The Stacks project

Lemma 27.8.11. Let $S$ be a graded ring. If $S$ is finitely generated as an algebra over $S_0$, then the morphism $\text{Proj}(S) \to \mathop{\mathrm{Spec}}(S_0)$ satisfies the existence and uniqueness parts of the valuative criterion, see Schemes, Definition 26.20.3.

Proof. The uniqueness part follows from the fact that $\text{Proj}(S)$ is separated (Lemma 27.8.8 and Schemes, Lemma 26.22.1). Choose $x_ i \in S_{+}$ homogeneous, $i = 1, \ldots , n$ which generate $S$ over $S_0$. Let $d_ i = \deg (x_ i)$ and set $d = \text{lcm}\{ d_ i\} $. Suppose we are given a diagram

\[ \xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[r] \ar[d] & \text{Proj}(S) \ar[d] \\ \mathop{\mathrm{Spec}}(A) \ar[r] & \mathop{\mathrm{Spec}}(S_0) } \]

as in Schemes, Definition 26.20.3. Denote $v : K^* \to \Gamma $ the valuation of $A$, see Algebra, Definition 10.50.13. We may choose an $f \in S_{+}$ homogeneous such that $\mathop{\mathrm{Spec}}(K)$ maps into $D_{+}(f)$. Then we get a commutative diagram of ring maps

\[ \xymatrix{ K & S_{(f)} \ar[l]^{\varphi } \\ A \ar[u] & S_0 \ar[l] \ar[u] } \]

After renumbering we may assume that $\varphi (x_ i^{\deg (f)}/f^{d_ i})$ is nonzero for $i = 1, \ldots , r$ and zero for $i = r + 1, \ldots , n$. Since the open sets $D_{+}(x_ i)$ cover $\text{Proj}(S)$ we see that $r \geq 1$. Let $i_0 \in \{ 1, \ldots , r\} $ be an index minimizing $\gamma _ i = (d/d_ i)v(\varphi (x_ i^{\deg (f)}/f^{d_ i}))$ in $\Gamma $. For convenience set $x_0 = x_{i_0}$ and $d_0 = d_{i_0}$. The ring map $\varphi $ factors though a map $\varphi ' : S_{(fx_0)} \to K$ which gives a ring map $S_{(x_0)} \to S_{(fx_0)} \to K$. The algebra $S_{(x_0)}$ is generated over $S_0$ by the elements $x_1^{e_1} \ldots x_ n^{e_ n}/x_0^{e_0}$, where $\sum e_ i d_ i = e_0 d_0$. If $e_ i > 0$ for some $i > r$, then $\varphi '(x_1^{e_1} \ldots x_ n^{e_ n}/x_0^{e_0}) = 0$. If $e_ i = 0$ for $i > r$, then we have

\begin{align*} \deg (f) v(\varphi '(x_1^{e_1} \ldots x_ r^{e_ r}/x_0^{e_0})) & = v(\varphi '(x_1^{e_1 \deg (f)} \ldots x_ r^{e_ r \deg (f)}/x_0^{e_0 \deg (f)})) \\ & = \sum e_ i v(\varphi '(x_ i^{\deg (f)}/f^{d_ i})) - e_0 v(\varphi '(x_0^{\deg (f)}/f^{d_0})) \\ & = \sum e_ i d_ i \gamma _ i - e_0 d_0 \gamma _0 \\ & \geq \sum e_ i d_ i \gamma _0 - e_0 d_0 \gamma _0 = 0 \end{align*}

because $\gamma _0$ is minimal among the $\gamma _ i$. This implies that $S_{(x_0)}$ maps into $A$ via $\varphi '$. The corresponding morphism of schemes $\mathop{\mathrm{Spec}}(A) \to \mathop{\mathrm{Spec}}(S_{(x_0)}) = D_{+}(x_0) \subset \text{Proj}(S)$ provides the morphism fitting into the first commutative diagram of this proof. $\square$

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