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Two rank-$n$ abstract rational normal scrolls are isomorphic exactly when their splitting degrees agree as multisets after a common integer shift. For $n\ge 3$, the primitive nef class $L$ of the ruling is intrinsically identified by $L^2=0$: restriction to a projective fibre forces every other divisor class with square zero to have zero coefficient of $M$. An isomorphism therefore preserves the ruling, and its effect on $M$ is a twist by $cL$. Pushforward gives an isomorphism of the defining bundles up to this twist; uniqueness in the Birkhoff–Grothendieck theorem recovers their splitting degrees. For surfaces the unique nef square-zero ray identifies the ruling except on ${\mathbb{P}}^1\times {\mathbb{P}}^1$, where exchanging the two rulings changes no normalized splitting type.

We prove the required duality pairing explicitly. This also identifies the word “canonical” in the final conclusion: the isomorphism will be the intrinsic residue trace on a smooth projective curve, not a choice of basis for a one-dimensional vector space.

Begin on the projective line. With $s=t^{-1}$, the equation $ds=-t^{-2}dt$ identifies its sheaf of Kähler differentials over a field with $\mathcal{O}(-2)$. The Laurent description in Question 3 pairs $H^1(\mathcal{O}(a))$ with $H^0(\mathcal{O}(-a-2))$ by multiplication and extraction of the coefficient of $t^{-1}dt$. For $a\le -2$ the respective bases areand their pairing matrix has entriesThis is an invertible antidiagonal matrix. For $a\ge -1$ both spaces vanish. The coefficient extraction is well defined on the Čech cohomology quotient: a differential regular on the first chart has no $t^{-1}dt$ term, and one regular on the second chart is a sum of terms $t^{-2-j}dt$ with $j\ge 0$, also with no such term.

To extend this calculation to every vector bundle $E$ on the projective line, we give the splitting argument. Every line bundle on ${\mathbb{P}}^1$ is $\mathcal{O}(a)$: choose a nonzero rational section to obtain a divisor, and use the divisors of $t-c$ to move each finite point to infinity. The denominator-clearing argument in Question 4 makes both $E(n)$ and $E^{\vee }(n)$ globally generated for large $n$. The first assertion ensures $H^0(E(-a))\ne 0$ for some integer $a$; the second gives an embedding $E\hookrightarrow \mathcal{O}(n{)}^{\oplus r}$, so this is impossible for $a>n$. Choose the largest such integer $a$.

A nonzero section gives $\mathcal{O}(a)\to E$. Saturating its image gives a line subbundle; if it were strictly larger, it would be $\mathcal{O}(a+d)$ for a positive integer $d$, recording the section's zeros, contrary to maximality. Thus the quotient $F$ is a vector bundle. Induct on rank and write $F={⨁}_j\mathcal{O}(b_j)$. Twist the sequence by $-a-1$. Since $H^0(E(-a-1))=0$ and $H^1(\mathcal{O}(-1))=0$, exactness forces $H^0(F(-a-1))=0$, whence every $b_j\le a$. Local lifts of a quotient summand differ on overlaps by a Čech cocycle in $\mathcal{O}(a-b_j)$. But $H^1(\mathcal{O}(a-b_j))=0$ by Question 3. Adjusting the lifts by a coboundary makes them glue, splitting the sequence. This proves the Birkhoff–Grothendieck theorem here. The direct-sum Laurent calculation consequently proves the perfect pairingAlthough a splitting proves nondegeneracy, the pairing itself uses only evaluation followed by the coefficient trace on $H^1({\Omega }_{{\mathbb{P}}^1}^1)$, so does not depend on that splitting. This is residue duality on the projective line.

Now choose a rational function $t$ on $C$ with $dt\ne 0$. Such a choice is available over the perfect field $k$: a local parameter at a smooth point has nonzero differential in the rank-one sheaf of Kähler differentials over a field. Since $dt$ spans the differentials of $k(C)/k$, the extension $k(C)/k(t)$ has zero relative differentials and is separable. Question 1 gives a finite generically separable morphism $\pi :C\to {\mathbb{P}}^1$. It is flat: locally its finite module over a discrete valuation ring is torsion-free, hence free. For the invertible sheaf $\mathcal{M}$, its pushforward $E={\pi }_{\ast }\mathcal{M}$ is therefore a vector bundle.

We next prove the differential-valued dual identification needed to transfer the pairing. At a point of the base, complete the local rings. Each branch is of the formIf $e$ is the order of $f$, formal division writes $B=A[u]$ with basis $1,u,\dots ,u^{e-1}$ and a monic degree-$e$ equation $P(u)=0$. In particular this argument does not assume tame ramification or the special equation $s=u^e$. For a polynomial $g$ the elementary interpolation identity isTo see it, pass to a field containing the distinct roots $u_i$ of $P$. Lagrange interpolation writes the remainder as ${\sum }_{i}g(u_i)P(T)/((T-u_i)P^{\prime }(u_i))$. Comparison of its leading coefficient gives the identity, and descent returns it to the original field. The top-coefficient pairing on $A[T]/(P)$ is perfect: in the power basis its matrix has zero entries for $i+j<e-1$ and ones for $i+j=e-1$, so its determinant is a unit. It follows that the trace-dual fractional module is exactly $P^{\prime }(u{)}^{-1}B$.

