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Use the quotient convention for projective space. To a scheme $T$, associate the set of pairswhere $\mathcal{L}$ is a line bundle, modulo isomorphisms $\mathcal{L}\to {\mathcal{L}}^{\prime }$ carrying $q$ to $q^{\prime }$. Pullback defines the contravariant functor. On $T=\mathrm{Spec}A$, this means rank-one projective module quotients of $A^{n+1}$; the quotient need not be free. Thus merely listing generating tuples modulo multiplication by a unit would omit points over rings with nontrivial Picard group.

This functor is represented by ${\mathbb{P}}_{\mathbb{Z}}^n$. For each $i$, the locus where $q(e_i)$ generates $\mathcal{L}$ is open. These loci cover $T$, since a surjection onto a rank-one free stalk must have some coefficient a unit in its local ring. On this locus, trivialize by $q(e_i)$ and writeThese functions specify a map to the standard chart $U_i=\mathrm{Spec}\mathbb{Z}[t_j:j\ne i]$. Their transition rules are exactly those of projective space, so the chart maps glue. Conversely the universal quotient on projective space pulls back to the given pair, and the construction is natural and unique. This proves that projective space represents invertible quotients.

Write its universal exact sequence asA first-order change of the quotient, with target trivialized locally, is a map $\delta q:V\otimes \mathcal{O}\to \mathcal{O}(1)$. Changing the target trivialization by $1+\epsilon h$ changes $\delta q$ by $hq$. Restriction to $\mathcal{K}$ removes this ambiguity, and every local map $\mathcal{K}\to \mathcal{O}(1)$ extends because the displayed sequence splits locally. Consequently the tangent sheaf of a scheme isThis identification can also be seen on $U_i$: the first-order change of $t_j$ is $\delta q(e_j)-t_j\delta q(e_i)$ after setting $q(e_i)=1$. Thus these maps are precisely the coordinate tangent directions.

Apply $\mathcal{H}om(-,\mathcal{O}(1))$ to the universal sequence. Since its terms are locally free and it splits locally, this gives the Euler sequenceIt remains exact after any base change. Taking its dual of a sheaf over a field $k$ givesFor $n=1$, the first term has rank one; taking determinants yieldsExplicitly, for $u=t^{-1}$ on the overlap of the two affine charts, $du=-t^{-2}dt$, the same transition as $\mathcal{O}(-2)$ after changing one local frame's sign.