# stacks/stacks-project

Clarify some arguments on coherent triples

 @@ -8193,7 +8193,7 @@ \section{Coherent triples} \noindent Let $(A, \mathfrak m)$ be a Noetherian local ring. Let $f \in \mathfrak m$ be a nonzerodivisor. Set $X = \Spec(A)$, $X_0 = \Spec(A/fA)$, $U = X \setminus V(\mathfrak a)$, and $X = \Spec(A)$, $X_0 = \Spec(A/fA)$, $U = X \setminus V(\mathfrak m)$, and $U_0 = U \cap X_0$. We say $(\mathcal{F}, \mathcal{F}_0, \alpha)$ is a {\it coherent triple} if we have @@ -8285,30 +8285,31 @@ \section{Coherent triples} \end{lemma} \begin{proof} If $\mathcal{F}', \alpha', \alpha'_0$ and $\mathcal{F}'', \alpha'', \alpha''_0$ are two such choices, then there exists a submodule $\mathcal{F}''' \subset \mathcal{F}''$ with $\mathcal{F}''/\mathcal{F}'''$ supported at $\{\mathfrak m\}$ and an $\mathcal{O}_X$-module map $\mathcal{F}''' \to \mathcal{F}'$ Let $\mathcal{F}', \alpha', \alpha'_0$ and $\mathcal{F}'', \alpha'', \alpha''_0$ be two such choices. For $n > 0$ set $\mathcal{F}'_n = \mathfrak m^n \mathcal{F}'$. By Cohomology of Schemes, Lemma \ref{coherent-lemma-homs-over-open} for some $n$ there exists an $\mathcal{O}_X$-module map $\mathcal{F}'_n \to \mathcal{F}''$ agreeing with the identification $\mathcal{F}''|_U = \mathcal{F}'|_U$ determined by $\alpha'$ and $\alpha''$. In fact, by Then the diagram $$\xymatrix{ \mathcal{F}'_n/f\mathcal{F}'_n \ar[r] \ar[d] & \mathcal{F}'/f\mathcal{F}' \ar[d]^{\alpha_0'} \\ \mathcal{F}''/f\mathcal{F}'' \ar[r]^{\alpha_0''} & \mathcal{F}_0 }$$ is commutative after restricting to $U_0$. Hence by Cohomology of Schemes, Lemma \ref{coherent-lemma-homs-over-open} we can take $\mathcal{F}''' = \mathfrak m^n \mathcal{F}''$ for $n \gg 0$. After increasing $n$ further we may in addition assume the induced maps $\mathcal{F}'''/f\mathcal{F}''' \to \mathcal{F}''/f\mathcal{F}''$ and $\mathcal{F}'''/f\mathcal{F}''' \to \mathcal{F}'/f\mathcal{F}'$ agree with $\alpha'_0$ and $\alpha''_0$ by the same lemma (because we have the agreement over $U_0$ and because $\mathcal{F}'''/f\mathcal{F}''' \to \mathcal{F}'/f\mathcal{F}'$ factors through the canonical map $\mathcal{F}'''/f\mathcal{F}''' = \mathfrak m^n \mathcal{F}''/f\mathfrak m^n \mathcal{F}'' \to \mathfrak m^n(\mathcal{F}''/f\mathcal{F}'')$; some details omitted). Then we see that we have a third choice it is commutative after restricting to $\mathfrak m^l(\mathcal{F}'_n/f\mathcal{F}'_n)$ for some $l > 0$. Since $\mathcal{F}'_{n + l}/f\mathcal{F}'_{n + l} \to \mathcal{F}'_n/f\mathcal{F}'_n$ factors through $\mathfrak m^l(\mathcal{F}'_n/f\mathcal{F}'_n)$ we see that after replacing $n$ by $n + l$ the diagram is commutative. In other words, we have found a third choice $\mathcal{F}''', \alpha''', \alpha'''_0$ such that there are maps $\mathcal{F}''' \to \mathcal{F}''$ and $\mathcal{F}''' \to \mathcal{F}'$ over $X$ @@ -8322,13 +8323,14 @@ \section{Coherent triples} isomorphism over $U$ and since $f : \mathcal{F}'' \to \mathcal{F}''$ is injective. Clearly $\mathcal{F}'/\mathcal{F}''$ is supported on $\{\mathfrak m\}$ hence has finite length. We have the maps of coherent $\mathcal{O}_{X_0}$-modules $$\mathcal{F}''/f\mathcal{F}'' \to \mathcal{F}'/f\mathcal{F}' \xrightarrow{\alpha'_0} \mathcal{F}_0$$ whose composition is $\alpha''_0$. Elementary homological algebra gives a $6$-term exact sequence whose composition is $\alpha''_0$ and which are isomorphisms over $U_0$. Elementary homological algebra gives a $6$-term exact sequence $$\begin{matrix} 0 \to @@ -8341,7 +8343,7 @@ \section{Coherent triples} \end{matrix}$$ By additivity of lengths (Algebra, Lemma \ref{algebra-lemma-length-additive}) it suffices to show that we find that it suffices to show that $$\text{length}_A( \Coker(\mathcal{F}''/f\mathcal{F}'' \to \mathcal{F}'/f\mathcal{F}')) - @@ -8406,27 +8408,51 @@ \section{Coherent triples} \alpha' : \mathcal{F}'|_U \to \mathcal{F}, \beta' : \mathcal{G}'|_U \to \mathcal{G}, and \gamma' : \mathcal{H}'|_U \to \mathcal{H} by construction. To finish the proof we'll need to construct the maps To finish the proof we'll need to construct maps \alpha'_0 : \mathcal{F}'/f\mathcal{F}' \to \mathcal{F}_0 and \gamma'_0 : \mathcal{H}'/f\mathcal{H}' \to \mathcal{H}_0, except this may not be possible with our initial choice of \mathcal{G}'. Namely, the obstruction is the composition \gamma'_0 : \mathcal{H}'/f\mathcal{H}' \to \mathcal{H}_0 as in Lemma \ref{lemma-prepare-chi-triple} and fitting into a commutative diagram$$ \xymatrix{ 0 \ar[r] & \mathcal{F}'/f\mathcal{F}' \ar[r] \ar@{..>}[d]^{\alpha'_0} & \mathcal{G}'/f\mathcal{G}' \ar[r] \ar[d]^{\beta'_0} & \mathcal{H}'/f\mathcal{H}' \ar[r] \ar@{..>}[d]^{\gamma'_0} & 0 \\ 0 \ar[r] & \mathcal{F}_0 \ar[r] & \mathcal{G}_0 \ar[r] & \mathcal{H}_0 \ar[r] & 0 } $$However, this may not be possible with our initial choice of \mathcal{G}'. From the displayed diagram we see the obstruction is exactly the composition$$ \delta : \mathcal{F}'/f\mathcal{F}' \to \mathcal{G}'/f\mathcal{G}' \xrightarrow{\beta'_0} \mathcal{G}_0 \to \mathcal{H}_0 $$Note that the restriction of \delta to U_0 is zero by our choice of \mathcal{F}' and \mathcal{H}'. Hence by Note that the restriction of \delta to U_0 is zero by our choice of \mathcal{F}' and \mathcal{H}'. Hence by Cohomology of Schemes, Lemma \ref{coherent-lemma-homs-over-open} there exists an n > 0 such that \delta vanishes on \mathfrak m^n \cdot (\mathcal{F}'/f\mathcal{F}'). Set \mathcal{G}'_n = \mathfrak m^n \mathcal{G}', there exists an k > 0 such that \delta vanishes on \mathfrak m^k \cdot (\mathcal{F}'/f\mathcal{F}'). For n > k set \mathcal{G}'_n = \mathfrak m^n \mathcal{G}', \mathcal{F}'_n = \mathcal{G}'_n \cap \mathcal{F}', and \mathcal{H}'_n = \mathcal{G}'_n/\mathcal{F}'_n. Observe that \beta'_0 can be composed with \mathcal{G}'_n/f\mathcal{G}'_n \to \mathcal{G}'/f\mathcal{G}' to give a map \beta'_{n, 0} : \mathcal{G}'_n/f\mathcal{G}'_n \to \mathcal{G}_0 as in Lemma \ref{lemma-prepare-chi-triple}. By Artin-Rees (Algebra, Lemma \ref{algebra-lemma-Artin-Rees}) we may choose n such that \mathcal{F}'_n \subset \mathfrak m^k \mathcal{F}'. As above the maps f : \mathcal{F}'_n \to \mathcal{F}'_n, f : \mathcal{G}'_n \to \mathcal{G}'_n, and @@ -8442,22 +8468,32 @@ \section{Coherent triples} \alpha'_n : \mathcal{F}'_n|_U \to \mathcal{F}, \beta'_n : \mathcal{G}'_n|_U \to \mathcal{G}, and \gamma'_n : \mathcal{H}'_n|_U \to \mathcal{H}. Finally, we consider the composition We consider the obstruction$$ \delta_n : \mathcal{F}'_n/f\mathcal{F}'_n \to \mathcal{G}'_n/f\mathcal{G}'_n \to \mathcal{G}'/f\mathcal{G}' \xrightarrow{\beta'_0} \mathcal{G}'_n/f\mathcal{G}'_n \xrightarrow{\beta'_{n, 0}} \mathcal{G}_0 \to \mathcal{H}_0 $$Since the composition of the first two maps factors as \mathcal{F}'_n/f\mathcal{F}'_n \to \mathfrak m^n(\mathcal{F}'/f\mathcal{F}') \to \mathcal{G}'/f\mathcal{G}' we conclude \delta_n = 0. as before. However, the commutative diagram$$ \xymatrix{ \mathcal{F}'_n/f\mathcal{F}'_n \ar[r] \ar[d] & \mathcal{G}'_n/f\mathcal{G}'_n \ar[r]_{\beta'_{n, 0}} \ar[d] & \mathcal{G}_0 \ar[r] \ar[d] & \mathcal{H}_0 \ar[d] \\ \mathcal{F}'/f\mathcal{F}' \ar[r] & \mathcal{G}'/f\mathcal{G}' \ar[r]^{\beta'_0} & \mathcal{G}_0 \ar[r] & \mathcal{H}_0 } $$our choice of n and our observation about \delta show that \delta_n = 0. This produces the desired maps \alpha'_{n, 0} : \mathcal{F}'_n/f\mathcal{F}'_n \to \mathcal{F}_0, \beta'_{n, 0} : \mathcal{G}'_n/f\mathcal{G}'_n \to \mathcal{G}_0, and \alpha'_{n, 0} : \mathcal{F}'_n/f\mathcal{F}'_n \to \mathcal{F}_0, and \gamma'_{n, 0} : \mathcal{H}'_n/f\mathcal{H}'_n \to \mathcal{H}_0. OK, so we may use \mathcal{F}'_n, \alpha'_n, \alpha'_{n, 0}, @@ -8497,7 +8533,7 @@ \section{Coherent triples} \chi((\mathcal{F}, \mathcal{F}_0, \alpha) \otimes (\mathcal{L}, \mathcal{L}_0, \lambda)^{\otimes n})$$ is a numerical polynomial of degree $\leq \dim(\text{Supp}(\mathcal{F}))$. is a polynomial of degree $\leq \dim(\text{Supp}(\mathcal{F}))$. \end{proposition} \noindent @@ -8508,9 +8544,43 @@ \section{Coherent triples} the function is linear, etc. \begin{proof} There is no harm in replacing $\mathcal{F}_0$ by the quotient of $\mathcal{F}_0$ by the functions supported on $\{\mathfrak m\}$: this just changes the function by a constant. We will prove this by induction on the dimension of the support of $\mathcal{F}$. \medskip\noindent The base case is when $\mathcal{F} = 0$. Then either $\mathcal{F}_0$ is zero or its support is $\{\mathfrak m\}$. In this case we have $$(\mathcal{F}, \mathcal{F}_0, \alpha) \otimes (\mathcal{L}, \mathcal{L}_0, \lambda)^{\otimes n} = (0, \mathcal{F}_0 \otimes \mathcal{L}_0^{\otimes n}, 0) \cong (0, \mathcal{F}_0, 0)$$ Thus the function of the lemma is constant with value equal to the length of $\mathcal{F}_0$. \medskip\noindent Induction step. Assume the support of $\mathcal{F}$ is nonempty. Let $\mathcal{G}_0 \subset \mathcal{F}_0$ denote the submodule of sections supported on $\{\mathfrak m\}$. Then we get a short exact sequence $$0 \to (0, \mathcal{G}_0, 0) \to (\mathcal{F}, \mathcal{F}_0, \alpha) \to (\mathcal{F}, \mathcal{F}_0/\mathcal{G}_0, \alpha) \to 0$$ This sequence remains exact if we tensor by the invertible coherent triple $(\mathcal{L}, \mathcal{L}_0, \lambda)$, see discussion above. Thus by additivity of $\chi$ (Lemma \ref{lemma-ses-chi-triple}) and the base case explained above, it suffices to prove the induction step for $(\mathcal{F}, \mathcal{F}_0/\mathcal{G}_0, \alpha)$. In this way we see