ratios therefore define a morphism of algebraic varieties into the corresponding affine chart of ${\mathbb{P}}^N$. Every homogeneous equation defining $Y$ vanishes on the original dense open, hence vanishes identically on this neighborhood. The local extension consequently lands in $Y$.

Two such extensions agree wherever the original rational map of algebraic varieties is defined. Their equalizer is closed, because $Y$ is a separated variety, and contains a dense open. Thus they agree throughout their overlap and glue. The rational map has a unique extension to a morphism on all of $X$. Projectivity is needed for the target; the source need not be projective.

For the function field correspondence, let $K/k$ be a finitely generated field extension of transcendence degree one. Choose a transcendence basis consisting of one element $t$. The extension $K/k(t)$ is finite: finitely many algebraic generators each have finite degree. Take the integral closures of $k[t]$ and $k[t^{-1}]$ in $K$. The finiteness assumption allowed in the paper makes both finite modules over their original coordinate rings. Integral closure commutes with localization, so these two rings agree over $k[t,t^{-1}]$ and glue to the finite normalization model of a one-variable function field $C_K\to {\mathbb{P}}^1$. This is a finite morphism, hence projective, and its function field is $K$. Its one-dimensional normal local rings are discrete valuation rings, so it is regular. Since the algebraically closed field $k$ is perfect, regularity here implies smoothness. Thus $C_K$ is a smooth projective curve.

Conversely the function field of a smooth projective curve is finitely generated over $k$ and has transcendence degree one. Two such curves with isomorphic function fields have mutually inverse birational maps. The extension argument just proved makes both maps morphisms; their composites equal the identity on a dense open, and hence everywhere by separatedness. This proves uniqueness of the smooth projective model. More precisely, pullback gives a contravariant correspondence between nonconstant morphisms of smooth projective curves and $k$-embeddings of their function fields. A field embedding defines the map on affine coordinates near the generic point, and the extension argument completes it. Constant morphisms are excluded from this correspondence because they do not induce embeddings of function fields.

Finally let $f:C\to D$ be nonconstant. Its image is dense, and pullback gives a finite extension $k(D)\subset k(C)$: both fields have transcendence degree one and the latter is finitely generated. Normalize $D$ in $k(C)$, applying the allowed finite-integral closure result on affine charts. This gives a finite morphism $C^{\prime }\to D$ with $C^{\prime }$ a smooth projective curve and function field $k(C)$. The identification of function fields gives an isomorphism $C\cong C^{\prime }$ by the preceding paragraph. Its composite with $C^{\prime }\to D$ equals $f$ at the generic point, hence equals $f$ everywhere. Every nonconstant morphism of these curves is finite.

Choose a nonconstant element $t$ of the function field $k(C)$. By Question 1 it gives a finite morphism $\pi :C\to {\mathbb{P}}^1$. The inverse images $U_0={\pi }^{-1}({\mathbb{A}}_t^1)$ and $U_1={\pi }^{-1}({\mathbb{A}}_{t^{-1}}^1)$ are affine, because the inverse image of an affine under a finite morphism is the spectrum of a finite algebra over its coordinate ring. They cover $C$. This proves the two-affine cover of a smooth projective curve assertion.

Their intersection is affine by Question 2. Thus the Čech cochain complex for any coherent sheaf $\mathcal{F}$ has just two nonzero terms,It follows that

On the projective line, put $t=X_1/X_0$ on the first chart. The twisting sheaf on projective space $\mathcal{O}(n)$ has frames $e_0=X_0^n$ and $e_1=X_1^n$, with $e_1=t^n{e}_0$ on the intersection; these are frame symbols also for negative $n$. In the $e_0$ frame the differential of the Čech cochain complex isA global section is therefore a Laurent polynomial in both $k[t]$ and $t^{n}k[t^{-1}]$. Its monomials have exponents $0\le j\le n$. For $n\ge 0$ the sections have basis $e_0,t{e}_0,\dots ,t^n{e}_0$; for $n<0$ there are none. The first sheaf cohomology isIts basis consists of the monomials $t^j{e}_0$ with $n<j<0$. These are precisely the exponents missed by the two chart spaces. Thus for every integer $n$,Both groups vanish at $n=-1$; all higher groups vanish for every $n$. This is also an explicit calculation of the Čech cohomology of twists on the projective line.

