Lemma 27.13.1. Let $S = \mathbf{Z}[T_0, \ldots , T_ n]$ with $\deg (T_ i) = 1$. The scheme

$\mathbf{P}^ n_{\mathbf{Z}} = \text{Proj}(S)$

represents the functor which associates to a scheme $Y$ the pairs $(\mathcal{L}, (s_0, \ldots , s_ n))$ where

1. $\mathcal{L}$ is an invertible $\mathcal{O}_ Y$-module, and

2. $s_0, \ldots , s_ n$ are global sections of $\mathcal{L}$ which generate $\mathcal{L}$

up to the following equivalence: $(\mathcal{L}, (s_0, \ldots , s_ n)) \sim (\mathcal{N}, (t_0, \ldots , t_ n))$ $\Leftrightarrow$ there exists an isomorphism $\beta : \mathcal{L} \to \mathcal{N}$ with $\beta (s_ i) = t_ i$ for $i = 0, \ldots , n$.

Proof. This is a special case of Lemma 27.12.3 above. Namely, for any graded ring $A$ we have

\begin{eqnarray*} \mathop{\mathrm{Mor}}\nolimits _{graded rings}(\mathbf{Z}[T_0, \ldots , T_ n], A) & = & A_1 \times \ldots \times A_1 \\ \psi & \mapsto & (\psi (T_0), \ldots , \psi (T_ n)) \end{eqnarray*}

and the degree $1$ part of $\Gamma _*(Y, \mathcal{L})$ is just $\Gamma (Y, \mathcal{L})$. $\square$

In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).