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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.