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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.

Work over $\mathbb{C}$, and let $C$ be a smooth projective curve of genus $g\ge 3$ and gonality three. A trigonal curve has a basepoint-free degree-three pencil, giving a finite morphism $f:C\to {\mathbb{P}}^1$ of degree three. It is not hyperelliptic, so its canonical map is an embedding. Let $D_t=f^{-1}(t)$, including its scheme multiplicities. The complete pencil has $h^0(D_t)=2$, and Riemann-Roch theorem givesThus the quotient of canonical sections evaluated on $D_t$ has dimension two, and $D_t$ spans a line in ${\mathbb{P}}^{g-1}$. This argument includes ramified fibres, rather than treating only three distinct points.

These two-dimensional quotients vary in a globally generated rank-two bundle $E$ on ${\mathbb{P}}^1$. More explicitly, the image ofhas constant rank two by the preceding dimension count, so it is a subbundle. The equality of the rank at every fibre justifies the bundle construction. On global sections it contains every canonical section, and therefore $H^0({\mathbb{P}}^1,E)=H^0(C,K_C)$. The Birkhoff–Grothendieck theorem givesGlobal generation forces nonnegative splitting degrees. Since $h^0(E)=a_1+a_2+2=g$,The natural map from $f^{\ast }E$ to $K_C$ is a surjection at every point, since the canonical system has no base points. By projectivization by quotients it gives a map $C\to F=\mathbb{P}(E)$ over ${\mathbb{P}}^1$, with $M{|}_C=K_C$. Its composition with the tautological map of $F$ is the canonical embedding of $C$, so $C\to F$ is itself a closed embedding. Each fibre of the ruling maps to the spanning line of $D_t$; their union is the canonical scroll of a trigonal curve, and the ruling restricts to the given $g_3^1$.

Put $a=a_2-a_1$. Tensoring $E$ by $\mathcal{O}(-a_2)$ identifies the abstract ruled surface with ${\mathbb{F}}_a=\mathbb{P}(\mathcal{O}(-a)\oplus \mathcal{O})$. The section coming from the quotient onto the smaller summand has self-intersection $-a$ and will be denoted by $B$. In this normalized presentation its tautological class is $B$; restoring the twist givesThis class gives the scroll map. It is an embedding if $a_1>0$. When $a_1=0$, it contracts $B$; thus the printed embedding description needs this positivity qualification. For example, in genus three the ruled model is ${\mathbb{F}}_1$ and its image is ${\mathbb{P}}^2$; in genus four the type $(0,2)$ has a quadric-cone image. In both cases the smooth curve and its canonical image are still described by the construction above.

As a divisor on $F$, the curve meets each fibre in degree three, so $C=3M+tL$ for an integer $t$. The scroll formula givesBy adjunction formula and $M{|}_C=K_C$,Taking degree on $C$ uses $\mathrm{deg}(L{|}_C)=3$ and gives $t=4-g$. ConsequentlyWriting $d=a+a_2+2$, this is

A divisor $rB+dL$ with $r\ge 0$ on ${\mathbb{F}}_a$ is basepoint-free when $d\ge ra$. Indeed, its pushforward along the ruling splits intoif each degree is nonnegative, sections generate these coefficient functions, and the fibre monomials generate at every point. In particular $3B+dL$ is basepoint-free when $d\ge 3a$. The Bertini smoothness theorem then makes its general member nonsingular, including the equality case, where the general member avoids $B$.

Conversely, if $d<3a$, then $(3B+dL)\cdot B=d-3a<0$, so every effective member contains $B$. Because the canonical scroll has $a_1\ge 0$,The residual system $|2B+dL|$ is basepoint-free. Its general member $R$ does not contain $B$ and intersects it inThus a general divisor in the original system is $B+R$ with intersecting components and is singular at an intersection point. The fixed component occurs exactly once for a general member, since $R$ is not forced to contain it. This proves both directions, with the canonical-scroll hypotheses retained:This is also the bound on the Maroni invariant. Merely observing a fixed component would not by itself prove singularity; the positive intersection with its residual divisor is essential.

The projective bundle of a real or complex vector bundle $E\to X$ has fiber the space of one-dimensional subspaces of $E_x$. It carries a tautological bundle $\lambda \subset p^{\ast }E$; in the real case this is the projectivization of a real vector bundle.

The same construction applies to an algebraic vector bundle over an algebraic variety: local trivializations glue products with projective space. In the lines convention its universal tautological bundle is a line bundle; the alternative projectivization by quotients convention parametrizes one-dimensional quotient spaces. Both conventions give dimension $\dim X+\mathrm{rank}(E)-1$ and a proper morphism to the base.

For this smooth projective bundle, put $L={\pi }^{\ast }{\mathcal{O}}_{{\mathbb{P}}^1}(1)$ and $M={\mathcal{O}}_{\mathbb{F}}(1)$, using projectivization by quotients. If every $a_i>0$, the complete system $|M|$ embeds it as a rational normal scroll. If every $a_i\ge 0$, the same system defines a scroll image which may be singular. Allowing arbitrary integers gives an abstract scroll; a common twist makes an embedding available. Its intersection numbers are $L^2=0$, $L{M}^{n-1}=1$, $M^n={\sum }_i{a}_i$, and its canonical divisor is $-nM+({\sum }_i{a}_i-2)L$.

On the projectivization by quotients, the fibre over a one-dimensional quotient of $E_x$ is that quotient line itself. These lines glue to the universal quotient. Its restriction to a projective fibre has degree one. It is the dual of the relative tautological line after identifying quotient projectivization of $E$ with line projectivization of $E^{\ast }$.