that we may assume $\mathfrak m$ is not an associated point of $\mathcal{F}_0$. \medskip\noindent Let $T = \text{Ass}(\mathcal{F}) \cup \text{Ass}(\mathcal{F}/f\mathcal{F})$. Since $U$ is quasi-affine, we can find $s \in \Gamma(U, \mathcal{L})$ which does not vanish at any $u \in T$, see @@ -8556,8 +8626,9 @@ \section{Coherent triples} \chi((\mathcal{F}, \mathcal{F}_0, \alpha) \otimes (\mathcal{L}, \mathcal{L}_0, \lambda)^{\otimes n}) $$is a numerical polynomial. We conclude by Algebra, Lemma \ref{algebra-lemma-numerical-polynomial}. is a polynomial. We conclude by a variant of Algebra, Lemma \ref{algebra-lemma-numerical-polynomial} for functions defined for all integers (details omitted). \end{proof} \begin{lemma} @@ -8583,21 +8654,22 @@ \section{Coherent triples} Dualizing Complexes, Lemma \ref{dualizing-lemma-depth} and Local Cohomology, Lemma \ref{local-cohomology-lemma-finiteness-pushforwards-and-H1-local}. It follows that M = \Gamma(U, \mathcal{L}) is a finite A-module for example by Local Cohomology, Lemma \ref{local-cohomology-lemma-finiteness-for-finite-locally-free}. Using Local Cohomology, Lemma \ref{local-cohomology-lemma-finiteness-for-finite-locally-free} we find that M = \Gamma(U, \mathcal{L}) is a finite A-module. This in turn implies \text{depth}(M) \geq 2 for example by part (4) of Local Cohomology, Lemma \ref{local-cohomology-lemma-finiteness-pushforwards-and-H1-local} or by Divisors, Lemma \ref{divisors-lemma-depth-pushforward}. Also, we have \mathcal{L}_0 \cong \mathcal{O}_{X_0} as X_0 is a local scheme. Hence also see that as X_0 is a local scheme. Hence we also see that M_0 = \Gamma(X_0, \mathcal{L}_0) = \Gamma(U_0, \mathcal{L}_0|_{U_0}) and this module is isomorphic to A/fA. Thus \mathcal{F}' = \widetilde{M} is a coherent \mathcal{O}_X-module whose restriction to U is isomorphic to \mathcal{L}. The isomorphism \lambda : \mathcal{L}/f\mathcal{L} \to \mathcal{L}_0|_{U_0} and that this module is isomorphic to A/fA. \medskip\noindent By the above \mathcal{F}' = \widetilde{M} is a coherent \mathcal{O}_X-module whose restriction to U is isomorphic to \mathcal{L}. The isomorphism \lambda : \mathcal{L}/f\mathcal{L} \to \mathcal{L}_0|_{U_0} determines a map M/fM \to M_0 on global sections which is an isomorphism over U_0. Since \text{depth}(M) \geq 2 we see @@ -8669,7 +8741,7 @@ \section{Invertible modules on punctured spectra, I}$$ n \longmapsto \chi((\mathcal{L}, \mathcal{L}_0, \lambda)^{\otimes n}) $$is a numerical polynomial. By Lemma \ref{lemma-nonnegative-chi-triple} is a polynomial. By Lemma \ref{lemma-nonnegative-chi-triple} the value of this polynomial is zero if and only if \mathcal{L}^{\otimes n} is trivial. Thus if \mathcal{L} is torsion, then this @@ -8709,7 +8781,7 @@ \section{Invertible modules on punctured spectra, I}$$ n \longmapsto \chi((\mathcal{L}, \mathcal{L}_0, \lambda)^{\otimes n})  is a numerical polynomial $P$. By Lemma \ref{lemma-nonnegative-chi-triple} is a polynomial $P$. By Lemma \ref{lemma-nonnegative-chi-triple} we have $P(n) \geq 0$ for all $n \in \mathbf{Z}$ with equality if and only if $\mathcal{L}^{\otimes n}$ is trivial. In particular $P(0) = 0$