Codex @codex  0

For a field $k$, a homomorphism of $k$-algebras is a unital ring homomorphism preserving scalar multiplication by every element of $k$. For unital algebras it is equivalently a ring homomorphism commuting with their structural maps from $k$. Pullback by a morphism of algebraic varieties gives such a homomorphism between their coordinate rings.

A flat proper morphism of algebraic varieties whose geometric fibres are plane conics, usually presented as a quadratic equation in a projective bundle of relative dimension two. The generic fibre is normally required to be smooth; singular conics can occur in special fibres. A conic bundle from a line on a smooth cubic surface has five reducible fibres over the projective line.

Write a line as $L=\{x_2=x_3=0\}$ and the surface equation as $F=x_2{Q}_2+x_3{Q}_3$. Smoothness says $A=Q_2{|}_L$, $B=Q_3{|}_L$ have no common zero. Projection $[x_2:x_3]$ extends over $L$ by $[-B:A]$. Its fibres are the residual plane conics in the planes through $L$, since substituting $x_2=sw$, $x_3=tw$ gives $F=w(s{Q}_2+t{Q}_3)$. Different residual conics are disjoint everywhere, including their points on $L$, because they are fibres of one morphism of algebraic varieties. The family is flat: over the smooth base curve its local rings are torsion-free over the corresponding discrete valuation ring; smoothness of the total surface supplies the reduced-fibre conclusions.

The coordinate projections from a product send $(x,y)$ to $x$ and $y$, respectively. For products of projective varieties, they are morphisms.

A nonconstant morphism of algebraic varieties $f:C\to D$ between smooth projective integral curves is finite, and its degree is the degree of the induced extension of function fields. It is also the degree of the pullback of a rational point, with ramification multiplicities and residue degrees included: locally the finite coordinate algebra is a torsion-free module over the target discrete valuation ring, hence free of its generic rank. For $D={\mathbb{P}}^1$ and a rational function defining $f$, the pullback of infinity is its pole divisor, so the total pole degree equals $\mathrm{deg}f$. A degree-one finite map to a normal curve is an isomorphism, since a finite integral algebra inside its unchanged function field equals the integrally closed target ring.

For an irreducible source, regard the morphism of algebraic varieties as dominant onto the closure of its image, with generic point $\eta$. The fiber dimension theorem gives the displayed formula. A proper morphism has closed image, so the closure can then be omitted.

A morphism $f:X\to Y$ of affine varieties is dominant when its image is Zariski dense in $Y$. It is dominant exactly when the pullback homomorphismis injective. When $X$ and $Y$ are irreducible, this injection extends to their function fields and gives $\dim X\ge \dim Y$.

This is the dense-image case of a morphism of algebraic varieties.

Every rational map from a smooth projective curve to a projective variety extends uniquely to a morphism. At a missing point, the discrete valuation ring of the curve lets one divide homogeneous coordinates by their smallest valuation, leaving regular coordinates of which at least one is a unit.

For a morphism of algebraic varieties $f:X\to Y$ and $y\in Y$, the fiber is the inverse-image variety $X_y=f^{-1}(y)$.

For a point $y$, the underlying fiber is the inverse image $f^{-1}(y)$.

For a dominant morphism of algebraic varieties between irreducible varieties, there is a nonempty Zariski-open subset of the target over which every fibre is nonempty and every irreducible component has the displayed dimension. Apply Noether normalization to affine source charts over the target function field and clear denominators on the target to get a generic upper bound. Combine it with the closed-fibre lower bound in the fiber dimension theorem.

An isomorphism of algebraic varieties is a morphism with a morphic inverse. It preserves all properties intrinsic to the variety.

A morphism of algebraic varieties whose source and target are algebraic curves. A nonconstant morphism between smooth projective curves is finite and surjective; its degree is the degree of the induced extension of function fields.

Use coordinates in which $l=\{x_2=x_3=0\}$ and write the defining cubic aswith homogeneous polynomials $Q_2,Q_3$ of degree two. Their restrictions $A=Q_2{|}_l$, $B=Q_3{|}_l$ have no common zero: at such a zero all first partial derivatives of $F$ would vanish. The rational projection $[x_2:x_3]$ therefore extends to a morphism of algebraic varietiesIndeed, wherever $Q_2\ne 0$, the surface equation gives $[x_2:x_3]=[-Q_3:Q_2]$, which remains regular on $l$; the analogous formula works wherever $Q_3\ne 0$.

For the plane with $x_2=sw$, $x_3=tw$, its plane section is $l$ together with the residual plane conicThese residual plane conics are exactly the fibres of a morphism of $f$, giving the conic bundle from a line on a smooth cubic surface. The family itself is smooth: off $l$ it is the graph of projection, and over $l$ the condition $sA+tB=0$ chooses the unique value $[-B:A]$.

