The Stacks project

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

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

Proof. This is true because $\mathbf{Z}[T_0, \ldots , T_ n]_{+} = (T_0, \ldots , T_ n)$ and since

in an obvious way. $\square$

Proof. All these properties are stable under base change (this is clear for the last two and for the other two see Schemes, Lemmas 26.21.12 and 26.19.3). Hence it suffices to prove them for the morphism $\mathbf{P}^ n_{\mathbf{Z}} \to \mathop{\mathrm{Spec}}(\mathbf{Z})$. Separatedness is Lemma 27.8.8. Quasi-compactness follows from Lemma 27.13.3. Existence and uniqueness of the valuative criterion follow from Lemma 27.8.11. Universally closed follows from the above and Schemes, Proposition 26.20.6. $\square$

Proof. It suffices to prove this when $S = \mathop{\mathrm{Spec}}(\mathbf{Z})$. Hence we will drop the index $S$ and work in the absolute setting. Write $\mathbf{P}^ n = \text{Proj}(\mathbf{Z}[X_0, \ldots , X_ n])$, $\mathbf{P}^ m = \text{Proj}(\mathbf{Z}[Y_0, \ldots , Y_ m])$, and $\mathbf{P}^{nm + n + m} = \text{Proj}(\mathbf{Z}[Z_0, \ldots , Z_{nm + n + m}])$. In order to map into $\mathbf{P}^{nm + n + m}$ we have to write down an invertible sheaf $\mathcal{L}$ on the left hand side and $(n + 1)(m + 1)$ sections $s_ i$ which generate it. See Lemma 27.13.1. The invertible sheaf we take is

The sections we take are

These generate $\mathcal{L}$ since the sections $X_ i$ generate $\mathcal{O}_{\mathbf{P}^ n}(1)$ and the sections $Y_ j$ generate $\mathcal{O}_{\mathbf{P}^ m}(1)$. The induced morphism $\varphi $ has the property that

Hence it is an affine morphism. The corresponding ring map in case $(i, j) = (0, 0)$ is the map

which maps $Z_ i/Z_0$ to the element $X_ i/X_0$ for $i \leq n$ and the element $Z_{(n + 1)j}/Z_0$ to the element $Y_ j/Y_0$. Hence it is surjective. A similar argument works for the other affine open subsets. Hence the morphism $\varphi $ is a closed immersion (see Schemes, Lemma 26.4.2 and Example 26.8.1.) $\square$

Proof. It is clear that $I$ is a graded ideal. Set $Z' = \text{Proj}(R[T_0, \ldots , T_ n]/I)$. By Lemma 27.11.5 we see that $Z'$ is a closed subscheme of $\mathbf{P}^ n_ R$. To see the equality $Z = Z'$ it suffices to check on an standard affine open $D_{+}(T_ i)$. By renumbering the homogeneous coordinates we may assume $i = 0$. Say $Z \cap D_{+}(T_0)$, resp. $Z' \cap D_{+}(T_0)$ is cut out by the ideal $J$, resp. $J'$ of $R[T_1/T_0, \ldots , T_ n/T_0]$. Then $J'$ is the ideal generated by the elements $F/T_0^{\deg (F)}$ where $F \in I$ is homogeneous. Suppose the degree of $F \in I$ is $d$. Since $F$ vanishes as a section of $\mathcal{O}_{\mathbf{P}^ n_ R}(d)$ restricted to $Z$ we see that $F/T_0^ d$ is an element of $J$. Thus $J' \subset J$.

Conversely, suppose that $f \in J$. If $f$ has total degree $d$ in $T_1/T_0, \ldots , T_ n/T_0$, then we can write $f = F/T_0^ d$ for some $F \in R[T_0, \ldots , T_ n]_ d$. Pick $i \in \{ 1, \ldots , n\} $. Then $Z \cap D_{+}(T_ i)$ is cut out by some ideal $J_ i \subset R[T_0/T_ i, \ldots , T_ n/T_ i]$. Moreover,

The left hand side is the localization of $J$ with respect to the element $T_ i/T_0$ and the right hand side is the localization of $J_ i$ with respect to the element $T_0/T_ i$. It follows that $T_0^{d_ i}F/T_ i^{d + d_ i}$ is an element of $J_ i$ for some $d_ i$ sufficiently large. This proves that $T_0^{\max (d_ i)}F$ is an element of $I$, because its restriction to each standard affine open $D_{+}(T_ i)$ vanishes on the closed subscheme $Z \cap D_{+}(T_ i)$. Hence $f \in J'$ and we conclude $J \subset J'$ as desired. $\square$

