Subfields and subrings of rational function fields

Written by GPT-6.1 Sol (OpenAI), in Codex at Ultra. Original course text and diagrams: CC0. No independent mathematical review is claimed.

Editable source: NOE-RAT-01.md

A field can have finitely many generators without having independent generators. This distinction is the starting point of rationality questions. We first show that finite generation survives passage to a subfield. We then prove that one variable has a stronger property: every nonconstant subfield has a single rational coordinate. Polynomial rings behave differently from fields, even in simple invariant examples.

We assume finite extensions, Galois correspondence, Gauss's lemma and Noetherian modules, as taught in Abstract Algebra II and Further algebra. The dimension theorem for finitely generated domains is used with its statement recalled below. The surface proof in §5 additionally uses normalization, coherent cohomology, and the exact open geometric proof providers specified there. Basic references are [Milne], [Noether, Rational functions] and [Stacks]. The next lesson, Noether's problem and generic polynomials, asks how rational coordinates produce Galois extensions.

Throughout, \(k\) is a field. A finitely generated extension is generated as a field, whereas a finitely generated algebra is generated using sums and products. An extension is rational if it is purely transcendental over its specified base field.

1. Finitely many generators inside a larger field

A transcendence basis of \(K/k\) is an algebraically independent set \(T\) for which \(K/k(T)\) is algebraic. If \(K/k\) is finitely generated, every transcendence basis is finite, and the algebraic extension over it is finite. Its cardinality is \(\operatorname{trdeg}_k K\). These facts are [Stacks, Tags 030D and 030F].

Lemma 1.1. Algebraic elements cannot accumulate indefinitely

If \(K/k\) is finitely generated, the elements of \(K\) algebraic over \(k\) form a finite extension of \(k\).

Proof. Choose a transcendence basis \(z_1,\ldots,z_r\) of \(K/k\), and put \(D=[K:k(z_1,\ldots,z_r)]\). For any finite algebraic subextension \(E/k\) inside \(K\), the \(z_i\) remain algebraically independent over \(E\). Moreover,

Indeed, a \(k\)-basis of \(E\) stays linearly independent after adjoining indeterminates: clear denominators in a proposed relation and compare polynomial coefficients. Thus \([E:k]\leq D\).

Among these finite subextensions choose one, \(E_0\), of largest degree. If \(a\in K\) is algebraic over \(k\), then \(E_0(a)\) is another finite subextension. Maximality gives \([E_0(a):k]=[E_0:k]\), so \(a\in E_0\). Hence \(E_0\) contains all the algebraic elements. \(\square\)

Reference: [Stacks, Tag 037J].

Theorem 1.2. Intermediate fields are finitely generated

Suppose \(K/k\) is finitely generated and \(k\subseteq L\subseteq K\). Then \(L/k\) is finitely generated.

Proof. Choose a transcendence basis \(T\) of \(L/k\). Its cardinality is bounded by \(\operatorname{trdeg}_k K\). The extension \(K/k(T)\) is finitely generated, since the original generators of \(K/k\) still generate after enlarging the base. Lemma 1.1 applied to this base shows that its relative algebraic closure in \(K\) is finite. Every element of \(L\) is algebraic over \(k(T)\), so \(L/k(T)\) is finite. A finite vector-space basis together with \(T\) generates \(L\) as a field. \(\square\)

This proof gives no claim that \(L\) is rational. A transcendence basis only removes transcendental dependence; a finite algebraic extension may remain.

Theorem 1.3. Separability gives one additional generator

If \(K/k\) is finitely generated and separable, every intermediate field \(L/k\) has a separating transcendence basis \(t_1,\ldots,t_s\), and

for some \(w\in L\). Here separable for a possibly transcendental extension means that all its finitely generated subextensions are separably generated. Equivalently, \(K\otimes_k k'\) is reduced for every extension \(k'/k\).

Proof. Theorem 1.2 gives finite generation of \(L/k\). In characteristic zero, any transcendence basis is separating. In characteristic \(p>0\), use the separability criterion: every finite \(k\)-linearly independent list \(a_i\) in \(K\) has \(k\)-linearly independent \(p\)-th powers \(a_i^p\). This exact characterization is [Stacks, Tag 0322]. It immediately holds for lists in \(L\).

We give the basis argument, so that existence is explicit. Among all transcendence bases \(T=(t_1,\ldots,t_s)\) of \(L/k\), choose one minimizing the inseparable degree \([L:k(T)]_i\). Suppose this degree exceeds one. Choose \(v\in L\) inseparable over \(k(T)\), and put \(E=k(T,v)\). The kernel of evaluation

is a height-one prime generated by an irreducible polynomial \(F\): localizing in the \(X_i\) gives the minimal polynomial of \(v\), and Gauss's lemma gives its primitive irreducible form.

At least one partial derivative of \(F\) is nonzero. Otherwise all exponents are divisible by \(p\). The relation \(F(T,v)=0\) would then be a dependence between \(p\)-th powers of monomials. The separability criterion would give a nonzero relation between the monomials with exponents divided by \(p\), of smaller total degree. It would belong to the same principal kernel, contradicting its degree.

Choose a variable with nonzero derivative and omit the corresponding element from the list \(t_1,\ldots,t_s,v\). The remaining \(s\) elements are algebraically independent: a relation involving only them would be in \((F)\), impossible since \(F\) has positive degree in the omitted variable. Write \(T'\) for this list. Gauss's lemma and the nonzero derivative show that \(E/k(T')\) is finite separable. Multiplicativity of inseparable degrees now gives

contradicting the choice of \(T\). Thus \(T\) is separating. Finally the primitive element theorem applied to the finite separable extension \(L/k(T)\) gives \(w\). If that extension is trivial, take \(w=0\). \(\square\)

The criterion used in the proof concerns the extension over \(k\). It does not assert that an arbitrarily chosen transcendence basis is separating. For instance \(k(t)/k(t^p)\) is purely inseparable even when \(k\) is perfect.

2. Measuring a rational map in one variable

For a nonconstant rational function \(u=f(t)/g(t)\), with relatively prime \(f,g\in k[t]\), define

This is unchanged by multiplying numerator and denominator by a nonzero scalar.

Proposition 1.4. The degree formula

For every \(u\in k(t)\setminus k\),

Proof. The polynomial \(f(X)-U g(X)\) is irreducible in \(k[X,U]\). A factorization would have one factor of degree zero in \(U\), since the total degree in \(U\) is one. Such a factor would divide both \(f(X)\) and \(g(X)\), hence would be constant. As a polynomial in \(X\) over \(k[U]\) it is primitive: a common nonconstant factor of its coefficients would force \(f\) and \(g\) to be proportional after evaluating a root of that factor. Gauss's lemma therefore makes it irreducible over \(k(U)\).

The element \(t\) satisfies \(f(t)-u g(t)=0\). Thus it is algebraic over \(k(u)\), and \(u\) must be transcendental over \(k\), since \(t\) is. Substitute \(u\) for \(U\) in the irreducible polynomial. Its degree in \(X\) is \(\max(\deg f,\deg g)\); equal leading degrees cannot cancel because \(u\notin k\). It is a scalar multiple of the minimal polynomial of \(t\) over \(k(u)\). \(\square\)

The formula counts inseparable degree as well. In characteristic \(p\), the map \(t\mapsto t^p\) has degree \(p\), although a geometric fibre has only one distinct point.

Theorem 1.5. Lüroth's theorem

If \(k\subsetneq L\subseteq k(t)\), there exists \(u\in k(t)\setminus k\) with \(L=k(u)\).

Proof. A nonconstant element of \(L\) makes \(k(t)/L\) finite by Proposition 1.4. Let its degree be \(n\), and let \(P(X)\in L[X]\) be the monic minimal polynomial of \(t\). Clear denominators and remove their common factor to obtain a polynomial \(Q(t,X)\in k[t,X]\), primitive as a polynomial in \(X\) over \(k[t]\), which is a nonzero scalar multiple of \(P\). Let \(m=\deg_t Q\).

Some coefficient \(u\) of \(P\) is nonconstant: otherwise \(P(t)=0\) would make \(t\) algebraic over \(k\). Write \(u=f(t)/g(t)\) in lowest terms. Every coefficient of \(P\) is a ratio of a coefficient of \(Q\) and its leading coefficient, so \(\deg u\leq m\).

Since \(f(X)-u g(X)\) vanishes at \(t\), \(P\), and hence \(Q\), divides

in \(k(t)[X]\). Primitivity and Gauss's lemma give \(R=QH\) in \(k[t,X]\). Since \(\deg_t R\leq\deg u\leq m=\deg_t Q\), the nonzero quotient \(H\) lies in \(k[X]\).

No nonconstant polynomial in \(k[X]\) divides \(R\). To see this, reduce modulo an irreducible factor \(h(X)\) of such a divisor. In the field \(k[X]/(h)\), the identity \(g(t)\overline f=f(t)\overline g\) and the linear independence of the two coefficient vectors of \(f\) and \(g\) force \(\overline f=\overline g=0\). This says that \(h\) divides both \(f\) and \(g\), a contradiction. Thus \(H\) is a nonzero constant. Comparing degrees in \(X\) gives \(n=\deg u\). Consequently

and the tower formula forces \(L=k(u)\). \(\square\)

Reference: [Milne, Theorem 9.19 and Lemma 9.24], for the classical elementary degree argument.

This proof applies over imperfect fields and to inseparable subextensions. It does not use a separability hypothesis.

Example 1.6. Generators need not retain their individual degrees

We have \(k(t^2,t^3)=k(t)\), because \(t=t^3/t^2\). On the other hand \([k(t):k(t^2)]=2\), in every characteristic. Also

If \(k\) contains a primitive \(n\)-th root of unity and \(\operatorname{char}k\nmid n\), the extension \(k(t)/k(t^n)\) is cyclic Galois. Its automorphisms are \(t\mapsto\zeta t\). For each divisor \(d\mid n\), the subgroup of order \(d\) fixes \(k(t^d)\); equality follows by the degree formula. These are all the intermediate fields.

3. Polynomial functions and their missing poles

Theorem 1.7. Polynomial intersections in one variable

For every \(k\subseteq L\subseteq k(t)\), the ring \(A=L\cap k[t]\) is a finitely generated \(k\)-algebra. It is either \(k\), or a domain of dimension one.

Proof. If \(A\) contains only constants, there is nothing to prove. Otherwise take a nonconstant polynomial \(p(t)\in A\), of degree \(d\). Its equation expresses \(t^d\) as a \(k[p]\)-linear combination of \(1,t,\ldots,t^{d-1}\). Repeated reduction shows that these powers generate \(k[t]\) as a \(k[p]\)-module.

The intersection \(A\) is a \(k[p]\)-submodule of this finite module. Since \(k[p]\cong k[U]\) is Noetherian, there are \(a_1,\ldots,a_r\in A\) generating it as a module. In particular \(A=k[p,a_1,\ldots,a_r]\) as an algebra. Its fraction field has transcendence degree one: it contains \(p\) and lies in \(k(t)\). The dimension theorem for finitely generated domains gives \(\dim A=1\). \(\square\)

If \(p\in k[t]\) is nonconstant, there is an especially precise answer:

Indeed, write an element as \(a(p)/b(p)\), with coprime \(a,b\in k[U]\). A Bézout identity shows that \(a(p),b(p)\) remain coprime in \(k[t]\). For the quotient to be polynomial, \(b(p)\) must be a unit, so \(b\) is constant.

Contrast this with \(L=k(t+t^{-1})\): here \(L\cap k[t]=k\). A rational function of \(u=t+t^{-1}\) with no finite \(u\)-poles must be a polynomial in \(u\), because every finite value of \(u\) has a nonzero finite preimage over an algebraic closure. A nonconstant polynomial in \(u\) has a pole at \(t=0\). It therefore cannot lie in \(k[t]\). This argument works in every characteristic.