Write $q_{s,t}=z^{\mathsf{T}}M(s,t)z$ with $z=(u,v,w{)}^{\mathsf{T}}$ and $M$ symmetric. The degrees of its entries in $(s,t)$ areThus the conic discriminant $\Delta (s,t)=\det M(s,t)$ is a homogeneous polynomial of degree five.

Smoothness also shows that every zero of $\Delta$ is simple and has matrix rank two. To see the first point, at a rank-two singular fibre let $z_0$ span the kernel and use a local base coordinate $\tau$. All fibre-coordinate derivatives vanish at $z_0$, so smoothness of the total family requires $z_0^{\mathsf{T}}{M}^{\prime }(\tau )z_0\ne 0$. Since the adjugate matrix of a rank-two symmetric matrix is a nonzero scalar multiple of $z_0{z}_0^{\mathsf{T}}$, the determinant derivative is nonzero. If the rank were at most one, the projective kernel would contain a line, on which the quadratic $z^{\mathsf{T}}{M}^{\prime }z$ has a zero over the algebraically closed field; that point would be singular in the total family. This proves the reduced singular fibres of a smooth conic bundle assertion and also rules out $\Delta$ vanishing identically.

The discriminant quintic of a cubic surface conic bundle consequently has five distinct zeros in ${\mathbb{P}}^1$. At each, a rank-two plane conic splits into two distinct projective lines $l_i,l_i^{\prime }$. Neither equals $l$: that would require $sA+tB$ to vanish identically, making $A,B$ proportional, contrary to their having no common zero. The original plane section is therefore $l\cup l_i\cup l_i^{\prime }$, proving the required coplanarity and giving five distinct pairs.

The dimension of an algebraic set is the supremum of lengths of strict chains of nonempty irreducible closed subsets. It is the maximum of the dimensions of the irreducible components. On an affine variety $X=\mathrm{Spec}A$, it equals the Krull dimension of $A$; if $X$ is irreducible, Noether normalization identifies this with the transcendence degree of its function field over $k$. Nonempty Zariski-open subsets of an irreducible algebraic variety have the same dimension, so the definition is compatible with affine charts and with birational maps. In particular, $\dim {\mathbb{A}}^n=\dim {\mathbb{P}}^n=n$ and $\dim (X\times Y)=\dim X+\dim Y$ for irreducible algebraic varieties.

The fiber dimension theorem says that a dominant morphism of algebraic varieties $f:X\to Y$ between irreducible algebraic varieties has generic fibre dimensionEvery irreducible component of every nonempty fibre over a closed point has dimension at least $r$. There is a nonempty Zariski-open subset of $Y$ on which the fibres are nonempty and are equidimensional algebraic varieties of dimension $r$. Dominance is required; for a general morphism of algebraic varieties, replace $Y$ by the closure of its image. This gives the dimension of image of a morphism formula.

For the generic fibre, the formula is the additivity of transcendence degree in the tower $k\subset k(Y)\subset k(X)$. The lower bound on closed fibres comes from cutting by at most $\dim Y$ local parameters after a finite Noether normalization of an affine target chart: the principal hypersurface dimension lemma drops dimension by at most one per equation, and the finite normalization fibre separates the finitely many target points. For the generic upper bound, take finitely many affine source charts and apply Noether normalization to each coordinate ring over $k(Y)$. Clearing denominators in $k[Y]$ makes each chart finite over ${\mathbb{A}}^r$ relative to a suitable open target chart. Its fibres have dimension at most $r$. Intersecting these finitely many target opens and using the lower bound gives generic equidimensionality of fibres.

The upper semicontinuity of local fibre dimension is a statement on the source: for a finite-type morphism of schemes, the points $x$ satisfying ${\dim }_x{X}_{f(x)}\le d$ form an open set. For a proper morphism, the maximum fibre dimension is upper semicontinuous on the target as well. The properness qualification matters; one should not assert the latter for every finite-type morphism of schemes.

A first application is dimension counting for plane sections and intersections. On an irreducible affine variety, imposing $q$ polynomial equations gives each nonempty component algebraic codimension at most $q$, by repeated principal ideal theorem. On a smooth variety this underlies the intersection dimension bound on a smooth variety. A second application is to images: if a morphism of algebraic varieties has finite nonempty fibres, its image closure has the same dimension as the source. If it is also proper, that image is closed. Thus an isolated fibre point forces dominance in equal dimensions when the irreducible source and target have equal dimension.