Formal division of $f(T)-s$ gives $f(T)-s=W(T,s)P(T,s)$ with $W$ a unit. Differentiate in $T$ and evaluate at $u$. This shows that $f^{\prime }(u)$ and $P^{\prime }(u)$ differ by a unit. Since $ds=f^{\prime }(u)du$, we obtainThis is the trace-dual module of a finite curve map identity. Its map sends a differential $\eta$ to the functional $b\mapsto \mathrm{Tr}(b\eta )$, where the trace of $hds$ is $\mathrm{Tr}(h)ds$. For several branches take the direct sum of these maps. Completion is faithfully flat, so their being isomorphisms proves the corresponding isomorphism of the original local modules. A local trivialization of $\mathcal{M}$ now gives, compatibly on overlaps,

The inverse images of the two standard affine charts and their intersection are affine. As in Question 4, the Čech cochain complexes identify $H^1(C,\mathcal{M})$ with $H^1({\mathbb{P}}^1,E)$. Global sections in the last display identify the other factor with $H^0(C,{\mathcal{M}}^{\vee }\otimes {\Omega }_C^1)$. Thus the perfect pairing already calculated on the projective line givesIn particular the stated vanishing of global sections forces $H^1(C,\mathcal{M})=0$. The proof has established the relevant case of Serre duality, rather than invoked it as the requested argument.

To identify its intrinsic trace, use the rational principal-parts resolution on an algebraic curve. The constant sheaf of rational sections and the sheaf of their finite-support principal parts are both flasque sheaves. Consequently a class of $H^1(C,\mathcal{M})$ is represented by finitely many rational principal parts $(m_P)$, modulo the family coming from one global rational section. For a global section $\eta$ of ${\mathcal{M}}^{\vee }\otimes {\Omega }_C^1$, evaluation gives the pairingThe algebraic residue of a rational differential is its $u^{-1}du$ coefficient, independent of the local parameter. Adding a regular representative changes no residue. The local trace calculation above also givesOne can check this directly in the monogenic description: expand a differential in the basis $1,u,\dots ,u^{e-1}$, use the displayed interpolation formula for its trace, and extract the coefficient of $s^{-1}ds$. Formal division of $f(u)-s$ makes that coefficient exactly the coefficient of $u^{-1}du$; for a product of branch fields the trace is the sum. Thus the identity remains valid in positive characteristic, including wild ramification. On ${\mathbb{P}}^1$, partial fractions show that a rational differential's residues at finite points sum to the negative of its residue at infinity. Tracing to ${\mathbb{P}}^1$ therefore proves that the sum of residues of every global rational differential on $C$ is zero. Adding a global rational section to $(m_P)$ consequently changes the pairing by zero. This defines the same pairing as the finite-pushforward calculation, intrinsically, without a chosen map to ${\mathbb{P}}^1$.

Finally take $\mathcal{M}={\Omega }_C^1$. Its differential-valued dual is ${\mathcal{O}}_C$, and Question 4 proved $H^0(C,{\mathcal{O}}_C)=k$. Perfectness identifies $H^1(C,{\Omega }_C^1)$ with the dual of this space. Evaluation at the distinguished constant section $1$ givesThis residue trace on a smooth projective curve uses neither a basis choice nor an auxiliary finite map. It is the canonical isomorphism.

Throughout, the braces denote a multiset, so repeated splitting degrees retain their multiplicities. If the multisets agree after a common shift, permuting summands gives $E_a\cong E_b\otimes \mathcal{O}(c)$. Tensoring all the fibres of a vector bundle by the same line bundle leaves their one-dimensional quotient spaces unchanged. HenceThis proves one implication in the isomorphism classification of abstract rational scrolls.

For the converse first take $n\ge 3$. Every divisor class on $F_a$ has the form $xL+yM$. If its square is zero, restriction to a fibre ${\mathbb{P}}^{n-1}$ giveswhere $h$ is the hyperplane class and $h^2\ne 0$ because $n-1\ge 2$. Thus $y=0$. It follows that $L$ is intrinsically characterized as the primitive nonzero nef class whose square is zero. An isomorphism $u:F_a\to F_b$ therefore carries $L_b$ to $L_a$; the sign is positive because both classes are nef and effective. The complete pencil $|L|$ determines the ruling, sofor an automorphism $\sigma$ of ${\mathbb{P}}^1$.

