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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.