Example 1.8. A rational invariant field with a singular invariant ring

Assume \(\operatorname{char}k\neq2\), and let \(\sigma(x,y)=(-x,-y)\). Set \(a=x^2\) and \(v=y/x\). Then

The displayed field is fixed, and \(k(x,y)=k(a,v)(x)\), with \(x^2=a\). The nontrivial automorphism proves that this extension has degree two, hence the displayed fixed field is exact. The generators \(a,v\) are algebraically independent, since the extension is finite and \(k(x,y)\) has transcendence degree two.

A polynomial is invariant precisely when each monomial has even total degree. Such monomials have both exponents even or both odd. Therefore

The presentation is exact: reduce to expressions with \(B\)-exponent zero or one, whose images are distinct monomials. This ring has dimension two, but its maximal ideal \((A,B,C)\) has \(\dim_k \mathfrak m/\mathfrak m^2=3\), since its relation has degree two. A polynomial ring of dimension two has tangent-space dimension two at every \(k\)-rational point. Thus the invariant ring is not a polynomial ring. Rationality of its fraction field did not imply polynomial coordinates for its whole ring.

Figure 1.1. A polynomial \(t^3+t^2\) has its only pole at infinity, whereas \(t+t^{-1}\) has poles at zero and infinity. The displayed intersections are proved in the preceding examples; the line is a projective schematic. Editable diagram.

4. Symmetry as a source of rational coordinates

Let \(e_j\) denote the elementary symmetric polynomials in \(x_1,\ldots,x_n\).

Theorem 1.9. Symmetric polynomial and rational invariants

For every commutative coefficient ring \(R\),

and the expression as a polynomial in the \(e_i\) is unique. For every field \(k\),

Proof. Work one homogeneous degree at a time and order monomials lexicographically with \(x_1>\cdots>x_n\). The largest monomial \(x_1^{a_1}\cdots x_n^{a_n}\) of a nonzero symmetric homogeneous polynomial has \(a_1\geq\cdots\geq a_n\); swapping an inverted pair would give a larger monomial with the same nonzero coefficient. The product

has this largest monomial with coefficient one. Subtract the original coefficient times this product. The leading monomial decreases, and there are only finitely many monomials of the given degree. Iteration proves existence over \(R\), without division. Distinct monomials in the \(e_i\) have distinct leading monomials in the \(x_i\), with coefficient one, so a nonzero relation cannot cancel its largest term. This proves uniqueness, even if \(R\) has zero divisors.

For a fixed rational function \(q=a/b\), form \(D=\prod_{\sigma\in S_n}\sigma(b)\). This is a nonzero symmetric polynomial. The product \(qD\) is a polynomial and is invariant. Both \(D\) and \(qD\) are polynomials in the \(e_i\), so \(q\in k(e_1,\ldots,e_n)\). The reverse inclusion is immediate. \(\square\)

The same denominator product proves \(\operatorname{Frac}(R)^G=\operatorname{Frac}(R^G)\) whenever a finite group acts on a domain \(R\). It requires neither averaging nor division by \(|G|\).

The polynomial

has distinct roots and splitting field \(k(x_1,\ldots,x_n)\) over the symmetric field. Its Galois group is \(S_n\): every permutation acts, and every automorphism permutes the roots. This holds in every characteristic. It is the first example of a polynomial whose coefficients parametrize a prescribed Galois group.

5. Rationality and finiteness beyond a line

Noether's Rationalbasis is a finite field-generating list; it need not be algebraically independent. Her Minimalbasis has exactly as many elements as the transcendence degree, and therefore is algebraically independent. Its existence means rationality. An Integritätsbasis is a finite algebra-generating list for the polynomial ring under discussion. Field generation, rationality and ring generation answer different questions.

Her Involutionsbasis has a more specific meaning [Noether, Rational functions, §5]. It is the coefficient list of the involution form: the monic equation of a generic linear combination of the ambient coordinates. This list can be redundant and is not required to be algebraically independent. A separating transcendence basis followed by one algebraic generator, as in Theorem 1.3, is another useful generating list; it should not be substituted for Noether's definition.

Proposition 1.10. The coefficient basis of a finite rational map

Let \(k\) have characteristic zero, let

and let \(U_1,\ldots,U_r\) be independent variables. Put \(z=\sum U_ix_i\). Its monic minimal polynomial over \(L(\mathbf U)\) is homogeneous of degree \(d\) in \(Z,\mathbf U\), has coefficients in \(L\), and those coefficients generate \(L\) over \(k\).

Proof. The \(d\) distinct \(L\)-embeddings of \(K\) into an algebraic closure give distinct linear combinations \(\sum U_i\sigma(x_i)\), since the \(x_i\) generate \(K\). Hence \(z\) is primitive for \(K(\mathbf U)/L(\mathbf U)\), and

is its minimal polynomial. Its coefficients belong to \(L\): a normal closure permutes the factors, and its fixed field is \(L\). The product also proves homogeneity.

Let \(B\subseteq L\) be the field over \(k\) generated by these coefficients. Specializing \(\mathbf U\) to the \(i\)-th coordinate vector gives a monic equation \(\Phi(Z,\mathbf e_i)\in B[Z]\) with root \(x_i\). Thus \(K/B\) is finite. Characteristic zero makes it separable. For any finite separable extension generated by \(x_1,\ldots,x_r\), the same distinct-embedding argument shows that \(\sum U_ix_i\) has degree \([K:B]\) over \(B(\mathbf U)\).

But \(\Phi\) is irreducible over \(L(\mathbf U)\), and therefore over the smaller field \(B(\mathbf U)\); its degree is \(d\). Consequently

The tower formula gives \([L:B]=1\), proving \(L=B\). \(\square\)

For \(L=k(t^3+t^2)\subset k(t)\), the involution form is

its nonconstant coefficient is the generator \(t^3+t^2\). For the sign action in Example 1.8, the two coordinate rows are \((x,y)\) and \((-x,-y)\), and

Its coefficients generate the invariant field. The same list generates the invariant polynomial ring in that example, but these are different assertions: the latter ring is singular and is not a polynomial ring.

Noether first reduces a system to one with as many ambient coordinates as its algebraic rank, then constructs this coefficient basis. The proposition treats the resulting finite-map situation directly. For arbitrary characteristics, the field-generation and separating-basis assertions proved earlier retain their own precise hypotheses; the historical coefficient argument here is stated in its characteristic-zero setting.

Polynomial generators and the absence of points at infinity

Noether's regular-system condition is a way to make a positive finiteness theorem possible. In an image with the same number of variables as the transcendence degree, choose polynomials whose highest homogeneous parts have no common nonzero geometric zero. Her §11 proves integrality using a resultant. The following proof gives the finite-module statement directly and works in every characteristic.

Proposition. Let \(f_1,\ldots,f_r\in k[t_1,\ldots,t_r]\) have positive degrees, and let \(H_i\) be their highest homogeneous parts. Suppose the \(H_i\) have no common nonzero zero over \(\overline k\). Then \(k[t_1,\ldots,t_r]\) is a finite module over \(k[f_1,\ldots,f_r]\). The \(f_i\) are algebraically independent. Every intermediate \(k\)-algebra containing \(k[f_1,\ldots,f_r]\) and contained in the polynomial ring is finitely generated over \(k\).

Proof. The strong Nullstellensatz, Theorem 2.2 implies that the radical of \((H_1,\ldots,H_r)\) over \(\overline k\) contains all the variables. Choose positive integers \(a_j\) such that \(t_j^{a_j}\) belongs to this ideal. For \(N=1+\sum_j(a_j-1)\), every degree-\(N\) monomial is divisible by at least one \(t_j^{a_j}\), so belongs to the ideal. Each resulting identity is a finite linear system over \(k\); solvability after a field extension implies solvability over \(k\), by Gaussian elimination. Taking homogeneous components, we may therefore write every degree-\(N\) monomial as \(\sum_i q_iH_i\) with \(q_i\) homogeneous of degree \(N-\deg f_i\); a negative degree means \(q_i=0\).

Any monomial \(M\) of degree \(n\geq N\) is a monomial multiple of one of degree \(N\). Hence \(M=\sum_i Q_iH_i\), where \(\deg Q_i=n-\deg f_i<n\). Write \(H_i=f_i-L_i\), with \(\deg L_i<\deg f_i\). We obtain

All polynomials \(Q_i\) and \(Q_iL_i\) have degree less than \(n\). Induction on degree expresses \(M\) as a \(k[f_1,\ldots,f_r]\)-linear combination of the finitely many monomials of degree less than \(N\). This proves the finite-module assertion.

A finite-module extension is integral: the determinant argument applied to multiplication by any element supplies a monic equation. The fraction field \(k(t_1,\ldots,t_r)\) is therefore algebraic over \(k(f_1,\ldots,f_r)\). Their transcendence degrees coincide and equal \(r\); since there are exactly \(r\) elements \(f_i\), they must be algebraically independent. Finally the base ring \(k[f_1,\ldots,f_r]\) is Noetherian. Each intermediate algebra is a submodule of a finite module, hence a finite module itself. Its module generators together with the \(f_i\) generate it as a \(k\)-algebra. \(\square\)

The same conclusion passes through a polynomial specialization that is injective on the algebra under consideration. If its image \(B^*\subseteq k[t_1,\ldots,t_r]\) contains these \(f_i\), apply the proposition to \(B^*\) and then pull its generators back through the isomorphism onto the image. This is the finite-generation mechanism in Noether's regular-system theorem. The injectivity and the condition on highest parts are hypotheses to verify; neither can be inferred merely from having a finitely generated fraction field.

For example, \(f_1=s^2+t\), \(f_2=t^3+s\) have highest parts \(s^2,t^3\), with only the origin as common zero. The proposition proves finiteness over \(k[f_1,f_2]\) in every characteristic. Here the six monomials \(1,s,t,st,t^2,st^2\) generate the module: repeatedly replace \(s^2\) by \(f_1-t\) and \(t^3\) by \(f_2-s\). Each replacement reduces total degree in \(s,t\), so it terminates.

In contrast, \(B=k[s,st,st^2,\ldots]\) is not finitely generated. Every element of \(B\) is a constant plus terms of positive \(s\)-degree. Subtract the constant from each proposed generator. Modulo \((s^2)\), all products of two such generators vanish. A finite list would therefore span the positive part of \(B/(B\cap(s^2))\) over \(k\), whereas the classes \(st^n\), \(n\geq0\), are linearly independent there. This is Noether's example following her regular-system theorem. It illustrates a nonregular subalgebra; the field-intersection counterexample below requires a further construction.

Source: [Noether, Rational functions], work 6, §§10–12, including the nonregular example in the footnote to Theorem IX. The geometric no-zero condition is equivalent to her nonvanishing homogeneous resultant criterion. The module argument here supplies the proof without an effective resultant bound.

Two variables: Castelnuovo's theorem

Theorem (Castelnuovo). If \(k\) is algebraically closed of characteristic zero, every intermediate field \(k\subseteq L\subseteq k(x,y)\) is rational over \(k\).

The geometric assertion is that a unirational surface is rational. Here unirational means that there is a dominant rational map from the plane to the surface. The map need not be birational: rationality is a conclusion about its target's function field.

The complex surface case: models and prerequisites

By Theorem 1.2, \(L/\mathbb C\) is finitely generated. In transcendence degree two, choose a transcendence basis \(u,v\) and a primitive element \(w\) for the finite separable extension over \(\mathbb C(u,v)\). The equation of \(w\) gives an integral affine surface with function field \(L=\mathbb C(u,v,w)\). The rational functions \(u(x,y),v(x,y),w(x,y)\) define a dominant generically finite rational map from the plane. A projective closure and resolution give a smooth projective integral surface \(S\) with the same function field, still dominated by \(\mathbb P^2\). We prove that \(S\) is rational.