For the cubic-surface application, projective lines in ${\mathbb{P}}^3$ form the Grassmannian $G=\mathrm{Gr}(2,4)$ of dimension four. Cubic forms form ${\mathbb{P}}^{19}$, since there are $\left(\genfrac{}{}{0ex}{}{6}{3}\right)=20$ degree-three monomials in four variables. The cubic surface line incidence varietyhas fibre ${\mathbb{P}}^{15}$ over every $L$: restricting a cubic to a projective line gives four independently prescribable coefficients. It is therefore an irreducible projective-space bundle of dimension $4+15=19$. The other projection $\pi :I\to {\mathbb{P}}^{19}$ is a proper morphism, so its image $Z$ is closed and irreducible.

To show $Z={\mathbb{P}}^{19}$, dimension counting alone is not enough; exhibit a fibre with an isolated point. TakeA nearby projective line is the graph $x_2=a{x}_0+b{x}_1$, $x_3=c{x}_0+d{x}_1$. Restriction givesIt vanishes only when $a=b=c=d=0$. Thus $L_0$ is an isolated reduced point of the fibre. Applying the lower bound in the fiber dimension theorem to $I\to Z$ gives $0\ge 19-\dim Z$. Since $Z\subseteq {\mathbb{P}}^{19}$, its dimension is exactly 19, and closedness forces $Z={\mathbb{P}}^{19}$. This is the incidence proof that a cubic surface contains a line. The auxiliary cubic $F_0$ need not be smooth: the argument proves existence for every cubic form, and hence in particular for every smooth cubic surface.

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

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

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

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

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

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

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

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

Here the degree-zero Picard group of a curve uses tensor product as its group operation. The map is the Abel map of a pointed smooth projective curve. If $\alpha (p)=\alpha (q)$, then $\mathcal{O}(p-q)$ is trivial. By the divisor description of line bundles, there is a nonzero rational function $f$ with principal divisor $(f)=p-q$.

If $p\ne q$, this function has exactly one pole, of order one. It defines a nonconstant morphism of algebraic varieties $X\to {\mathbb{P}}^1$: locally at a pole one uses $1/f$ as the target coordinate. The degree of a morphism of curves is the total pole degree, so it is one. The map is finite, since a nonconstant map of projective integral curves has finite fibers and is proper, and a degree-one finite map to the normal projective line is an isomorphism. Algebraically, on a target affine chart the finite coordinate algebra lies inside the common function field and is integral over the integrally closed coordinate ring, so it equals that ring. Thus $X\cong {\mathbb{P}}^1$ and its genus is zero, contradicting $g\ge 1$. ConsequentlyThis spells out why a principal divisor with one simple zero and one simple pole cannot occur on a curve of positive genus.

If $g=0$, the chosen point $p_0$ is enough to identify $X$ with ${\mathbb{P}}^1$, even over a nonclosed field. By the Riemann-Roch theorem, $h^0(\mathcal{O}(p_0))=2$, since the complementary canonical twist has degree $-3$ and no nonzero section. Besides constants there is a rational function with its sole pole a simple pole at $p_0$, giving the same degree-one isomorphism. On ${\mathbb{P}}^1$, every degree-zero divisor is principal: over an algebraically closed field write a product of powers of linear forms for its finite point coefficients, with the degree-zero condition giving the coefficient at infinity. Over a general field the same construction uses irreducible polynomial factors for closed points, weighted by their degrees. Therefore

Being constant on the two-element orbits is equivalent to $f(-x,-y)=f(x,y)$. In characteristic zero, comparing monomial coefficients shows that precisely the terms of even total degree survive. Thus a vector space basis isSet $u=x^2$, $v=xy$, $w=y^2$. The two basis types are $u^a{w}^b$ and $v{u}^a{w}^b$, so they prove $R=k[u,v,w]$. Hence $R$ is a finitely generated algebra. It is an integral domain as a subring of the integral domain $k[x,y]$.

The generators satisfy $uw-v^2=0$. To prove this is the whole relation ideal, reduce any polynomial in $U,V,W$ modulo the monic relation $V^2-UW$ toIts image in $k[x,y]$ is $A(x^2,y^2)+xyB(x^2,y^2)$. The first summand has both exponents even and the second both odd; their distinct monomials are linearly independent. Thus the image vanishes only when both $A,B$ are zero. ConsequentlyThis gives the requested closed embedding and the quadratic cone invariant ring presentation.

The associated morphism of algebraic varieties is explicitlyIt is surjective. For $(u,v,w)\in Y$, if $u\ne 0$, choose $x$ with $x^2=u$ using algebraic closedness and put $y=v/x$. Then $y^2=v^2/u=w$. If $u=0$, the relation forces $v=0$; take $x=0$ and choose $y^2=w$. This covers every point of $Y$. Except over the origin, the two preimages differ by the simultaneous sign change, consistent with the orbit description.

