= Solution
Use the quotient convention for <projective space>. To a <scheme> $T$, associate the set of pairs
$$
(q,\mathcal L),\qquad q:\mathcal O_T^{\oplus(n+1)}\twoheadrightarrow\mathcal L,
$$
where $\mathcal L$ is a <line bundle>, modulo <isomorphisms> $\mathcal L\to\mathcal L'$ carrying $q$ to $q'$. Pullback defines the contravariant <functor>. On $T=\operatorname{Spec}A$, 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 $\mathbb P^n_{\mathbb Z}$. 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 write
$$
t_j=\frac{q(e_j)}{q(e_i)},\qquad j\ne i.
$$
These functions specify a map to the standard chart $U_i=\operatorname{Spec}\mathbb Z[t_j:j\ne i]$. 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 as
$$
0\longrightarrow\mathcal K\longrightarrow V\otimes\mathcal O\xrightarrow q\mathcal O(1)\longrightarrow0,\qquad V=\mathbb Z^{n+1}.
$$
A 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 $1+\varepsilon h$ 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> is
$$
\mathcal T_{\mathbb P^n/\mathbb Z}\cong\mathcal H om(\mathcal K,\mathcal O(1)).
$$
This 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 sequence>
$$
\boxed{0\longrightarrow\mathcal O\xrightarrow{(X_0,\ldots,X_n)}\mathcal O(1)^{\oplus(n+1)}\longrightarrow\mathcal T_{\mathbb P^n/\mathbb Z}\longrightarrow0.}
$$
It remains exact after any base change. Taking its <dual of a sheaf> over a <field> $k$ gives
$$
0\longrightarrow\Omega^1_{\mathbb P^n_k/k}\longrightarrow\mathcal O(-1)^{\oplus(n+1)}\longrightarrow\mathcal O\longrightarrow0.
$$
For $n=1$, the first term has rank one; taking determinants yields
$$
\boxed{\Omega^1_{\mathbb P^1_k/k}\cong\mathcal O(-2).}
$$
Explicitly, 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.
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