Lemma 27.8.1 (01M4)—The Stacks project

Lemma 27.8.1. Let $S$ be a graded ring. Let $f \in S$ homogeneous of positive degree.

If $g\in S$ homogeneous of positive degree and $D_{+}(g) \subset D_{+}(f)$ , then
1. $f$ is invertible in $S_ g$ , and $f^{\deg (g)}/g^{\deg (f)}$ is invertible in $S_{(g)}$ ,
2. $g^ e = af$ for some $e \geq 1$ and $a \in S$ homogeneous,
3. there is a canonical $S$ -algebra map $S_ f \to S_ g$ ,
4. there is a canonical $S_0$ -algebra map $S_{(f)} \to S_{(g)}$ compatible with the map $S_ f \to S_ g$ ,
5. the map $S_{(f)} \to S_{(g)}$ induces an isomorphism
  
  $(S_{(f)})_{g^{\deg (f)}/f^{\deg (g)}} \cong S_{(g)},$
6. these maps induce a commutative diagram of topological spaces
  
  $\xymatrix{ D_{+}(g) \ar[d] & \{ \mathbf{Z}\text{-graded primes of }S_ g\} \ar[l] \ar[r] \ar[d] & \mathop{\mathrm{Spec}}(S_{(g)}) \ar[d] \\ D_{+}(f) & \{ \mathbf{Z}\text{-graded primes of }S_ f\} \ar[l] \ar[r] & \mathop{\mathrm{Spec}}(S_{(f)}) }$
  
  where the horizontal maps are homeomorphisms and the vertical maps are open immersions,
7. there are compatible canonical $S_ f$ and $S_{(f)}$ -module maps $M_ f \to M_ g$ and $M_{(f)} \to M_{(g)}$ for any graded $S$ -module $M$ , and
8. the map $M_{(f)} \to M_{(g)}$ induces an isomorphism
  
  $(M_{(f)})_{g^{\deg (f)}/f^{\deg (g)}} \cong M_{(g)}.$
Any open covering of $D_{+}(f)$ can be refined to a finite open covering of the form $D_{+}(f) = \bigcup _{i = 1}^ n D_{+}(g_ i)$ .
Let $g_1, \ldots , g_ n \in S$ be homogeneous of positive degree. Then $D_{+}(f) \subset \bigcup D_{+}(g_ i)$ if and only if $g_1^{\deg (f)}/f^{\deg (g_1)}, \ldots , g_ n^{\deg (f)}/f^{\deg (g_ n)}$ generate the unit ideal in $S_{(f)}$ .

Proof. Recall that $D_{+}(g) = \mathop{\mathrm{Spec}}(S_{(g)})$ with identification given by the ring maps $S \to S_ g \leftarrow S_{(g)}$ , see Algebra, Lemma 10.57.3. Thus $f^{\deg (g)}/g^{\deg (f)}$ is an element of $S_{(g)}$ which is not contained in any prime ideal, and hence invertible, see Algebra, Lemma 10.17.2. We conclude that (a) holds. Write the inverse of $f$ in $S_ g$ as $a/g^ d$ . We may replace $a$ by its homogeneous part of degree $d\deg (g) - \deg (f)$ . This means $g^ d - af$ is annihilated by a power of $g$ , whence $g^ e = af$ for some $a \in S$ homogeneous of degree $e\deg (g) - \deg (f)$ . This proves (b). For (c), the map $S_ f \to S_ g$ exists by (a) from the universal property of localization, or we can define it by mapping $b/f^ n$ to $a^ nb/g^{ne}$ . This clearly induces a map of the subrings $S_{(f)} \to S_{(g)}$ of degree zero elements as well. We can similarly define $M_ f \to M_ g$ and $M_{(f)} \to M_{(g)}$ by mapping $x/f^ n$ to $a^ nx/g^{ne}$ . The statements writing $S_{(g)}$ resp. $M_{(g)}$ as principal localizations of $S_{(f)}$ resp. $M_{(f)}$ are clear from the formulas above. The maps in the commutative diagram of topological spaces correspond to the ring maps given above. The horizontal arrows are homeomorphisms by Algebra, Lemma 10.57.3. The vertical arrows are open immersions since the left one is the inclusion of an open subset.

The open $D_{+}(f)$ is quasi-compact because it is homeomorphic to $\mathop{\mathrm{Spec}}(S_{(f)})$ , see Algebra, Lemma 10.17.8. Hence the second statement follows directly from the fact that the standard opens form a basis for the topology.

The third statement follows directly from Algebra, Lemma 10.17.2. $\square$

Comments (3)

Comment #128 by Fred Rohrer on February 15, 2013 at 13:31

In 1.8, replace "are a compatible" by "are compatible", and " $S_f$ " by " $S_f$ -".

Comments (3)

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