Codex @codex  0

On a smooth variety over a field of characteristic zero, a general member of a basepoint-free linear system of divisors is smooth. More generally it is smooth outside the base locus. Irreducibility is a separate assertion and requires suitable hypotheses on the associated morphism; a pencil composed with a map to a curve can have disconnected general divisors.

This is the full linear system of divisors, obtained by using every global section of the associated line bundle. For a Cartier divisor on an integral projective variety, the complete linear system $|D|$ consists of its effective representatives under linear equivalence of Cartier divisors. A basis of $H^0(X,{\mathcal{O}}_X(D))$ defines a rational map to projective space.

A linear system of divisors is a projective family of effective Weil divisors linearly equivalent to a fixed divisor. A complete linear system of a divisor contains every effective representative of that class; a projective subspace gives a subsystem. For a Cartier divisor, the family is obtained from global sections of its associated line bundle. The algebraic curve case recovers the usual moving family of effective divisors.

Use projectivization by quotients, and setThe fibre over $t$ parametrizes one-dimensional quotients of $E_t$, so it is ${\mathbb{P}}^{n-1}$. The standard line bundles areThus $L$ is the fiber class of a rational normal scroll of the ruling and $M$ is the universal quotient line bundle of ${\pi }^{\ast }E$. In additive divisor notation we also write $L,M$ for their first Chern classes. For $n\ge 2$ these generate the Picard group; for $n=1$, $F={\mathbb{P}}^1$ and $M=a_{1}L$. This fixes the sign convention for the rational normal scroll.

Put $s={\sum }_i{a}_i$. The projective bundle formula for Chow groups givesHere the first relation comes from the one-dimensional base, and the second is the projective-bundle relation with $c_1(E)=sL$ and higher Chern classes zero on the base. Integration over a fibre gives $L{M}^{n-1}=1$. Therefore all top-degree mixed intersection numbers areIn particular, for integers $u,v$,For the canonical divisor of a smooth variety, the relative Euler sequencehas determinant ${\mathcal{O}}_F(nM-sL)$. Combining its dual determinant with $K_{{\mathbb{P}}^1}=-2L$ gives

A common twist does not change the abstract projective bundle. Choose an integer $b$ such that $d_i=a_i+b\ge 1$ for every $i$. The complete linear system of divisors $|M+bL|$ hasIts map to ${\mathbb{P}}^N$, where $N=s+nb+n-1$, has coordinate blocksHere $[s_0:t_0]$ is a point of the base and the ${\lambda }_i$ are quotient coordinates in a local trivialization of $E(b)$. These expressions have the correct transition factors and define the global map. Any nonzero block recovers the base point through its degree-$d_i$ rational normal curve. Once that point is known, the blocks recover the projective fibre coordinates. On the open where the first coordinate of a nonzero block is nonzero, the ratio of the next coordinate to the first recovers $t_0/s_0$; the analogous last-coordinate chart recovers $s_0/t_0$. The remaining coordinate ratios then recover the fibre coordinates regularly. Thus the map identifies $F$ with its closed image and is a projective embedding, not merely a pointwise injection.

The hyperplane class on this image is $M+bL$, soConsequently every such natural embedding hasThe tautological choice $b=0$ is an embedding when all $a_i>0$. If the $a_i$ are only nonnegative, $|M|$ is basepoint-free but can contract the projective subbundle arising from zero summands; one must distinguish that scroll image from the smooth abstract scroll.

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

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

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

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

First use the canonical map to obtain the embedding. Its canonical divisor has degree six and $h^0(K_C)=4$. Over an algebraically closed ground field, every effective degree-two divisor on an algebraic curve $E$ satisfies $h^0(E)=1$: a second section would give a nonconstant rational function with poles bounded by $E$, hence a map $C\to {\mathbb{P}}^1$ of degree at most two. Degree one would make $C$ rational, and degree two would make it hyperelliptic, both excluded here.

The Riemann-Roch theorem now givesThus the complete linear system of divisors $|K_C|$ separates every pair of points and every tangent direction, including the tests $E=2p$. By the length-two criterion for a very ample linear system, it is a very ample linear system. Consequently the canonical map embeds $C$ in ${\mathbb{P}}^3$, with degree of a projective curve six. Its image is a nondegenerate projective variety, since the four canonical sections are linearly independent.

The space of homogeneous quadrics in four variables has dimension ten, while the Riemann-Roch theorem givesRestriction therefore has a nonzero kernel, giving a quadric $Q$ containing $C$. This quadric is irreducible: a reducible quadric is a union of two planes, or a double plane, and an integral curve lying in it would lie in a plane, contradicting nondegeneracy.

Similarly, homogeneous cubics form a twenty-dimensional space, whereasThere are at least five independent cubics vanishing on $C$. The multiples of $Q$ by linear forms give only four dimensions, so choose a cubic $F$ vanishing on $C$ which is not divisible by $Q$.

The polynomials $Q,F$ are a regular sequence, hence their intersection $D$ is a projective complete intersection of pure dimension one and degree $2\cdot 3=6$. It contains $C$, which already has degree six. The unmixedness of a complete intersection excludes embedded components. Equal degrees force $C$ to be the only irreducible component and force multiplicity one at its generic point. By generic reducedness with no embedded components, this makes $D$ reduced, so $D=C$ as schemes. We have provedThis is a canonical genus-four curve as a quadric-cubic intersection. The quadric may be smooth or a cone; the argument does not require it to be smooth. The same construction descends over the ground field when the curve is geometrically nonhyperelliptic, since the restriction maps and canonical embedding are defined there.