Here a divisor is an integer combination of curves on a smooth surface. Its line bundle is \(\mathcal O_S(D)\), and \(D\cdot C\) is the degree of its restriction to a curve \(C\), extended bilinearly. A complete linear system \(|D|\) consists of the nonzero sections of this line bundle up to scalar. It is empty precisely when there is no nonzero section. We write \(h^i=\dim H^i\), \(\chi=\sum_i(-1)^ih^i\), and \(K\) for a canonical divisor, so \(\mathcal O_S(K)=\omega_S\). A smooth rational curve of square \(-1\) is called an exceptional curve of the first kind.

The geometric and topological foundations have the following exact earlier programme proofs. The reductions from them to surface rationality are written below.

The tagged geometric proofs are part of the programme’s separately linked core/fork collection. The AI Integrated Stacks edition is pinned at revision 565b10e987aba5969b21145a0833f42d69f96790; the labels used above are in varieties.tex, duality.tex, curves.tex, resolve.tex and more-morphisms.tex. Their source expression retains GFDL 1.2. Links to these proofs do not place their expression under this lesson’s CC0 dedication. Smoothness makes the surface Cohen–Macaulay and Gorenstein, so the proper duality theorem applies with its invertible canonical sheaf. Normalization of curves is finite by Normalization, Theorem 5.1; regular local rings are factorial by the programme’s core proof Tag 0AG0, so every curve on the smooth surface is Cartier.

The route is to prove that plane domination forces \(q=P_2=0\), contract the exceptional curves, and produce a smooth rational curve with nonnegative square by adjunction. Its linear system then gives either a birational map to the plane or a genus-zero pencil, which Tsen splits over \(\mathbb C(t)\).

What plane domination forces on a smooth surface

Let \(S\) be a smooth projective integral surface over \(\mathbb C\), and suppose that \(\mathbb P^2\dashrightarrow S\) is dominant. We prove \[ H^1(S,\mathcal O_S)=0, \qquad H^0(S,\omega_S^{\otimes m})=0\quad(m\geq1). \tag{C1} \] These statements concern the target. They make no claim that the given dominant map has degree one.

We use the written resolution course’s elimination of indeterminacy to obtain a diagram \[ \mathbb P^2\xleftarrow{\ b\ }T\xrightarrow{\ f\ }S, \] where \(T\) is smooth and projective, \(b\) is a sequence of blow-ups at points, and \(f\) is a dominant generically finite morphism. Indeed, elimination uses smooth centres. Each one-dimensional connected component of a proper smooth centre in a smooth surface is a Cartier divisor, whose blow-up is an isomorphism. The zero-dimensional components over \(\mathbb C\) are finitely many points. The components are disjoint, so they may be blown up separately, and the nontrivial steps are point blow-ups. The generic degree \(d=[\mathbb C(T):\mathbb C(S)]\) is positive and finite.

First we compute the effect of a point blow-up \(\pi:\widetilde X\to X\) on \(H^1(\mathcal O)\), for any smooth integral surface. Away from its centre it is an isomorphism. On a sufficiently small affine neighbourhood \(U=\operatorname{Spec}A\) of the centre, its ideal has generators \(x,y\) forming a regular system of parameters at that point. The blow-up has the two affine charts \[ B_1=A[t]/(y-xt),\qquad B_2=A[s]/(x-ys), \] After shrinking \(U\), the two generators generate the point’s ideal throughout \(U\) and form a regular sequence, justifying these Rees-chart presentations. Their intersection has coordinate ring \[ C=A[t,t^{-1}]/(y-xt),\qquad s=t^{-1}. \] Every Laurent polynomial in \(C\) is the sum of the image of a polynomial with nonnegative powers of \(t\), hence an element of \(B_1\), and one with negative powers of \(t\), hence an element of \(B_2\). Consequently the degree-one Cech quotient \(C/(B_1+B_2)\) is zero. Since the charts and their intersection are affine, this calculates \(H^1(\pi^{-1}U,\mathcal O)\). It also shows that all higher Cech groups vanish. Shrinking such neighbourhoods proves \(R^1\pi_*\mathcal O_{\widetilde X}=0\).

Furthermore, \(\pi_*\mathcal O_{\widetilde X}=\mathcal O_X\). A section upstairs is a rational function downstairs regular off the centre; extension across codimension two on the normal surface \(X\) makes it regular at the centre. Conversely a downstairs section pulls back. The low-degree Leray sequence therefore gives \[ H^1(\widetilde X,\mathcal O_{\widetilde X}) \cong H^1(X,\mathcal O_X). \tag{C2} \]

For completeness, \(H^1(\mathbb P^2,\mathcal O)=0\) follows directly from its three standard affine charts. Their Cech complex splits into Laurent monomials \(X_0^{a_0}X_1^{a_1}X_2^{a_2}\) with \(a_0+a_1+a_2=0\). Such a monomial appears on an intersection indexed by \(I\subseteq\{0,1,2\}\) exactly when \(I\) contains every index with negative exponent. If all exponents are nonnegative the monomial is the constant 1, and its complex is the simplex complex, with zero positive-degree cohomology. Otherwise the set of negative indices is nonempty and proper: all three exponents cannot be negative when their sum is zero. Choosing an index outside that set contracts its Cech complex by inserting or deleting that index, with the usual alternating sign. Thus its degree-one cohomology also vanishes. Applying (C2) along \(b\) yields \(H^1(T,\mathcal O_T)=0\).

Now consider \(f_*\mathcal O_T\). Proper coherent direct images make this a finite \(\mathcal O_S\)-module and algebra. On an affine open \(U=\operatorname{Spec}A\subseteq S\), write \(B=\Gamma(f^{-1}U,\mathcal O_T)\). This is a finite \(A\)-algebra contained in \(\mathbb C(T)\). Every \(b\in B\) is integral over \(A\); its field trace is a sum of its conjugates and is therefore integral over \(A\). That trace lies in \(\mathbb C(S)=\operatorname{Frac}A\), and \(A\) is integrally closed because \(S\) is smooth. Thus the field trace defines an \(A\)-linear map \(B\to A\). These maps agree on overlapping opens, since they are restrictions of the same field trace. They give a sheaf map \[ \operatorname{Tr}_f:f_*\mathcal O_T\longrightarrow\mathcal O_S, \qquad \operatorname{Tr}_f(f^*a)=d a. \] As \(d\) is invertible in \(\mathbb C\), \(d^{-1}\operatorname{Tr}_f\) splits the inclusion \(\mathcal O_S\to f_*\mathcal O_T\). Hence \(H^1(S,\mathcal O_S)\) injects into \(H^1(S,f_*\mathcal O_T)\). The low-degree Leray sequence injects the latter into \(H^1(T,\mathcal O_T)=0\). This proves the first equality in (C1).

To prove the second, the differential of the generically finite separable map \(f\) gives a morphism of canonical bundles \[ f^*\omega_S\longrightarrow\omega_T \] which is an isomorphism at the generic point. Its \(m\)-th tensor power therefore makes pullback injective on global sections: \[ H^0(S,\omega_S^{\otimes m}) \hookrightarrow H^0(T,\omega_T^{\otimes m}). \tag{C3} \] The morphism \(b\) is an isomorphism over \(\mathbb P^2\setminus Z\) for a finite set \(Z\). A pluricanonical section on \(T\) consequently restricts to a section of \(\omega_{\mathbb P^2}^{\otimes m}\) there. Extension of a section of an invertible sheaf across codimension two on a normal scheme extends it uniquely to \(\mathbb P^2\). The canonical bundle is \(\omega_{\mathbb P^2}=\mathcal O_{\mathbb P^2}(-3)\), as follows: on an overlap, \(u'=1/u,v'=v/u\) gives \(du'\wedge dv'=-u^{-3}du\wedge dv\); the other transitions follow by permuting the three homogeneous coordinates. A global section of \(\mathcal O(-3m)\) would be a homogeneous polynomial of negative degree, so is zero. This homogeneous-section calculation is also proved in Cohomology of projective space, §2. Thus the original section on \(T\) is zero on a dense open and hence zero. Combining this with (C3) proves (C1).

In particular \(q(S)=0\), \(p_g(S)=0\), \(P_2(S)=0\), and \(\chi(\mathcal O_S)=1\). The last equality uses Serre duality on the smooth projective surface, which identifies \(H^2(S,\mathcal O_S)\) with the dual of \(H^0(S,\omega_S)\), and \(H^0(S,\mathcal O_S)=\mathbb C\).

Two numerical formulas on a smooth surface

Let \(S\) be a smooth projective integral complex surface, let \(K\) be a canonical divisor, and write \(D\cdot E\) for the intersection of Cartier divisors. The quadratic Euler-polynomial construction and proper Serre duality give the full surface Riemann–Roch formula \[ \chi(\mathcal O_S(D))=\chi(\mathcal O_S) +\frac{D\cdot(D-K)}2. \tag{1} \] Here is the bridge between those exact inputs and (1). The Euler polynomial in two divisor classes has total degree at most two, with quadratic part \(\tfrac12(mD+nE)^2\). Its constant term is \(\chi(\mathcal O_S)\). Define \[ \ell(D)=\chi(\mathcal O_S(D))-\chi(\mathcal O_S)-\tfrac12D^2. \] Comparison of the linear coefficients of that two-variable polynomial shows \(\ell(D+E)=\ell(D)+\ell(E)\) and \(\ell(nD)=n\ell(D)\) for every integer \(n\). Serre duality on the two-dimensional smooth projective surface gives \[ \chi(\mathcal O_S(D))=\chi(\mathcal O_S(K-D)). \] At \(D=0\) this says \(\ell(K)+K^2/2=0\). At general \(D\), expansion of the right-hand side therefore gives \[ \ell(D)+D^2/2=-\ell(D)-K\cdot D+D^2/2. \] Thus \(2\ell(D)=-K\cdot D\), proving (1) for every divisor, with no ampleness or effectivity restriction.

For an integral curve \(C\subset S\), the divisor sequence and (1) now give numerical adjunction: \[ \chi(\mathcal O_C) =\chi(\mathcal O_S)-\chi(\mathcal O_S(-C)) =-\frac{C^2+K\cdot C}{2}, \qquad 2p_a(C)-2=C^2+K\cdot C, \tag{2} \] where \(p_a(C)=1-\chi(\mathcal O_C)\). On every plane-dominated surface the vanishings just proved and Serre duality give \[ \chi(\mathcal O_S)=1. \tag{3} \] Normalization \(\nu:\widetilde C\to C\) is finite, and its quotient in \[ 0\longrightarrow\mathcal O_C\longrightarrow \nu_*\mathcal O_{\widetilde C}\longrightarrow Q\longrightarrow0 \] has finite support. Consequently \[ p_a(C)=g(\widetilde C)+\dim_{\mathbb C}H^0(Q)\geq0. \tag{4} \] Equality forces \(Q=0\) and genus zero. Then \(\nu\) is an isomorphism, and \(C\) is smooth. To identify a smooth genus-zero curve over \(\mathbb C\), choose a point \(p\). Curve Riemann–Roch gives \(h^0(\mathcal O(p))=2\). This degree-one system has no base point: one would leave a degree-zero line bundle with two independent sections, whereas a nonzero section of a degree-zero line bundle is nowhere zero and makes it trivial. The resulting morphism to \(\mathbb P^1\) has degree one and is an isomorphism, since both curves are smooth and the target is normal.

In particular, if \(K\cdot C<0\) and \(C^2<0\), (2) forces \[ C^2=K\cdot C=-1,\qquad C\cong\mathbb P^1. \tag{C6} \] Indeed the two negative integers have sum at most \(-2\), while \(p_a(C)\geq0\) makes their sum at least \(-2\). Both integers are therefore \(-1\), and \(p_a(C)=0\). Its normal bundle has degree \(-1\), so this is precisely an exceptional curve of the first kind, to which the existing contraction theorem applies.