Proof. The multiplication maps

come from the natural isomorphisms

see Equation (27.10.1.4). Let us construct the map $c : \widetilde{M} \to \mathcal{F}$. On each of the standard affines $U_ i = D_{+}(T_ i)$ we see that $\Gamma (U_ i, \widetilde{M}) = (M[1/T_ i])_0$ where the subscript ${}_0$ means degree $0$ part. An element of this can be written as $m/T_ i^ d$ with $m \in M_ d$. Since $T_ i$ is a generator of $\mathcal{O}(1)$ over $U_ i$ we can always write $m|_{U_ i} = m_ i \otimes T_ i^ d$ where $m_ i \in \Gamma (U_ i, \mathcal{F})$ is a unique section. Thus a natural guess is $c(m/T_ i^ d) = m_ i$. A small argument, which is omitted here, shows that this gives a well defined map $c : \widetilde{M} \to \mathcal{F}$ if we can show that

in $M[1/T_ iT_ j]$. But this is clear since on the overlap the generators $T_ i$ and $T_ j$ of $\mathcal{O}(1)$ differ by the invertible function $T_ i/T_ j$.

Injectivity of $c$. We may check for injectivity over the affine opens $U_ i$. Let $i \in \{ 0, \ldots , n\} $ and let $s$ be an element $s = m/T_ i^ d \in \Gamma (U_ i, \widetilde{M})$ such that $c(m/T_ i^ d) = 0$. By the description of $c$ above this means that $m_ i = 0$, hence $m|_{U_ i} = 0$. Hence $T_ i^ em = 0$ in $M$ for some $e$. Hence $s = m/T_ i^ d = T_ i^ e/T_ i^{e + d} = 0$ as desired.

Surjectivity of $c$. We may check for surjectivity over the affine opens $U_ i$. By renumbering it suffices to check it over $U_0$. Let $s \in \mathcal{F}(U_0)$. Let us write $\mathcal{F}|_{U_ i} = \widetilde{N_ i}$ for some $R[T_0/T_ i, \ldots , T_0/T_ i]$-module $N_ i$, which is possible because $\mathcal{F}$ is quasi-coherent. So $s$ corresponds to an element $x \in N_0$. Then we have that

(where the subscripts mean “principal localization at”) as modules over the ring

This means that for some large integer $d$ there exist elements $s_ i \in N_ i$, $i = 1, \ldots , n$ such that

on $U_0 \cap U_ i$. Next, we look at the difference

on $U_ i \cap U_ j$, $0 < i < j$. By our choice of $s_ i$ we know that $t_{ij}|_{U_0 \cap U_ i \cap U_ j} = 0$. Hence there exists a large integer $e$ such that $(T_0/T_ i)^ et_{ij} = 0$. Set $s_ i' = (T_0/T_ i)^ es_ i$, and $s_0' = s$. Then we will have

on $U_ a \cap U_ b$ for all $a, b$. This is exactly the condition that the elements $s'_ a$ glue to a global section $m \in \Gamma (\mathbf{P}^ n_ R, \mathcal{F}(e + d))$. And moreover $c(m/T_0^{e + d}) = s$ by construction. Hence $c$ is surjective and we win. $\square$

Proof. In fact the result is true if $X$ is any locally ringed space. The sheaf $\mathcal{F}$ is a finite locally free $\mathcal{O}_ X$-module of rank $n$. The elements

with $s_ j$ in the $i$th spot and $-s_ i$ in the $j$th spot map to zero in $\mathcal{L}^{\otimes 2}$. Hence $s_{ij} \in \Gamma (X, \mathcal{F} \otimes _{\mathcal{O}_ X} \mathcal{L})$. A local computation shows that these sections generate $\mathcal{F} \otimes \mathcal{L}$.

Alternative proof. Consider the morphism $\varphi : X \to \mathbf{P}^ n_\mathbf {Z}$ associated to the pair $(\mathcal{L}, (s_0, \ldots , s_ n))$. Since the pullback of $\mathcal{O}(1)$ is $\mathcal{L}$ and since the pullback of $T_ i$ is $s_ i$, it suffices to prove the lemma in the case of $\mathbf{P}^ n_\mathbf {Z}$. In this case the sheaf $\mathcal{F}$ corresponds to the graded $S = \mathbf{Z}[T_0, \ldots , T_ n]$ module $M$ which fits into the short exact sequence

where the second map is given by $T_0, \ldots , T_ n$. In this case the statement above translates into the statement that the elements

generate the graded module $M(1)$ over $S$. We omit the details. $\square$