The printed vanishing claim needs the restriction $i\ge 1$. On $C={\mathbb{P}}^1$ with $\mathcal{F}=\mathcal{O}$ and $D=n\{\infty \}$, Question 3 gives $h^0(\mathcal{F}(D))=n+1$ for every $n\ge 0$. Thus no bound on the degree of a divisor can force vanishing in degree zero. We prove finiteness in every degree and the corrected vanishing in every positive degree, uniformly over all divisors on an algebraic curve of large degree.

First take a finite morphism $\pi :C\to {\mathbb{P}}^1$ as in Question 3 and put $E={\pi }_{\ast }\mathcal{F}$. The direct image sheaf $E$ is coherent: on an affine chart its module is finitely generated over a finite algebra and hence over the chart's coordinate ring. The two affine-cover Čech cochain complexes for $\mathcal{F}$ on $C$ and for $E$ on ${\mathbb{P}}^1$ are identical. Therefore

We need an elementary presentation of any coherent sheaf $E$ on the projective line. Choose finitely many generators on each of its two standard affine charts. A generator on the first chart restricts to an element of the second chart's module localized at $t^{-1}$. Multiplication by a sufficiently large power of $t^{-1}$ brings that element into the second chart's module. Since the transition of $\mathcal{O}(n)$ is $t^n$, this makes the original generator extend to a global section of $E(n)$. A generator on the second chart similarly extends after clearing its denominator at $t=0$. Use one common exponent for all the finitely many generators. The extended sections generate $E(n)$ on both charts. Untwisting gives a surjectionwith coherent kernel $K$. The long exact sequence in sheaf cohomology and Question 3 show that $H^1(E)$ is a quotient of the finite-dimensional $H^1(P)$, since $H^2(K)=0$. Apply the same presentation argument to $K$ to obtain finiteness of $H^1(K)$. Then the segment $H^0(P)\to H^0(E)\to H^1(K)$ proves finiteness of $H^0(E)$. Higher groups vanish by Question 3. Hence every $H^i(C,\mathcal{F})$ is finite-dimensional.

Next establish a useful degree estimate without using the duality requested later. A global regular function on $C$ is constant: a nonconstant one would give a finite morphism $C\to {\mathbb{P}}^1$ by Question 1. Such a morphism is surjective, because its closed image contains the generic point, but a regular function misses $\infty$, a contradiction. Thus $H^0(C,{\mathcal{O}}_C)=k$. Write $g=h^1(C,{\mathcal{O}}_C)$, the genus of a smooth projective curve. For any divisor on an algebraic curve $A$ and point $P$, the local discrete valuation ring givesThe last term is a length-one skyscraper sheaf. Additivity of the Euler characteristic of a coherent sheaf and repeated addition or subtraction of points giveThe inequality follows because $h^1$ is nonnegative, and is sufficient for the following argument.

Fix $P\in C$. For $n\ge g+1$ this estimate gives at least two independent sections of ${\mathcal{O}}_C(nP)$, one of them the constant section. A nonconstant section is a rational function on an algebraic variety whose poles are confined to $P$. It must have a pole there, since a globally regular function is constant. Its extension $C\to {\mathbb{P}}^1$ is finite, and its inverse image of ${\mathbb{A}}^1$ is exactly $C\setminus \{P\}$. Consequently $U=C\setminus \{P\}$ is affine.