Contracting the exceptional curves

Apply the projective contraction theorem whenever \(S\) contains a smooth rational curve of square \(-1\). The new surface is smooth and projective, and the contraction is the inverse of a point blow-up. The plane still dominates it: compose the resolved plane map with the contraction. Therefore the preceding proof of \(q=0\) and vanishing of every pluricanonical system applies afresh to each target.

This procedure terminates. A point blow-up leaves \(b_1\) unchanged and increases \(b_2\) by one: remove a small four-ball at its centre and replace it by the disk bundle of \(\mathcal O_{\mathbb P^1}(-1)\), whose boundary is \(S^3\) and which retracts to \(\mathbb P^1\). The Mayer–Vietoris sequence, with \(H^1(S^3;\mathbb Q)=H^2(S^3;\mathbb Q)=0\), gives the claimed change in degrees one and two. The Betti numbers of a compact manifold are finite. Each contraction thus decreases the nonnegative integer \(b_2\). We reach a surface \(X\) with no smooth rational curve of square \(-1\), still with \(q=0\), \(h^0(2K)=0\), and a generically finite morphism \(Y\to X\), where \(Y\) is a sequence of point blow-ups of the plane. All these contractions preserve the function field.

We now prove that \(X\) contains a smooth rational curve of nonnegative square. This is the step that distinguishes the surface theorem from Lüroth’s one-variable theorem.

A terminal divisor produces the curve

Call an effective divisor \(D\) terminal when \(|K+D|=\varnothing\). Suppose also that \(K\cdot D<0\). Some integral component \(C\) of \(D\) satisfies \(K\cdot C<0\). Since \(D-C\) is effective, a section of \(K+C\) would give a section of \(K+D\), so \(|K+C|=\varnothing\).

There are no sections of \(-C\), and surface duality gives

\[ h^2(\mathcal O_X(-C))=h^0(\mathcal O_X(K+C))=0. \]

Equations (1)–(3) therefore imply

\[ p_a(C)=\chi(\mathcal O_X(-C))=-h^1(\mathcal O_X(-C))\leq0. \]

Equation (4) now proves \(C\simeq\mathbf P^1\). Adjunction gives \(C^2+K\cdot C=-2\). Since \(K\cdot C<0\), it follows that \(C^2\geq-1\). The square \(-1\) is excluded by minimality, so \(C^2\geq0\).

We will produce such a terminal divisor in all three cases for the sign of \(K^2\).

The cases of nonpositive canonical square

If \(K^2=0\). Apply (1) to \(-K\). Duality gives \(h^2(-K)=h^0(2K)=0\), hence \(h^0(-K)\geq1\). Choose \(A\in|-K|\). It is nonzero: if \(-K\sim0\), then \(2K\sim0\), contradicting \(h^0(2K)=0\). Fix an effective ample divisor \(H\). Then \(K\cdot H=-A\cdot H<0\), so \((H+nK)\cdot H<0\) for large \(n\). Thus \(|H+nK|\) eventually becomes empty. Choose its last nonempty value, with an effective representative \(D\sim H+nK\). It is terminal and

\[ K\cdot D=K\cdot H+nK^2=K\cdot H<0. \]

If \(K^2<0\). First find an effective divisor \(E\) with \(K\cdot E<0\). Take an effective very ample divisor \(H\), and put \(a=K\cdot H\), \(b=K^2<0\).

If \(a<0\), take \(E=H\). If \(a=0\), take an effective representative of \(K+nH\) for sufficiently large \(n\); its existence follows from ample twisting, and its \(K\)-degree is \(b<0\).

If \(a>0\), put \(r_0=-a/b>0\). The real class \(H+r_0K\) has zero \(K\)-degree, while

\[ (H+r_0K)^2=H^2-a^2/b>0, \qquad (H+r_0K)\cdot H=H^2-a^2/b>0. \]

Choose a rational \(r>r_0\) close enough that both inequalities persist. A positive integral multiple \(L\) of \(H+rK\) is an integral Cartier class satisfying

\[ L^2>0,\qquad L\cdot H>0,\qquad L\cdot K<0. \]

For large \(m\), the divisor \(K-mL\) has negative \(H\)-degree, so \(h^2(mL)=h^0(K-mL)=0\). Formula (1) gives

\[ \chi(mL)=1+\tfrac12(m^2L^2-mL\cdot K)>0. \]

Therefore \(h^0(mL)>0\), and an effective \(E\sim mL\) has negative \(K\)-degree.

Choose an integral component \(C\) of \(E\) with \(K\cdot C<0\). Adjunction and \(p_a(C)\geq0\) show \(C^2\geq-1\). If \(C^2=-1\), then necessarily \(K\cdot C=-1\) and \(p_a(C)=0\), which contradicts minimality. Thus \(C^2\geq0\). Such a curve is nef: every other integral curve meets it nonnegatively, and its own square is nonnegative.

Now \((C+nK)\cdot C=C^2+nK\cdot C\) is eventually negative, so \(|C+nK|\) eventually becomes empty. Its last nonempty value gives a terminal effective \(D\sim C+nK\) with

\[ K\cdot D=K\cdot C+nK^2<0. \]

This completes both nonpositive cases without Hodge index or a cone theorem.

The positive case: integral anticanonical curves

Suppose \(K^2>0\), and assume for contradiction that no terminal effective divisor has negative \(K\)-degree. Formula (1) and \(h^0(2K)=0\) give

\[ h^0(-K)\geq1+K^2\geq2. \tag{5} \]

Every anticanonical divisor is integral and reduced. Otherwise an effective \(A\in|-K|\) can be written \(A=B+C\) with \(B,C\ne0\) effective. Since \(K\cdot A=-K^2<0\), one summand, say \(B\), has negative \(K\)-degree. But \(K+B\sim-C\), which has no sections. Thus \(B\) is a prohibited terminal divisor.

The only effective terminal divisor is zero. Let a nonzero terminal \(D\) be given, and choose a point \(p\in\operatorname{Supp}D\). Because of (5), evaluation at \(p\) has a nonzero kernel, so there is \(A\in|-K|\) passing through \(p\). This \(A\) is integral. If it is a component of \(D\), then \(K+D\sim D-A\) is effective, contradicting terminality. Otherwise \(A\cdot D>0\), by positive local intersection at \(p\), so \(K\cdot D=-A\cdot D<0\), again a contradiction.

Choose any effective divisor \(H\). A fixed ample divisor has negative intersection with \(K\), because \(-K\) is nonzero effective. Hence \(|H+nK|\) is empty for large \(n\). Its last effective representative is terminal and must be zero. Thus every effective divisor class is an integer multiple of \(-K\). Every line bundle is represented by a difference of effective divisors, after sufficiently positive ample twisting. We have proved

\[ \operatorname{Pic}(X)=\mathbb Z[K]. \tag{6} \]

The notation really denotes an infinite cyclic group: \(K^2>0\) prevents any nonzero multiple of \(K\) from being linearly trivial.

Two-point jets give \(K^2\leq5\). By adjunction, every integral \(A\in|-K|\) has \(p_a(A)=1\). Formula (4) implies that such a curve has at most one singular point: at each singular point the normalization quotient has positive length. A dimension-one normal local ring is regular, so a singular point cannot contribute length zero.

Choose distinct points \(x,y\in X\). First-order jet evaluation is a linear map

\[ H^0(X,-K)\longrightarrow (-K)\otimes\mathcal O_{X,x}/\mathfrak m_x^2 \oplus (-K)\otimes\mathcal O_{X,y}/\mathfrak m_y^2. \]

The target has dimension \(3+3=6\), since \(X\) is smooth of dimension two. If \(h^0(-K)>6\), a nonzero section would vanish to order at least two at both points. Its divisor is integral and singular at both, contrary to its arithmetic genus. Therefore \(h^0(-K)\leq6\). Combining this with (5) gives

\[ K^2\leq5. \tag{7} \]

Only the domination assumption remains to rule out (6).

Plane domination rules out the exceptional positive case

The complex orientation makes \(f:Y\to X\) a map of closed oriented four-manifolds of positive degree \(d=[\mathbf C(Y):\mathbf C(X)]\). At a point outside the branch and exceptional loci, its \(d\) inverse images have nonsingular complex derivative and hence local degree \(+1\).

Degree identities. Poincaré duality defines

\[ f_!:H^r(Y;\mathbb Q)\longrightarrow H^r(X;\mathbb Q) \]

by transport of homology pushforward. The cap-product identity gives the projection formula. Since \(f_*[Y]=d[X]\), it follows that

\[ f_!f^*(a)=d a. \tag{8} \]

One may also verify (8) by pairing both sides with arbitrary complementary classes and using the perfect Poincaré pairing. In particular \(f^*\) is injective, and \(f_!\) is surjective over \(\mathbb Q\).

Cohomology of a point blow-up. A point blow-up replaces a small real four-ball with the disk bundle of \(\mathcal O_{\mathbf P^1}(-1)\), with boundary \(S^3\). The new disk bundle retracts onto its exceptional \(\mathbf P^1\). Mayer–Vietoris, using \(H^1(S^3)=H^2(S^3)=0\), shows that \(H^1\) is unchanged and \(H^2\) acquires the one generator represented by the exceptional curve. Starting from the usual cell decomposition of \(\mathbf P^2\), we obtain

\[ H^1(Y;\mathbb Q)=0, \qquad H^2(Y;\mathbb Q) =\langle [L],[E_1],\ldots,[E_N]\rangle_{\mathbb Q}, \tag{9} \]

where \(L\) is the pullback of a line and the \(E_i\) may be taken as total-transform exceptional classes. Strict-transform curve classes also span, since the change from total transforms is triangular with diagonal one.

The algebraic intersection number used here equals the oriented pairing of divisor Chern classes. For very ample classes choose general smooth transverse divisors using the proved Bertini statement. Each intersection point has local algebraic multiplicity one and positive complex-oriented topological multiplicity one. Thus both pairings agree for these classes. Every divisor class is a difference of two sufficiently positive very ample classes; bilinearity proves the equality for all divisors.

All these degree-two classes are algebraic curve classes. The Gysin pushforward of a curve class under \(f\) is its degree times the class of its image curve, or zero when it is contracted. This follows directly by pushing the fundamental cycle of its normalization; over the smooth locus of an image curve the map has its ordinary covering degree, and finitely many omitted points carry no two-dimensional fundamental cycle.

The divisor class of an integral complex curve equals the first Chern class of its divisor line bundle. On its smooth locus this is the defining Thom-class calculation for the zero section, with its complex orientation. The singular locus is finite; degree-two supported cohomology at these real codimension-four points vanishes, so the identification extends over it. Thus (6), (8), and (9) imply that \(H^2(X;\mathbb Q)\) is spanned by \(c_1(K)\). Since its square is \(K^2>0\), it is nonzero. Also (8) and (9) give \(H^1(X;\mathbb Q)=0\). Hence

\[ b_1(X)=0,\qquad b_2(X)=1, \qquad \chi_{\mathrm{top}}(X)=2-2b_1+b_2=3. \tag{10} \]

The one-dimensional intersection form is positive, so \(\sigma(X)=1\).

For a complex rank-two bundle \(V\), the underlying real bundle has

\[ p_1(V_{\mathbb R})=c_1(V)^2-2c_2(V). \tag{11} \]

Indeed its complexification is \(V\oplus\overline V\); the Chern conjugation and Whitney rules give \(c_2(V\oplus\overline V)=2c_2(V)-c_1(V)^2\), and \(p_1=-c_2\) of the complexification. For \(V=T_X\), we have \(c_1(T_X)=-c_1(K)\), while the top Chern class is the Euler class and its evaluation is \(\chi_{\mathrm{top}}(X)\). The four-dimensional signature identity therefore gives

\[ 3\sigma(X) =\langle p_1(T_X),[X]\rangle =K^2-2\chi_{\mathrm{top}}(X). \]

Substitution of (10) and \(\sigma=1\) yields

\[ K^2=3+6=9, \tag{12} \]

contradicting (7). The assumed failure of a negative-\(K\) terminal divisor is impossible. The terminal-divisor argument produces the required smooth \(\mathbf P^1\) with nonnegative square.

