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. $\square$


Comments (1)

Comment #9873 by on

Lean says no. On further inspection, it looks like a missing d (there's a gamma_i on the last but one line of the displayed calculation, and the definition of gamma_i involves d, but there's no d on the line before). It can be fixed simply by multiplying the first three terms in the displayed calculation by d.

Independent of this, we didn't need to do the case splitting on whether things were <= r or > r in Lean because valuations coming from valuation rings are defined on all of K in Lean and take value +infinity on 0; introducing this convention means that you don't need to do the case split just before the calculation because v(0) makes sense. But probably you're not bothered by this. In fact the truth is that in Lean valuations are multiplicative not additive (i.e. they're norms), and we use a (totally ordered) "group with 0" for the target of the valuation (a concept apparently invented by Artin: it's a monoid which is the disjoint union of a multiplicative group with {0}; fields are good examples), with v(0)=0. Not that I'm suggesting you change it, everything else in the proof seems to be fine.

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