Let $T$ be the torsion subsheaf of $\mathcal{F}$ and let $V=\mathcal{F}/T$. The support of $T$ is finite; such a coherent sheaf is a finite direct sum of sheaves supported at points and has zero higher sheaf cohomology. The torsion-free sheaf $V$ is locally free: a finitely generated torsion-free module over a discrete valuation ring is free. Tensoring by any invertible sheaf is exact, so

The inclusions $V(nP)\subset V((n+1)P)$ have point-supported quotients. Their long exact sequences in sheaf cohomology therefore give surjections on $H^1$. Moreovera section regular away from $P$ has some finite pole order at $P$. Compute this direct limit on a finite affine cover of $C$. All its intersections with $U$ are affine by Question 2, and localization commutes with filtered direct limits. The Čech cochain complex consequently givesThe maps from $H^1(C,V)$ to these groups are surjective. Choose a finite basis of this starting vector space. Each basis vector becomes zero at some stage in the direct limit; one common stage kills the entire basis. Thus there exists $n_0\ge 0$ with $H^1(C,V(n_{0}P))=0$.

Now let $D$ be any divisor on an algebraic curve with $\mathrm{deg}D\ge n_0+g$. The estimate above gives $h^0({\mathcal{O}}_C(D-n_{0}P))\ge 1$. A nonzero section identifies $D-n_{0}P$ with an effective divisor $E$ up to linear equivalence of divisors. Hence $V(D)\cong V(n_{0}P+E)$. Adding $E$ gives a point-supported quotient of $V(n_{0}P+E)/V(n_{0}P)$, so its first sheaf cohomology is a quotient of $H^1(C,V(n_{0}P))=0$. We have proved uniform high-degree vanishing on a smooth projective curve:This bound concerns all high-degree divisors, not only multiples of a previously chosen divisor.

An isogeny of elliptic curves over $k$ is a nonconstant $k$-morphism $\phi :E_1\to E_2$ which sends $O_1$ to $O_2$. A nonconstant morphism of smooth projective curves is finite and surjective. Its compatibility with the elliptic curve group law from Riemann-Roch can be seen without assuming it in the definition.

Identify a point $P$ with the degree-zero divisor class $[P-O_1]$. A finite morphism pushes forward divisors, and pushes a principal divisor to the principal divisor of the norm of its rational function. It therefore induces a group homomorphism on degree-zero Picard groups of a curve. On geometric points, the pushforward of $[P-O_1]$ is $[\phi (P)-O_2]$. Since point addition is addition of these divisor classes, this provesThe equality is defined over $k$, so in particular it holds on the groups of $k$-rational points and after every field extension. The argument also covers inseparable isogenies of elliptic curves; separability is not part of the definition.

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 $P\in X(k)$ 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.

For a divisor on an algebraic curve $D$, let $L(D)=H^0(V,\mathcal{O}(D))$ and choose a basis of dimension $r+1$. Regarding the basis as sections of the divisor line bundle, evaluation gives ${\varphi }_D:V\to {\mathbb{P}}^r$. If the linear system of divisors has no base points, at least one section is nonzero at each point, and these homogeneous coordinates define a morphism. If the effective divisor has a fixed part $F$, first divide all sections by their common vanishing factor. The moving divisor $E=D-F$ is base-point-free, has the same section space, and gives the uniquely extended morphism on the smooth curve. Thus effectiveness alone does not make the raw evaluations everywhere nonzero, but cancellation supplies the usual map.

A morphism from a projective integral curve is a finite morphism onto its image precisely when it is nonconstant, equivalently $r\ge 1$: a positive-dimensional fiber of a curve would be the whole curve, while a nonconstant proper map with finite fibers is finite. It is an isomorphism onto its image precisely when its moving complete linear system is very ample. In section-space terms, putting $h=h^0(E)$, the criteria arewhere $Q=P$ is allowed. These express absence of base points, separation of distinct points, and separation of tangent directions. In the base-point-free convention $E=D$.