The dimension-four signature identity

The four-dimensional identity has the following precise proof route. Signature and the first Pontryagin number are additive oriented-bordism invariants. For signature, relative Poincaré duality makes the restriction image in the middle cohomology of an oriented boundary an isotropic subspace of half the dimension. Pair a basis of that subspace with complementary vectors and subtract half their mutual pairings: this gives a hyperbolic basis, of signature zero. For a Pontryagin number, \(TW|_{\partial W}=T\partial W\oplus\mathbb R\), while the boundary fundamental cycle maps to zero in \(H_4(W)\). Naturality and stability therefore make that number zero on boundaries. These are the full boundary arguments in Signature, §§B–C.

The earlier rational bordism proof, Theorems J.4–J.5, supplies the exact rank theorem: \(\Omega^{SO}_4\otimes\mathbb Q\) has dimension one. Its finite-stage comparison uses a canonical oriented rank-\(k\) bundle over \(\widetilde G_k(\mathbb R^{k+p})\). In dimension four take \(k=6,p=5\). The first omitted Thom cell has dimension \(k+p+1=12\), so the finite and infinite Thom homotopy groups agree in degree \(4+k=10\). Pontryagin–Thom identifies that group with oriented bordism. The Thom space is \((k-1)\)-connected, and the proved rational Hurewicz comparison applies since \(10<2k-1=11\). Thom isomorphism and the stabilized degree-four universal oriented cohomology ring then give a one-dimensional rational group. These bounds state exactly the scope of the earlier proof; no general classification of surfaces enters.

The class of \(\mathbb {CP}^2\) is a rational basis because its signature is one. Its tangent Euler sequence gives \(c(T\mathbb {CP}^2)=(1+h)^3\), and the square of the positive hyperplane class evaluates to one. Thus its first Pontryagin number is \(9-2\cdot3=3\). The two rational bordism functionals agree on a basis after multiplying signature by three, proving \(\langle p_1(TM),[M]\rangle=3\sigma(M)\) for every closed oriented four-manifold. This is also the dimension-four specialization of the earlier full signature proof.

A smooth rational curve with nonnegative square

Let \(S\) be a smooth projective integral complex surface with \(H^1(S,\mathcal O_S)=0\). If it contains a smooth curve \(C\cong\mathbb P^1\) with \(C^2=n\geq0\), then \(S\) is rational.

The divisor sequence \[ 0\longrightarrow\mathcal O_S\longrightarrow\mathcal O_S(C) \longrightarrow\mathcal O_C(C)\longrightarrow0 \] and the assumed vanishing show that restriction on sections is surjective. The last line bundle is \(\mathcal O_{\mathbb P^1}(n)\), whose sections are the binary monomials of degree \(n\). The only global functions on this projective integral complex surface are scalars: proper coherence makes its ring of functions a finite integral \(\mathbb C\)-algebra inside its function field. Therefore \(h^0(S,\mathcal O_S(C))=n+2\). Moreover \(|C|\) has no base points: its defining section is nonzero off \(C\), and the restricted sections generate \(\mathcal O_C(C)\) at every point of \(C\).

If \(n\geq1\), blow up \(n-1\) distinct points on \(C\). Its strict transform \(C'\) is still a smooth rational curve, and its square is 1. At a smooth point of the curve the strict-transform normal bundle is the old normal bundle tensored by \(\mathcal O_C(-p)\), as the blow-up chart extracts the single vanishing factor at \(p\). Each point therefore lowers the square by one. The calculation (C2) preserves \(H^1(\mathcal O)=0\). Apply the preceding divisor sequence to \(C'\). Its complete linear system has three sections and no base points, so defines a morphism \[ \varphi:S'\longrightarrow\mathbb P^2, \qquad \varphi^*\mathcal O_{\mathbb P^2}(1)=\mathcal O_{S'}(C'). \] Its image has dimension two. Otherwise two general lines of \(\mathbb P^2\) can be chosen to meet outside its curve image; their inverse images are disjoint, which would give \((C')^2=0\). Since \(\varphi\) is proper, its two-dimensional image is all of \(\mathbb P^2\). It is generically finite. If its degree is \(e\), choose the intersection point of two lines in the open over which it is finite étale. Such an open exists by properness, generic finiteness and characteristic-zero separability, after shrinking the target. Their inverse-image divisors meet transversely in exactly \(e\) points. Consequently \[ 1=(C')^2=e. \] Thus the morphism induces an isomorphism of function fields, and both \(S'\) and \(S\) are rational. No classification of surfaces of minimal degree is used.

Suppose instead that \(n=0\). The two sections of \(|C|\) give a morphism \(f:S\to\mathbb P^1\) with the reduced smooth curve \(C\) as one fibre. Its Stein factorization has a smooth projective integral curve \(B\), a morphism \(g:S\to B\) with connected fibres, and a finite map \(h:B\to\mathbb P^1\). Over the point whose fibre is \(C\), the fibre of \(h\) has just one point: different points would give disjoint nonempty curves in \(C\). Write the pulled-back local parameter as \(a u^e\), where \(u\) is a uniformizer at this point of \(B\), \(a\) is a unit, and \(e\) is the ramification index. The divisor of \(g^*u\) is \(mC\) for a positive integer \(m\). Since the original fibre is the reduced divisor \(C\), \(em=1\), so \(e=m=1\). The sum of the ramification indices over that point equals the degree of \(h\), since the residue fields are \(\mathbb C\). Hence \(\deg h=1\), and \(B\cong\mathbb P^1\). In particular the generic fibre of \(f\) has no enlarged constant field.

Set \(K=\mathbb C(t)\) and let \(F/K\) be that generic fibre. It is a smooth projective geometrically integral curve. Indeed its local rings are localizations of those of the smooth surface, hence regular; a regular curve over the perfect field \(K\) is smooth. The Stein factorization says that \(K\) is algebraically closed in its function field, giving geometric irreducibility in characteristic zero. The morphism to the smooth base curve is flat: locally a uniformizer acts without torsion on the integral surface’s local rings, and a torsion-free module over a discrete valuation ring is flat. The proved Grothendieck-complex theorem and Euler-constancy corollary apply to this proper morphism and its flat structure sheaf. Locally they compute fibre cohomology by a finite complex of finite projective modules, whose alternating rank is constant. Since the base is connected, its fibres have the same Euler characteristic. The fibre \(C\cong\mathbb P^1\) has Euler characteristic 1, so \(\chi(F,\mathcal O_F)=1\). As its only global functions are \(K\), it has genus zero.

We explain why this genus-zero curve is rational over \(K\), rather than merely after algebraic closure. Its anticanonical line bundle has degree 2 and three sections. The curve Riemann–Roch formula and Serre duality give this directly: \(\deg\omega_F=-2\), and \(H^1(F,\omega_F^{-1})\) is dual to \(H^0(F,\omega_F^{\otimes2})=0\). Over an algebraic closure, a point \(p\) on the genus-zero curve satisfies \(h^0(\mathcal O(p))=2\); the base-point and degree-one argument already given after (4) makes it isomorphic to \(\mathbb P^1\). Hence the anticanonical system becomes the degree-two Veronese embedding. Flat base change of sections, Theorem 2.1, and the programme core proof of faithful descent of a closed immersion, Tag 02L6, show that the anticanonical system already embeds \(F\) over \(K\) as a smooth plane conic. Its degree-two equation is defined over \(K\): after field extension, the vector space of quadratic equations vanishing on the image has dimension one; flat base change of the ideal sheaf gives a one-dimensional equation space already over \(K\). A nonzero generator becomes the defining equation of the conic, and faithful field extension detects equality of the two closed subschemes.

Tsen’s theorem, with its full programme proof, says that \(\mathbb C(t)\) is a \(C_1\) field. A homogeneous quadratic equation in three variables therefore has a nonzero \(K\)-solution. This gives a point on the conic. Projection from that point, or the degree-one system just described, identifies its function field with \(K(u)\). Thus \[ \mathbb C(S)=\mathbb C(t,u), \] and \(S\) is rational also in the case \(C^2=0\).

Every contracted surface has the same function field as the original model. The produced curve and the last argument therefore give \(\mathbb C(S)=\mathbb C(s,t)\). Since \(\mathbb C(S)=L\), this proves the complex surface case. The argument is independently expressed using the free comparisons [Kumar, Lecture 9], [Mustață, Lecture 5] and [Freed, Lectures 11–12], with the exact earlier programme proofs above. [Castelnuovo] is the original freely readable treatment of the theorem.

Passage to an arbitrary algebraically closed field of characteristic zero. Write finitely many field generators of \(L/k\) as rational functions of \(x,y\). Let \(k_0\subseteq k\) be the field generated over \(\mathbb Q\) by their finitely many coefficients, and let

where \(h\) is a common denominator. For every field extension \(K/k_0\), tensoring this inclusion gives an inclusion \(B_0\otimes_{k_0}K\subseteq K[x,y,1/h]\), because field extension is flat. Hence every such base change is integral. Its dimension is two, and its function field over \(k\) is exactly \(L\).

The finitely generated characteristic-zero field \(k_0\) embeds in \(\mathbb C\): send a transcendence basis to algebraically independent complex numbers, then extend across its finite algebraic extension. The resulting complex surface is still dominated by the plane. The complex result supplies rational inverse maps between \((S_0)_{\mathbb C}\) and \(\mathbb A^2_{\mathbb C}\).

Existence of these maps descends to an algebraic closure of \(k_0\). Here is the finite-equation argument, which also explains why algebraic closedness matters. Write the two maps as ratios of polynomials. Retain their particular finite monomial supports and regard their coefficients as unknowns. Choose a finite generating set of the ideal defining \(S_0\). The equations saying that the maps land in \(S_0\) and that both compositions are identities become finitely many polynomial equations after clearing denominators. When an identity is taken modulo that ideal, include the finitely many coefficient unknowns of the polynomial multipliers witnessing its ideal membership; use the supports of the witnesses furnished by the complex maps.

Require the finitely many denominators to be nonzero in the corresponding coordinate rings. For a polynomial of a fixed bounded degree, its class lies in a finite-dimensional \(k_0\)-vector space of polynomial classes. Express it in a fixed basis of that space. A denominator is nonzero precisely when at least one coordinate is nonzero. Select a coordinate that is nonzero for the complex maps and impose its nonvanishing by adjoining an unknown \(z\) with the equation \(z c=1\). Apply the same procedure to denominators after substitution in the compositions. This ensures that the resulting identities hold as rational maps on nonempty open sets.

We have obtained a finite system of polynomial equations over \(k_0\) with a complex solution. Its coordinate algebra is nonzero; faithful field extension keeps it nonzero over \(\overline{k_0}\). The weak Nullstellensatz, as proved in The Nullstellensatz and Jacobson rings, gives a solution over \(\overline{k_0}\). Since \(k\) is algebraically closed, the inclusion \(k_0\subseteq k\) extends to an embedding of this algebraic closure in \(k\). The two rational inverse maps therefore exist over \(k\), and their identities give \(L=k(s,t)\).

Smaller transcendence degree. If \(\operatorname{trdeg}_kL=1\), the rational map from the plane to an integral curve model of \(L\) restricts dominantly to a suitable line. To check this, choose a nonconstant generator \(f\): a general vertical line works when \(f\) depends on \(y\), and otherwise a general horizontal line works. Avoid the finitely many choices on which the needed denominators vanish identically. The induced map of function fields embeds \(L\) in \(k(t)\), so Theorem 1.5 gives \(L=k(s)\). In degree zero, algebraic closedness gives \(L=k\). This proves the theorem. \(\square\)

For example, the degree-two map \((x,y)\mapsto(x^2,y)\) has rational target field \(k(x^2,y)\). Castelnuovo's conclusion does not turn this particular map into a degree-one map. Algebraic closedness and characteristic zero are part of the theorem; this proof makes no assertion about arbitrary ground fields or inseparable plane maps.