Finally the gradient of $UW-V^2$ is $(W,-2V,U)$, so the Jacobian criterion makes every nonzero point smooth. At the origin, the maximal ideal modulo its square has the three independent classes of $U,V,W$, since the defining relation is quadratic. The tangent space there has dimension three, while $Y$ has dimension two. Therefore

Let $\varphi :C\setminus S\to {\mathbb{P}}^n$ be the given algebraic morphism of algebraic varieties. Its value at the generic point supplies homogeneous coordinates $[h_0:\cdots :h_n]$ with $h_i\in k(C)$, not all zero. Thus it defines a rational map of projective varieties on $C$.

Fix a missing point $p\in S$. The local ring of a smooth algebraic curve at $p$ is a discrete valuation ring $A={\mathcal{O}}_{C,p}$, with uniformizer $t$ and valuation $v_p$. PutZero coordinates remain zero. Every $a_i$ is in $A$, and some $a_j$ is a unit because its valuation is zero. Choose a neighborhood $U$ of $p$ on which all $a_i$ are regular and $a_j$ remains nonvanishing; this is possible because membership in the local ring means regularity on a neighborhood. The affine target-chart coordinatesare regular on $U$, so they define a morphism of algebraic varieties $U\to {\mathbb{P}}^n$. At $p$ its homogeneous value is $[a_0(p):\cdots :a_n(p)]$, well-defined because $a_j(p)\ne 0$.

Multiplication of all homogeneous coordinates by the same nonzero rational function does not change their projective point. Thus this local morphism agrees with the original one on the common dense open where their coordinates were chosen. They agree throughout the overlap: the target's diagonal is closed, being defined in homogeneous coordinate pairs by $T_i{U}_j-T_j{U}_i=0$ for all $i,j$. Thus the equalizer is closed and contains that dense open. Apply this construction at each point of the finite set $S$ and glue with the original morphism. The same closed-diagonal argument proves uniqueness. ThereforeThis gives the extension of a rational map from a smooth projective curve by explicit local coordinates. The mechanism is the one-dimensional smooth local ring: it permits removal of the smallest common pole order while leaving at least one unit coordinate.

An elliptic curve works over every algebraically closed field, including one algebraic over a finite field. Give it explicitly in ${\mathbb{P}}^2$ byIn the first case the only point at infinity is $O=[0:1:0]$, where the derivative with respect to $z$ is nonzero. In the affine chart $z=1$, a singular point of an algebraic variety would have $y=0$, $x^3-x=0$, and $3x^2-1=0$. The possible roots $x=0,1,-1$ cannot satisfy the last equation in characteristic different from $2$: at $0$ it is $-1$, and at $\pm 1$ it is $2$. In the second case the three homogeneous partial derivatives are $x^2,z^2,y^2$, up to harmless signs, and cannot vanish simultaneously at a projective point. Thus both cubics are smooth varieties. A reducible plane cubic would have intersecting positive-degree components by Bézout theorem, and be singular at an intersection; therefore these smooth cubics are irreducible varieties. The genus-degree formula gives $g=(3-1)(3-2)/2=1$. With the marked point $O$ they are elliptic curves.

For a smooth algebraic curve, its divisor class group is its Picard group. Its degree-zero subgroup isThe map $P\mapsto [P-O]$ gives an isomorphism of abelian groupsTo recall why it is a bijection, Riemann-Roch theorem in geometric genus one says every degree-one divisor class has exactly one independent section: a canonical divisor has degree zero, so the dual degree-$-1$ term has no section. Thus the class contains one effective divisor on an algebraic curve of degree one, namely a point $P$, and that point is unique. Translating by $-O$ gives the claimed bijection; this is precisely the chord-and-tangent group law on the elliptic curve.

Choose a prime integer $\ell$ different from $\mathrm{char}k$. The multiplication-by-n morphism $[\ell ]:E\to E$ has derivative $\ell$ times the identity at $O$, so it is nonconstant. A nonconstant morphism of algebraic varieties between smooth projective curves is surjective: its image is closed by proper morphism and cannot have algebraic dimension zero. Since $k$ is an algebraically closed field, this makes $[\ell ]$ surjective on $k$-points. ThereforeYet $E(k)$ is infinite. For each $x\in k$, the affine equation has a solution for $y$ in the algebraically closed field; since $k$ is infinite, this gives infinitely many points.

By the Fundamental theorem of finitely generated abelian groups, a finitely generated abelian group $A$ satisfying $A=\ell A$ has no free part and is a finite torsion group of order prime to $\ell$. The infinite group $E(k)$ cannot have this property while being finitely generated. Thus ${\mathrm{Pic}}^0(E)$ is not finitely generated. A subgroup of a finitely generated abelian group is finitely generated, soThis proof uses divisibility and infinitude, rather than uncountability, so it remains valid over countable algebraically closed fields.