Now suppose the genus is two. The Riemann-Roch theorem gives $\mathrm{deg}K=2$ and $h^0(K)=2$. There cannot be a nonconstant function with only a single simple pole on a positive-genus smooth projective curve: it would give a degree-one map to ${\mathbb{P}}^1$, an isomorphism. Hence $h^0(P)=1$ for every point $P$. Applying Riemann-Roch theorem to $K-P$ gives $h^0(K-P)=h^0(P)=1$, so the canonical linear system has no base point. The resulting map ${\varphi }_K:V\to {\mathbb{P}}^1$ has degree $\mathrm{deg}K=2$.

Set $D=K+P_1+P_2$, of degree four. The Riemann-Roch theorem gives $h^0(D)=3$ and $h^0(D-P)=2$ for every $P$, since the residual divisors have negative degree. Thus ${\varphi }_D$ is a base-point-free map to ${\mathbb{P}}^2$ with nondegenerate image $C$. Degree of the pullback of a general line givesThe image cannot be a line because the three sections are independent. The only alternatives are a birational quartic or a double cover of a conic. In the latter case identify the smooth conic with ${\mathbb{P}}^1$. Its hyperplane bundle has degree two, so $D\sim 2A$, where $A$ is a degree-two divisor with $h^0(A)\ge 2$. Again Riemann-Roch theorem implies $h^0(K-A)=h^0(A)-1\ge 1$. Since $K-A$ has degree zero, it must be linearly equivalent to zero; hence $A\sim K$. This would force $P_1+P_2\sim K$, contrary to the hypothesis. Therefore the image is a plane quartic and the map is birational. A smooth plane quartic has genus $(4-1)(4-2)/2=3$, while its normalization here has genus two. Consequently $\boxed{{\varphi }_D\text{ is birational onto a singular plane quartic}}$.

Use the graph morphism of schemesIt is a closed immersion: the square with ${\Delta }_g:Y\to Y{\times }_{Z}Y$ and the map $X{\times }_{Z}Y\to Y{\times }_{Z}Y$, $(x,y)\mapsto (f(x),y)$, is a fibre product of schemes square. The separatedness of $g$ makes ${\Delta }_g$ a closed immersion, and base change preserves a closed immersion.

The projection $\pi :X{\times }_{Z}Y\to Y$ is the base change of the proper morphism $gf:X\to Z$, so it is a proper morphism. The closed immersion ${\Gamma }_f$ is a finite morphism, hence a proper morphism by part (i). Since $f=\pi {\Gamma }_f$, part (ii) provesThis argument supplies finite type for $f$ as well; it need not be separately assumed.

A finite morphism is a finite type morphism. It is also a separated morphism: on an affine scheme chart $\mathrm{Spec}B\to \mathrm{Spec}A$, its diagonal morphism is represented by the surjective multiplication ring homomorphism $B{\otimes }_{A}B\to B$, and is therefore a closed immersion by Question 1.

We prove the lifting condition in the valuative criterion for properness. Choose an affine open subset of $Y$ containing the image of the closed point of $\mathrm{Spec}V$. Every point of $\mathrm{Spec}V$ is a generalization of its closed point, so the whole map factors through this affine open subset. Its inverse image under the finite morphism is $\mathrm{Spec}B$, with $B$ a finitely generated $A$-module. Every $b\in B$ is integral over $A$: apply the determinant trick to multiplication by $b$ on a finite set of module generators to obtain a monic polynomial which annihilates all generators and hence annihilates $1$.

In the given valuative square, we have compatible ring homomorphisms $A\to V$ and $B\to K$. The image in $K$ of each $b\in B$ satisfies a monic polynomial over $V$. To see directly why it belongs to $V$, suppose $x\in K\setminus V$. The defining property of a valuation ring gives $x^{-1}\in V$, and it lies in the maximal ideal since it cannot be a unit. A monic equation for $x$, multiplied by $x^{-n}$, would then express $1$ as a sum of elements in that maximal ideal, a contradiction. This proves that valuation rings are integrally closed and that $B\to K$ factors through $B\to V$. The factorization is unique because $V\hookrightarrow K$ is injective. Thus the lift exists uniquely, proving