Intersections in transcendence degree at most two

Hilbert's fourteenth problem asks whether \(A\cap L\) is finitely generated when \(A=k[x_1,\ldots,x_n]\) and \(k\subseteq L\subseteq\operatorname{Frac}A\). Normality, rather than the particular polynomial presentation, is the useful hypothesis in the positive result.

Theorem (Zariski). Let \(A\) be a normal finitely generated domain over any field \(k\), and let \(L\subseteq\operatorname{Frac}A\) contain \(k\). If \(\operatorname{trdeg}_kL\leq2\), then \(A\cap L\) is a finitely generated \(k\)-algebra.

We prove the intersection statement by realizing it as the ring of regular functions on a normal curve or surface. The surface argument uses coherent cohomology and resolution. These are substantial geometric prerequisites, whose exact open proof providers are specified below; rationality of the surface is not required.

For this argument, a normal variety has integrally closed local rings, and a regular variety has regular local rings. A quasi-projective variety is an open subvariety of a projective variety. On a regular surface a divisor is an integer combination of integral curves; it defines a line bundle \(\mathcal O_S(E)\). The intersection \(E\cdot C\) is the degree of that line bundle on the curve \(C\), extended bilinearly to divisors. A divisor is nef if these degrees are nonnegative on all curves. We write \(h^i(\mathcal F)=\dim_kH^i(S,\mathcal F)\) and \(\chi=\sum_i(-1)^ih^i\). Stein factorization replaces a proper map by a map whose direct image of the structure sheaf is the structure sheaf, followed by a finite map. These definitions explain the geometric tools appearing in the proof.

Geometric tools used in the proof. On a normal finite-type scheme, regular functions are the rational functions belonging to all codimension-one local rings; see the exact open proof Stacks, Tag 031T. Removing a closed subset of codimension at least two therefore preserves global functions. Normalization in a finite function-field extension is finite by Normalization, Theorem 5.1. For cohomology we use the written proofs Proper morphisms and coherent direct images, Theorem 4.1 and Corollary 5.1, and Serre's theorems, Theorem 2.2.

The remaining geometric providers are the exact full open proofs surface resolution, Tag 0BGP, Stein factorization, Tag 03H0, numerical Euler polynomials, Tag 0BEM, intersection bilinearity, Tag 0BER, the pullback formula, Tag 0BET, Cartier-divisor restriction, Tag 0BEU, duality for proper Cohen–Macaulay schemes, Tag 0FVZ, and curve ampleness by component degrees, Tag 0B5Y. They are read in the AI Integrated Stacks Project edition and retain the GNU FDL 1.2; the tagged links also provide the official proofs. Their conditions apply because every scheme used here is of finite type over a field, hence excellent, and our resolved surface is regular and projective. A regular scheme is Gorenstein, so its dualizing module is invertible; this is Tag 0AWX. All fields are allowed; regularity below is a property of local rings and does not mean geometric smoothness.

One recurring consequence has a short proof here. For a proper birational map \(b:X\to Y\) of normal integral schemes, take an affine normal open \(\operatorname{Spec}B\subseteq Y\). Coherent direct images make \(\Gamma(b^{-1}(\operatorname{Spec}B),\mathcal O_X)\) a finite \(B\)-module inside \(\operatorname{Frac}B\). Each of its elements is integral over \(B\), so normality gives equality with \(B\). Thus \(b_*\mathcal O_X=\mathcal O_Y\).

Functions on a normal surface

Surface lemma. If \(U\) is a normal quasi-projective integral surface over \(k\), then \(\Gamma(U,\mathcal O_U)\) is finitely generated over \(k\).

Take a projective closure, normalize it, and resolve its surface singularities. We obtain a regular projective integral surface \(S\) and a proper birational map from an open subset \(U'\subseteq S\) to \(U\). Proper birational invariance gives equality of their rings of functions. Ignore the finitely many isolated points of the boundary using normality. Thus it suffices to consider \(U'=S\setminus D\), where \(D\) is an effective divisor. On a regular surface every curve is Cartier, by the full open proof that regular local rings are factorial, Tag 0AG0.

We first establish the numerical fact needed to treat this boundary. For Cartier divisors, write \(E\cdot F\) for their intersection number. The polynomial Euler-characteristic theorem gives

Intersection is symmetric and bilinear, is unchanged by field extension, and distinct integral curves have nonnegative intersection. An ample divisor has positive intersection with every nonzero effective curve. These facts follow from the numerical Euler-polynomial construction and the exact sequence of an effective Cartier divisor.

Here is the Hodge-index consequence, including its proof. Fix a very ample divisor \(H\). To bound sections on \(H\), we may extend the field purely transcendently and choose a generic hyperplane. This hyperplane is integral: its universal incidence variety is a projective bundle over the integral surface, and its generic fibre over the dual projective space is integral. Field extension preserves cohomology dimensions and intersections. On an integral projective curve, sections of a line bundle of fixed degree have a uniform bound: choose a regular closed point of degree \(a\), subtract it repeatedly, and use the exact sequences whose quotients have dimension \(a\). Negative-degree line bundles have no sections.

If \(E\cdot H=0\), then \(h^0(\mathcal O_S(nE))\) is bounded independently of \(n\): an effective divisor of intersection zero with \(H\) is zero. Duality bounds \(h^2(\mathcal O_S(nE))\) as well, since it equals \(h^0(\omega_S\otimes\mathcal O_S(-nE))\). Repeated restriction to \(H\) bounds this last dimension by finitely many curve-section bounds; the degrees of these restrictions are \(\omega_S\cdot H-jH^2\), independent of \(n\), and after finitely many subtractions the degree against \(H\) is negative. Since \(\chi=h^0-h^1+h^2\), the displayed quadratic polynomial implies \(E^2\leq0\).

If also \(E^2=0\), apply the same inequality to \(E+t(F-(F\cdot H/H^2)H)\) for rational \(t\). The linear coefficient in \(t\) must vanish, so \(E\cdot F=0\) for every divisor \(F\). Thus, after quotienting numerical zero classes, the form has one positive direction and is negative definite on the perpendicular space to any class of positive square. In particular, two disjoint effective divisors of positive square cannot exist, and a nonzero effective divisor perpendicular to a positive-square class has negative square.

We shall also need a boundary contraction that follows directly from cohomology. Suppose an effective divisor \(C\) satisfies \(C\cdot C_i>0\) on every integral component of its support. Then \(\mathcal O_C(C)\) is ample by the curve-degree criterion. In

Serre vanishing makes \(H^1(C,\mathcal O_C(nC))=0\) for large \(n\). Consequently the finite dimensions \(h^1(S,\mathcal O_S(nC))\) eventually decrease monotonically and hence stabilize. The maps between these groups are then isomorphisms. The restriction of global sections to \(C\) is therefore surjective for all sufficiently large \(n\). Since \(\mathcal O_C(nC)\) is globally generated, these lifted sections generate along \(C\); the canonical section of \(nC\) generates off \(C\). Thus \(\mathcal O_S(nC)\) is globally generated.

When \(C^2>0\), its resulting projective map has two-dimensional image: a pullback of a line bundle from a curve has square zero. Its Stein factorization yields a proper birational map \(b:S\to Y\) onto a normal projective surface, with \(\mathcal O_S(nC)=b^*Q\) for an ample line bundle \(Q\) on \(Y\). The canonical section descends because \(b_*\mathcal O_S=\mathcal O_Y\). Its complement \(W\subseteq Y\) is affine, and \(S\setminus C=b^{-1}(W)\). Every curve disjoint from \(C\) is contracted: the pullback of \(Q\) has degree zero on it, whereas a nonconstant map to \(Y\) would give positive degree.

Now put \(R=\Gamma(S\setminus D,\mathcal O)\) and \(K=\operatorname{Frac}R\). If \(\operatorname{trdeg}_kK=0\), it is contained in the finite relative algebraic closure of \(k\) in \(k(S)\), so \(R\) is finite-dimensional over \(k\).

Suppose \(\operatorname{trdeg}_kK=2\). Choose algebraically independent \(f,g\in R\). The map \([1:f:g]\) is defined on \(S\setminus D\). Resolve its indeterminacy, enlarging the projective surface by a proper birational modification; global functions on the inverse image are unchanged. Clearing the boundary poles gives a linear system whose moving line bundle \(M\) is globally generated and whose map has two-dimensional image. Hence \(M\) is nef and \(M^2>0\). The divisor \(M_0\) of its section corresponding to \(1\) is effective and supported on the boundary.

The support of \(M_0\) is connected. Otherwise split it into disjoint nonzero effective divisors \(B_1,B_2\). Nefness gives \(B_i^2=M\cdot B_i\geq0\). Equality contradicts the strict negative-square conclusion above, since \(B_i\) is effective and perpendicular to \(M\); positivity for both contradicts the single positive direction. Let \(C_{\mathrm{red}}\) be the connected component of the entire boundary containing \(M_0\). Its intersection matrix has nonnegative off-diagonal entries and contains a vector of positive square, namely the multiplicities of \(M_0\), with zeros on the remaining components.

This matrix has a positive eigenvalue with an eigenvector all of whose coordinates are positive. Indeed, maximize its quadratic form on the unit sphere, replace coordinates by their absolute values, and use connectedness: a zero coordinate adjacent to a positive coordinate contradicts the eigenvector equation. Rational approximation and scaling give positive integer multiplicities \(c_i\) such that \(C=\sum c_iC_i\) satisfies \(C\cdot C_i>0\) for all components. In particular \(C^2>0\).

Use the contraction just proved. The remaining boundary \(E=D_{\mathrm{red}}\setminus C_{\mathrm{red}}\) is disjoint from \(C\), so it maps to finitely many points \(Z\subset W\). Restriction identifies a function on \((S\setminus C)\setminus E\) with a function on \(W\setminus Z\): outside \(Z\), proper birational invariance applies. Normality extends that function across \(Z\). Pulling it back gives the original function by equality on a dense open. Thus

which is finitely generated because \(W\) is an affine surface of finite type.

Finally suppose \(\operatorname{trdeg}_kK=1\), and choose a nonconstant \(f\in R\). Resolve the map \(S\dashrightarrow\mathbb P^1\) defined by \(f\), and take its Stein factorization \(S\to T\to\mathbb P^1\). Here \(T\) is a normal projective integral curve. Localization of global sections, computed on a finite affine cover, gives \(R\otimes_{k[f]}k(f)=\Gamma(U'_\eta,\mathcal O)\). If a boundary component were horizontal, \(U'_\eta\) would be a punctured proper regular curve, hence affine with function ring of transcendence degree one over \(k(f)\). This would give \(\operatorname{trdeg}_kR=2\), a contradiction. The boundary is therefore vertical, and localization gives the generic fibre's field of constants, namely \(k(T)\). Consequently \(K=k(T)\).

A rational function in \(k(T)\) is regular on \(U'\) exactly when it has no pole at a point of \(T\) whose fibre has a component meeting \(U'\). The order along such a component is the order on \(T\) multiplied by its positive fibre multiplicity. Hence \(R=\Gamma(T\setminus B,\mathcal O_T)\), where \(B\) is the finite set of points whose entire fibre was removed. If \(B\) is empty this ring is finite-dimensional by proper cohomology. Otherwise \(T\setminus B\) is affine: the effective divisor of these points has positive degree, is ample by the curve criterion, and its complement is affine. This proves the surface lemma. \(\square\)

All three transcendence degrees occur. On \(\mathbb P^1\times\mathbb P^1\), removing one fibre gives \(\mathbb A^1\times\mathbb P^1\), with function ring \(k[t]\); removing that fibre and a fibre of the other projection gives \(\mathbb A^2\), with function ring \(k[t,u]\). Blowing up a rational point of \(\mathbb P^2\) and removing the exceptional curve gives \(\mathbb P^2\) minus that point, whose functions are just \(k\), by the codimension-two extension property. Thus finite generation concerns the actual functions, and does not assert that every open surface is affine.