On a fibre, the isomorphism preserves the positive generator ${\mathcal{O}}_{{\mathbb{P}}^{n-1}}(1)$ of its Picard group. Consequentlyfor some integer $c$. Push forward along the ruling, using ${\pi }_{\ast }\mathcal{O}(M)=E$ and the projection formula for sheaves:Every automorphism of ${\mathbb{P}}^1$ preserves each $\mathcal{O}(d)$. The uniqueness of the splitting degrees in the Birkhoff–Grothendieck theorem now implies equality of the two multisets up to a shift. Replacing this $c$ by $-c$ gives the sign convention in the final formula. One can see that uniqueness numerically: for a split bundle,The first difference counts the degrees with $a_i+t\ge 0$, and the second difference counts the multiplicity of $a_i=-t$. Hence these dimensions determine every splitting degree, with multiplicity.

For $n=2$, subtract the smaller degree and write the normalized bundle as $\mathcal{O}\oplus \mathcal{O}(a)$, $a\ge 0$. Its surface is the Hirzebruch surface ${\mathbb{F}}_a$. Let $B$ be its negative section and $L$ its fibre, so $B^2=-a$, $BL=1$, $L^2=0$. The nef cone is generated by $L$ and $B+aL$: a nef divisor $xL+yB$ must have $y\ge 0$ from intersection with a fibre and $x-ay\ge 0$ from intersection with $B$, and these two inequalities also express it as a nonnegative combination of those generators. If $a>0$, a class $uL+v(B+aL)$ with $u,v\ge 0$ has squarewhich vanishes only when $v=0$. Thus the ruling is again intrinsic, and the same bundle argument applies. If $a=0$, the surface is ${\mathbb{P}}^1\times {\mathbb{P}}^1$ and has two nef square-zero rays, its two rulings. This distinguishes it from every $a>0$ surface. Isomorphisms may exchange its rulings, but both normalized splitting degrees are zero, so the stated classification is unchanged.

For $n=1$, every scroll is ${\mathbb{P}}^1$, and any two single splitting degrees differ by a common shift. Thus all ranks are covered:Reading the braces as ordinary sets would lose information and make the assertion false when degrees repeat. For example, $(0,0,1,2)$ and $(0,1,1,2)$ have the same underlying set but not the same splitting multiset; their minima are both zero, so no nonzero common shift can reconcile them.

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.

We prove the Birkhoff–Grothendieck theorem by induction on the rank of the vector bundle $E$. Rank zero is immediate; rank one is the classification $\mathrm{Pic}({\mathbb{P}}_k^1)\cong \mathbb{Z}$ established above.

First there is a largest integer $a$ with $H^0(E(-a))\ne 0$. To see existence and boundedness directly, trivialize $E$ on the two standard affine charts: the associated finite projective modules over the principal ideal domains $k[t]$ and $k[t^{-1}]$ are free. Let $A(t)\in G{L}_r(k[t,t^{-1}])$ be its transition matrix, so a section of $E(-a)$ is a pair satisfyingIf $a$ exceeds the largest exponent in any entry of $A$, the right side has only strictly negative powers; the left side has nonnegative powers. Hence both vanish. For $a$ sufficiently negative, take a nonzero constant vector $p$ and put $q=t^{a}A(t{)}^{-1}p$; all its exponents are nonpositive, so it supplies a nonzero section. Thus a largest $a$ exists.

Such a section gives a nonzero map $\mathcal{O}(a)\to E$. It is injective, since it is nonzero generically and the source is torsion-free. Saturate its image. The resulting line subbundle is $\mathcal{O}(a+d)$ for the degree $d\ge 0$ of the effective divisor of zeros of the original map. If $d>0$, its inclusion in $E$ would give a nonzero section of $E(-a-d)$, contradicting maximality. Therefore $d=0$ andhas a locally free quotient. By induction, $Q\cong {⨁}_j\mathcal{O}(b_j)$.

Twist this exact sequence by $\mathcal{O}(-a-1)$. The outer groups $H^0(\mathcal{O}(-1))$ and $H^1(\mathcal{O}(-1))$ are zero by the Čech cohomology of twists on the projective line. The long exact sequence in sheaf cohomology therefore givesThe first is zero by the choice of $a$, so each $H^0(\mathcal{O}(b_j-a-1))$ vanishes. Thus $b_j\le a$ for every $j$. The extension class lies insince $a-b_j\ge 0$. The identification of this Ext functor with cohomology can be seen without additional duality: local splittings of a vector-bundle extension differ by a Čech cocycle in $\mathcal{H}om(Q,\mathcal{O}(a))$, and a coboundary changes them to compatible splittings. Vanishing therefore gives a global splitting. Consequentlycompleting the induction over any field $k$.

The degrees can be arranged in nonincreasing order and their multiset is unique. Indeed, for any such decomposition,so the dimensions of the twisted global sections recover the number of summands of every degree.