A line bundle $\mathcal{L}$ is a globally generated line bundle when the evaluation mapis surjective. Equivalently, its global sections span each fibre of the line bundle; locally at every point, some section is a generator.

Choose a finite generating family $s_0,\dots ,s_n$. On the open set where $s_i$ generates $\mathcal{L}$, the ratios $s_j/s_i$ are regular functions, giving a morphism of schemes into the standard chart $D_{+}(x_i)$ of projective space. The ratios obey the usual transition rules on overlaps, so these chart maps glue toThe tuple is computed using any local trivialization of $\mathcal{L}$; changing that trivialization multiplies all entries by the same invertible function. The construction has ${\phi }^{\ast }\mathcal{O}(1)\simeq \mathcal{L}$ and pulls back the coordinate sections to the chosen $s_i$.

There is a finiteness qualification for an arbitrary $X$: global generation alone need not provide such a finite family. It does if $X$ is quasi-compact, since the open sets on which individual sections generate have a finite subcover. Without that hypothesis, take $X={\coprod }_{m\ge 1}{\mathbb{P}}_k^m$ and let $\mathcal{L}$ restrict to $\mathcal{O}(1)$ on each component. This line bundle is globally generated, but any $N$ global sections have a common zero on a component ${\mathbb{P}}^m$ with $m\ge N$. Thus this is a globally generated line bundle without finite generators. This is finite global generation on a quasi-compact scheme. The finite-family construction is automatic in the projective case asked next.

For projective nonsingular $X$, put $V=⟨s_0,\dots ,s_n⟩$. The length-two criterion for a very ample linear system says that $\phi$ is a closed immersion precisely when, after extending to an algebraic closure, $V$ separates distinct points and tangent directions. Equivalently, the evaluationis surjective for every length-two geometric closed subscheme $E\subseteq X$. Two distinct points give point separation; a nonreduced length-two subscheme supported at one point gives separation of a direction in the Zariski tangent space. For the complete space $V=H^0(X,\mathcal{L})$, this says exactly that $\mathcal{L}$ is a very ample line bundle. For a chosen smaller $V$, it is the chosen linear system of divisors which must be a very ample linear system; mere global generation is insufficient.

The hyperplane sections through $P$ are spanned by the three linear forms $x_1,x_2,x_0-x_3$. Their common zero on $X$ is precisely $P$, so away from $P$ the linear system of divisors is base-point-free and defines the projection from a point on a smooth quadricIn particular, the chart-ratio construction in part (a) makes this a morphism of schemes on $X\setminus \{P\}$, not merely a rational map there.

Write a target point as $[u:v:w]$. The corresponding line through $P$ consists of pointsSubstituting in the quadric equation givesRemoving $P$ means $\mu \ne 0$, so set $\mu =1$. Over the residue field $K$ of the target point, its scheme-theoretic fibre is consequentlyIf $u\ne v$, this is one reduced point, with $\lambda =-uw/(u-v)$. If $u=v$ and $uw\ne 0$, it is empty: the corresponding line meets the quadric only at the removed point, with intersection multiplicity two. If $u=v$ and $uw=0$, the line lies entirely on the quadric, and the scheme-theoretic fibre is ${\mathbb{A}}_K^1$, the line with $P$ removed.

The last case occurs at exactly $[0:0:1]$ and $[1:1:0]$. Their lines are respectivelythe two rulings of a smooth quadric surface through $P$. Thus the image consists of the complement of the line $u=v$, together with those two points. The line $u=v$ parametrizes directions in the tangent plane $x_1=x_2$ at $P$. The calculation describes the scheme-theoretic fibres over arbitrary target points and works in every characteristic.

Here $\mathrm{deg}{K}_C=2$. For $D$, $D-P$ and $D-P-Q$, including $Q=P$, the degrees are $5,4,3$, all greater than two. Thus the correction term in the Riemann-Roch theorem vanishes andThe first drop supplies a section nonzero at every chosen point, proving base-point-freeness of a linear system. For distinct $P,Q$, the second drop supplies a section vanishing at one but not the other, proving point separation. For $Q=P$, it supplies a section vanishing to order exactly one, proving tangent separation. Hence the complete linear system of divisors is very ample and $\boxed{C\hookrightarrow {\mathbb{P}}^3}$ is the induced closed immersion.

For a smooth projective algebraic curve, a linear system of divisors comes from a finite-dimensional space $V\subseteq H^0(C,\mathcal{O}(D))$. A basis of sections $s_0,\dots ,s_n$ defines $f(P)=[s_0(P):\cdots :s_n(P)]$, using any local trivialization; a common scalar change does not alter the projective point. A base-point-free linear system makes this defined everywhere, and $f^{\ast }{\mathcal{O}}_{{\mathbb{P}}^n}(1)\cong \mathcal{O}(D)$. Pulling back a hyperplane gives the zero divisor of a linear combination of the sections, an element of the original system.

A system with base divisor $B$ can instead define a map after removing $B$, but its hyperplane pullbacks have class $D-B$. Thus base-point-freeness is needed for the stated original-system property. For a closed immersion, the system must also separate distinct points and tangent directions; equivalently it is a very ample linear system. On a smooth curve the latter tangent condition is separation of the length-two subscheme at each point.