Realizing the intersection

Return to \(R=A\cap L\), put \(F=\operatorname{Frac}A\), and set \(K=\operatorname{Frac}R\subseteq L\). Theorem 1.2 makes \(K/k\) finitely generated. Choose a finitely generated subalgebra \(B\subseteq R\) with fraction field \(K\), by writing field generators as fractions of elements of \(R\). Normalize \(B\) in \(K\). Its normalization is finite and still lies in \(A\), because \(A\) is integrally closed. We may therefore suppose \(B\) normal. Also \(A\cap K=R\).

By generic freeness, Stacks Tag 051R, there is a nonzero \(h\in B\) such that \(A_h\) is a free \(B_h\)-module. Here \(B\) is a Noetherian domain, \(A\) is a finite-type \(B\)-algebra, and we apply the lemma to the finite \(A\)-module \(A\). It is nonzero, so this free module is faithfully flat. Over this open every height-one point of \(\operatorname{Spec}B\) is the contraction of a height-one point of \(\operatorname{Spec}A\). Indeed, localize at the base point, a discrete valuation ring. Faithful flatness makes its uniformizer a nonunit upstairs; a minimal prime above that uniformizer has height one by the principal ideal theorem, and contracts to the maximal ideal of the base discrete valuation ring. The corresponding prime of \(A_h\) also has height one, since all smaller primes survive this localization. Conversely, going down for flat maps, Tag 00HS, shows that a height-one prime upstairs has contraction of height at most one.

Only finitely many height-one primes of \(A\) lie outside this open, namely those dividing \(h\). Restrict their discrete valuations to \(K\), discarding trivial restrictions. Every remaining restricted valuation is divisorial. Here is the required check. If \(v\) is such a valuation, residue elements algebraically independent over its residue subfield in \(K\) lift to units algebraically independent over \(K\): divide a putative polynomial relation by a coefficient of least valuation and reduce. Therefore

Since the upstairs valuation is a divisor, its residue field has transcendence degree \(\operatorname{trdeg}_kF-1\). The restricted residue field thus has transcendence degree at least \(\operatorname{trdeg}_kK-1\). The reverse inequality follows by adjoining an element of positive valuation to lifts of a residue transcendence basis: distinct powers have distinct valuations, so these elements are algebraically independent. Equality follows. The restricted value group is a nonzero subgroup of \(\mathbb Z\), and its residue field is finitely generated by Theorem 1.2.

These finitely many valuations can all be represented by divisors on one normal projective modification \(X\to\operatorname{Spec}B\). To construct it, for each valuation choose a uniformizer \(\pi\) in \(K\) and lifts \(z_i\) of a residue transcendence basis. They generate a polynomial algebra \(k[\pi,z_i]\) whose fraction field has finite extension \(K\). Normalize that algebra in \(K\). The normalization lies in the valuation ring, and its centre is a height-one prime above \((\pi)\). Take the closure of the graph of the rational maps from \(\operatorname{Spec}B\) to projective closures of these finitely many models, then normalize. This is projective over \(\operatorname{Spec}B\), and hence quasi-projective over \(k\). The valuative criterion supplies a centre on \(X\). Its projection lies in the auxiliary affine model: the coordinate algebra of that model lies in the valuation ring, so the map of its spectrum factors through this affine open. The auxiliary centre has residue transcendence degree \(\operatorname{trdeg}_kK-1\). Its residue field injects into that of the centre on \(X\), forcing this centre to have codimension one by the dimension theorem for finite-type domains.

Every other nontrivial restricted valuation was already a height-one valuation of \(B_h\). Its centre on \(X\) still has codimension one: the residue field of that base divisor injects into the centre's residue field, forcing dimension at least \(\operatorname{trdeg}_kK-1\), and a nontrivial valuation cannot centre at the generic point. Thus all required valuations are represented after this one finite graph construction. Choose a nonzero \(q\in B\) such that \(X\) is an isomorphism over \(\operatorname{Spec}B_q\) and this open lies in \(\operatorname{Spec}B_h\). Such an open exists because the modification is birational. Shrinking the open introduces no missing valuations: the preceding argument has already represented every restricted height-one valuation of \(A\).

Delete from \(X\) precisely the prime divisors outside this common open which do not represent a restricted height-one valuation of \(A\). Only finitely many prime divisors lie outside it, since all are components of the zero divisor of \(q\). On the common open every base divisor is represented, by faithful flatness. For a represented divisor the restricted valuation ring equals its height-one local ring: a discrete valuation ring in \(K\) dominated by a valuation ring of the same field has this property, as every element is a unit times an integral power of a uniformizer. The resulting normal quasi-projective open \(V\) has

by the codimension-one characterization. Its dimension is \(\operatorname{trdeg}_kK\leq2\). The surface lemma, or the corresponding projective-curve argument, proves finite generation. For dimension zero, the initial field argument already gives a finite-dimensional algebra. This proves Zariski's theorem. \(\square\)

The field \(K=\operatorname{Frac}(A\cap L)\) need not equal \(L\). For instance \(A=k[x,y]\) and \(L=k(x/y)\) give \(A\cap L=k\). To see this over any field, adjoin an independent variable \(t\). A polynomial belonging to \(k(x/y)\) is unchanged by \((x,y)\mapsto(tx,ty)\). Comparing its homogeneous parts in the polynomial identity \(P(tx,ty)=P(x,y)\) leaves only the constant part. The proof works with \(K\) because it records exactly the fractions supplied by the intersection.

Sources and rights. Zariski's original 1954 theorem is credited in the bibliography. Schröer, On contractible curves on normal surfaces, §§3 and 6, was consulted as a protected guide; its expression is not imported and its free PDF is not treated as a reuse licence. The proof written here replaces the boundary semi-ampleness step with the displayed cohomology argument. Bodzenta–Pełka–Weißmann, Saturation of algebraic surfaces, §4A of arXiv:2601.20621v2 (CC BY 4.0), explicitly recalls finite generation from Schröer; it is not used circularly as that theorem's proof. The stated Stacks geometric prerequisites have accessible GFDL proofs; the course supplies links to those proofs rather than copying GFDL prose into its CC0 exposition.

The positive result ends at transcendence degree two. The following linear action gives a negative answer in higher dimension.

Theorem 1.11. A Nagata-type linear counterexample

There is a linear action of the six-dimensional additive group on an eighteen-dimensional vector space over \(\mathbb C\) whose polynomial invariant ring is not finitely generated. Consequently Hilbert's fourteenth problem has a negative answer, even for linear unipotent actions in characteristic zero.

We prove the nine-point version of Nagata's construction. The connection with the total coordinate ring is due to Nagata and Mukai; [Totaro, Theorem 2.1 and Lemma 2.2] describe it. The proof below supplies the invariant-ring calculation and an explicit sequence of divisor classes, so no infinite Weyl-group assertion is needed.

Step 1: choose points at which every required transformation is defined. Choose eighteen complex numbers \(a_i,b_i\), \(1\leq i\leq9\), algebraically independent over \(\mathbb Q\), and put

Such numbers exist: adjoining finitely many numbers to the countable field \(\mathbb Q\) leaves its algebraic closure in \(\mathbb C\) countable, so one can choose the next number outside that closure. In particular the points are distinct and no three are collinear.

For three noncollinear points, change coordinates to make them the coordinate points and apply the quadratic transformation

Keep the three distinguished points as the coordinate points in the new plane, and transform the other six. This is an involutive rational operation on the space of ordered configurations, after undoing the coordinate change. The operation and its inverse are defined over \(\mathbb Q\); they are birational on nonempty open subsets. For any fixed finite word of these operations, its domain and the condition that the resulting configuration have no collinear triple contain a nonempty open subset defined over \(\mathbb Q\). This follows successively by pulling back nonempty opens under the birational operations. Its complement is contained in the zero set of a nonzero polynomial over \(\mathbb Q\) in the eighteen affine coordinates. Algebraic independence avoids every such polynomial. Thus every finite word is allowed for our configuration. This property is called Cremona general position.

Step 2: the geometry and its elementary numerical rules. Let \(S\) be the plane blown up at the nine points, let \(H\) be the class of a line pulled back from the plane, and let \(E_i\) be the exceptional curves. Then

Here divisors and their intersections have a direct interpretation. Every prime curve other than an exceptional curve is the strict transform of a plane curve. A degree-\(d\) plane curve with multiplicities \(m_i\) has class \(dH-\sum m_iE_i\): in local coordinates the blow-up substitutes \((x,y)=(x,xt)\), and extracts the factor \(x^{m_i}\). Plane divisors modulo rational functions are classified by their degree, by unique factorization of homogeneous polynomials. Hence \(H,E_i\) generate the Picard group. Two general lines intersect once; a line avoiding the points is disjoint from the \(E_i\); distinct exceptional curves are disjoint. The two blow-up charts show that the normal bundle of an exceptional curve is \(\mathcal O_{\mathbb P^1}(-1)\), giving its self-intersection \(-1\). These intersection numbers also prove independence of the stated generators. Intersections of distinct effective prime curves are nonnegative: they are sums of local intersection multiplicities, the lengths of the corresponding local quotients.

Resolving (N1) at its three coordinate points gives an isomorphism between their blow-up and the blow-up of the new plane at its three coordinate points. Indeed the three coordinate lines become the three exceptional curves for the inverse map, and the same two local charts resolve that inverse. The other six points avoid these lines, so the isomorphism extends after blowing them up. For the operation at a triple \(I\), its action on marked divisor classes is

To check this formula, the new line pulls back to \(2H-\sum_{i\in I}E_i\), and the new exceptional curve indexed by \(i\in I\) pulls back to the line through the other two points, of class \(H-\sum_{j\in I\setminus\{i\}}E_j\). The other exceptional classes stay fixed. These rules agree with (N3) on the basis in (N2).

Step 3: infinitely many negative curves. Write \(r_A,r_B,r_C\) for the triples \(123,456,789\), and set \(T=r_Cr_Br_A\), \(U=T^2\). All finite words realizing \(U^n\) are allowed by Step 1. Pulling an exceptional curve back through the resulting isomorphism gives an effective irreducible curve of class \(U^n(E_1)\), with self-intersection \(-1\).

Its degree grows without bound. Here is a complete calculation. For \(D=dH+\sum e_iE_i\), put \(A=e_1+e_2+e_3\), \(B=e_4+e_5+e_6\), \(C=e_7+e_8+e_9\). Formula (N3) makes \(T\) act on \((d,A,B,C)^t\) by

Let \(N=W^2-I\). Direct multiplication gives

Thus \(W^{2n}=I+nN+\binom n2N^2\). Since \(E_1\) has aggregate vector \((0,1,0,0)^t\), the degrees of these curves are

They are therefore distinct.

A prime curve \(C\) with \(C^2=-1\) is the unique effective divisor in its linear equivalence class. An effective divisor \(D\) in that class must contain \(C\), because otherwise \(D\cdot C\geq0\), whereas \(D\cdot C=-1\). Then \(D-C\) is effective and linearly equivalent to zero. An effective principal divisor on a projective integral variety is zero: its rational function has no poles, hence is a global regular function and is constant. Consequently the class of \(C\) cannot be a sum of two nonzero effective divisor classes. The infinitely many curves just constructed give infinitely many indecomposable elements of the monoid of effective classes on \(S\).

Step 4: identify the polynomial invariants. Let

The rank of \(P\) is three. The additive group \(G\cong\mathbb G_a^6\) acts linearly by

After inverting \(x_1\cdots x_9\), set \(v_i=y_i/x_i\) and \(w=Pv\). Choosing a linear complement to \(G\) changes the \(v_i\) into three quotient coordinates \(w_j\) and six coordinates translated independently by \(G\). A polynomial fixed by every translation is independent of those six coordinates: differentiating the translation identity gives zero partial derivatives, and characteristic zero forces independence. Hence

Give \(R\) the grading by the nine pair degrees \(\deg_i x_i=\deg_i y_i=1\), together with the total \(y\)-degree. These gradings preserve invariant homogeneous components: scaling each pair commutes with (N5), and scaling all the \(y_i\) normalizes \(G\). Thus every invariant is a finite sum of homogeneous invariants. By (N6), a component of pair degrees \(a_i\) and \(y\)-degree \(d\) is uniquely

where \(F\) is a homogeneous ternary form of degree \(d\).

Expression (N7) belongs to \(R\) precisely when \(F\) vanishes at \(p_i\) to multiplicity at least \(m_i=d-a_i\), whenever \(m_i>0\). To verify this assertion, expand \(F(Pv)\) as a polynomial in \(v_i\). Its highest \(v_i\)-power is \(d-\operatorname{mult}_{p_i}F\). In coordinates making \(p_i=[1:0:0]\), this follows by writing \(F=\sum_j z_0^{d-j}F_j(z_1,z_2)\), whose first nonzero \(F_j\) has index equal to the multiplicity. The other columns of \(P\) span the plane, so a nonzero coefficient polynomial does not become zero on substitution. Replacing \(v_i\) by \(y_i/x_i\) now shows that no negative \(x_i\)-power remains exactly when \(a_i\) is at least that highest power. Checking every \(i\) checks membership in the polynomial ring, since the expression already belongs to its localization at the product of the \(x_i\).

On the blow-up, forms of degree \(d\) with these multiplicities are exactly the global sections of \(\mathcal O_S(dH-\sum m_iE_i)\). For negative \(m_i\), the extra exceptional component is allowed and imposes no vanishing condition. This follows from the local factor extraction in Step 2 and pushforward of the effective divisor to the plane. There are no nonzero sections for \(d<0\), and a nonzero section always has \(m_i\leq d\). Multiplication of forms adds degrees and multiplicities. We have therefore proved an isomorphism of graded rings

where multiplication on the right uses the fixed divisor basis in (N2). This is the total coordinate ring, also called the Cox ring, of \(S\).

Step 5: conclude failure of finite generation and the intersection statement. If the ring in (N8) were generated by finitely many elements, taking their finitely many homogeneous components would give finitely many homogeneous generators. Every class with a nonzero section would then be a nonnegative integer sum of their degrees: express a nonzero homogeneous section as a polynomial in the generators and take its homogeneous component. This would generate the monoid of effective classes by finitely many elements, contradicting Step 3. Thus \(R^G\) is not finitely generated.

Finally put \(L=\operatorname{Frac}(R)^G\). It is a subfield of the rational function field in eighteen variables, and \(L\cap R=R^G\), directly from the definition of fixed elements. This supplies the required field-intersection counterexample to Hilbert's problem. \(\square\)

The construction proves the negative existence assertion assigned here. It does not identify this eighteen-variable version with Nagata's original thirty-two-variable example. Nagata's counterexample of 1959 is presented in [Nagata, Fourteenth problem]; [Totaro] explains the subsequent Nagata–Mukai formulation used above.

6. Exercises

1. Easy. Determine \(k(t^2,t^3)\), and compute the degrees of \(k(t)/k(t^2)\) and \(k(t)/k(t^3+t^2)\). Explain what changes about separability in characteristics two and three.
2. Medium. For relatively prime \(f,g\in k[t]\), with \(g\neq0\) and \(f/g\) nonconstant, prove directly that \([k(t):k(f/g)]=\max(\deg f,\deg g)\). Deduce that every generator of \(k(t)\) over \(k\) is \((at+b)/(ct+d)\) with \(ad-bc\neq0\).
3. Medium. Under the roots-of-unity hypotheses of Example 1.6, describe every intermediate field of \(k(t)/k(t^n)\) for an arbitrary positive integer \(n\), with its degree over each endpoint. Explain why the characteristic hypothesis matters.
4. Medium. Assume \(\operatorname{char}k\neq2\). For the sign action on \(k(x,y)\), compute both the fixed field and its intersection with \(k[x,y]\). Prove that the intersection is not a polynomial ring.
5. Hard. Reconstruct the finite-module proof of Theorem 1.7. Then compute \(k(t^3+t^2)\cap k[t]\) and \(k(t+t^{-1})\cap k[t]\).
6. Medium. Over an arbitrary commutative ring, express \(x^3+y^3\) and \(x^2y+xy^2\) using elementary symmetric polynomials. Explain why the calculation does not require the integer three to be invertible.

7. Solutions

1. The ratio \(t^3/t^2\) gives \(t\), so the first field is \(k(t)\). Proposition 1.4 gives degrees two and three for the other extensions. For \(t^2\), the minimal equation has derivative \(2X\), so the extension is inseparable precisely in characteristic two. For \(t^3+t^2\), the derivative is \(3X^2+2X\), which is nonzero in every characteristic; the degree-three equation is irreducible by Proposition 1.4, hence separable in every characteristic. In characteristic three its derivative is \(-X\), which still does not vanish identically.
2. The polynomial \(f(X)-Ug(X)\) has degree one in \(U\); a factor independent of \(U\) would divide \(f,g\). It is primitive over \(k[U]\), hence irreducible over \(k(U)\). Since \(f/g\) is transcendental, substitution yields the minimal equation and the degree formula. A generator gives degree one, so both numerator and denominator are linear. Nonconstancy says exactly \(ad-bc\neq0\); conversely the inverse fractional linear transformation expresses \(t\) in terms of it.
3. For each divisor \(d\mid n\), the field is \(k(t^d)\), with degree \(d\) below \(k(t)\) and degree \(n/d\) above \(k(t^n)\). The cyclic group of order \(n\) has exactly one subgroup of each divisor order, so these are all the intermediate fields. For \(n=12\), the divisors are \(1,2,3,4,6,12\). If the characteristic divides \(n\), \(X^n-t^n\) has repeated roots, and this Galois argument does not apply; a primitive root of unity of that order cannot exist either.
4. Put \(a=x^2,v=y/x\). The field \(k(a,v)\) is fixed and has index two, because adjoining \(x\) recovers both \(x,y\) and the sign automorphism is nontrivial. An invariant polynomial has only even-total-degree monomials, so the intersection is \(k[x^2,xy,y^2]\). The exact presentation \(k[A,B,C]/(AC-B^2)\) has dimension two and tangent dimension three at the origin, precluding a polynomial ring in two variables. A polynomial ring in any other finite number of variables has the wrong transcendence degree.
5. For nonconstant \(p\) of degree \(d\), reduce powers of \(t\) using its degree-\(d\) equation over \(k[p]\). The intersection is a submodule of the finite module \(k[t]\), so Noetherianity supplies finite generators. In the first example, a reduced fraction \(a(p)/b(p)\) stays reduced after substitution, forcing constant \(b\); the answer is \(k[t^3+t^2]\). In the second, write a reduced rational function \(a(u)/b(u)\). Any nonconstant \(b\) has a root over an algebraic closure; the equation \(t^2-u t+1=0\) gives a finite nonzero preimage and therefore a pole. Thus \(b\) is constant. A nonconstant polynomial in \(u=t+t^{-1}\) has a pole at zero, so only constants remain.
6. With \(e_1=x+y,e_2=xy\), expansion gives \(x^3+y^3=e_1^3-3e_1e_2\) and \(x^2y+xy^2=e_1e_2\). These are integral polynomial identities, so they hold over every ring. No division is used.

Prerequisites and scope

The general characterizations of separable field extensions, the primitive element theorem, and the dimension theorem for finite-type domains are prerequisites. Castelnuovo's surface theorem is proved in §5 by the plane-domination, adjunction, characteristic-number and rational-pencil arguments, using the exact earlier programme proofs listed there. The passage to every algebraically closed characteristic-zero field and the smaller-transcendence-degree cases are also proved here. Zariski's finiteness theorem is proved in §5 for every ground field, using the explicitly linked geometric prerequisites; the numerical and boundary arguments are supplied here. The Nagata-type linear counterexample is proved in Theorem 1.11. The finite-generation, separating-basis, Lüroth, one-variable intersection and symmetric-invariant assertions taught above have proofs here.

References

• [Milne] J. S. Milne, Fields and Galois Theory, Chapter 9, especially Theorem 9.19 and Lemma 9.24; Chapter 5, symmetric polynomials and the general polynomial.

• [Stacks] The Stacks Project, official project, consulted in AI Integrated Stacks Project, an edition with AI-proposed corrections and AI-written additions that have not been reviewed by the Stacks Project's maintainers. Transcendence, relative algebraic closure, separability, and primitive elements. The source editions retain their GFDL licence.

• [Noether, Rational functions] Emmy Noether, Rationale Funktionenkörper, Jahresbericht der Deutschen Mathematiker-Vereinigung 22 (1913), 316–319; Körper und Systeme rationaler Funktionen, Mathematische Annalen 76 (1915), 161–196, especially §§3–6 and §§8–12. German and English editions.

• [Castelnuovo] Guido Castelnuovo, Sulla razionalità delle involuzioni piane, full original PDF, Mathematische Annalen 44 (1894), 125–155, DOI 10.1007/BF01446977. Complete article, Chapter I, nos.1–10; Chapter II, nos.11–21. Original public-domain work; linked deposit marked CC0.

• [Liedtke] Christian Liedtke, Algebraic Surfaces in Positive Characteristic, Section 4, Theorem 4.6 (the rationality criterion of Castelnuovo and Zariski), and Section 9.1, Theorem 9.2 (separably unirational surfaces are rational).

• [Zariski, Interprétations] Oscar Zariski's finiteness theorem of 1954 is proved, for transcendence degree at most two, in Masayoshi Nagata, Lectures on the Fourteenth Problem of Hilbert, Tata Institute of Fundamental Research, 1965, Chapter 5, Theorem 4.

• [Schröer] Stefan Schröer, On contractible curves on normal surfaces, Journal für die reine und angewandte Mathematik 524 (2000), 1–15; author PDF dated 16 March 2005, §§3 and 6. Consulted as a protected guide for functions on normal surfaces; no source prose imported.

• [Bodzenta–Pełka–Weißmann] Agnieszka Bodzenta, Tomasz Pełka and Dario Weißmann, Saturation of algebraic surfaces, version 2, 29 May 2026, §4A, pp.14–15; CC BY 4.0. This passage recalls finite generation from Schröer rather than supplying that theorem's proof.

• [Nagata, Fourteenth problem] Masayoshi Nagata, Lectures on the Fourteenth Problem of Hilbert, Tata Institute of Fundamental Research, 1965, Chapter 3, which presents the counterexample of 1959.

• [Totaro] Burt Totaro, Hilbert's fourteenth problem over finite fields, and a conjecture on the cone of curves, Compositio Mathematica 144 (2008), 1176–1198, §2, Theorem 2.1 and Lemma 2.2. Consulted as a proof guide; the invariant-ring calculation and explicit class sequence in Theorem 1.11 are written out independently here.

• [Kumar, Lecture 9] Abhinav Kumar, MIT 18.727, Algebraic Surfaces, Lecture 9, Spring 2008, pp.1–4: adjunction and the anticanonical genus-one obstruction. Freely accessible lecture notes.
• [Mustață, Lecture 5] Mircea Mustață, Rationality and unirationality for surfaces, §2, Lemma 2.5 and Proposition 2.6. Freely accessible lecture notes. The terminal-adjunction argument is written above; the exceptional positive case uses plane domination and the proved signature identity.
• [Freed, Lectures 11–12] Daniel S. Freed, Bordism: Old and New, Lecture 11, Theorems 11.31, 11.48, 11.66 and equation (11.55), and Lecture 12, rational bordism and projective generators, 2012. Freely accessible author notes; the earlier programme proofs and the dimension-four reduction above supply the results used.