From a98acf98c6b802be1a8b0866430bb753f8927a02 Mon Sep 17 00:00:00 2001 From: Tim Hosgood Date: Sun, 27 Aug 2023 01:31:32 +0200 Subject: [PATCH] a few pages of 6.iv --- sga-1/sga-1-ii/sga-1-ii-2.Rmd | 6 +-- sga-6/index.Rmd | 4 ++ sga-6/sga-6-iv/sga-6-iv-1.Rmd | 69 ++++++++++++++++++++++++++++++++++- 3 files changed, 75 insertions(+), 4 deletions(-) diff --git a/sga-1/sga-1-ii/sga-1-ii-2.Rmd b/sga-1/sga-1-ii/sga-1-ii-2.Rmd index fba52bd..59e5bf2 100644 --- a/sga-1/sga-1-ii/sga-1-ii-2.Rmd +++ b/sga-1/sga-1-ii/sga-1-ii-2.Rmd @@ -10,7 +10,7 @@ b. $f^{-1}(y)$ be smooth over $k(y)$ at $x$. ::: {.proof} Since the composite of two flat morphisms is flat, and since $Y[t_1,\ldots,t_n]\to Y$ is a flat morphism, we see that smooth implies flat; -taking [(1.3), (ii)](#II.1.3) into account, this proves necessity. +taking [1.3, (ii)](#II.1.3) into account, this proves necessity. Now suppose that (a) and (b) are satisfied, and let $V$ be an affine neighbourhood of $y$ of ring $A$, and $U$ an affine neighbourhood of $x$ over $V$ of ring $B$. Taking $U$ to be small enough, we can suppose, by (b), that there exists an *étale* $k(y)$-morphism $$ @@ -26,7 +26,7 @@ $$ $$ (after possibly multiplying the $g_i$ by a single non-zero element of $k$). But $U$ is flat over $Y$ by (a), and so too is $Y[t_1,\ldots,t_n]$; -we also know that $g$ induces an étale morphism between the fibres over $y$, and so $g$ is étale at $x$ by [(I 5.8)](#I.5.8). +we also know that $g$ induces an étale morphism between the fibres over $y$, and so $g$ is étale at $x$ by [I 5.8](#I.5.8). ::: ::: {.itenv #II.2.2 title="Corollary 2.2" latex="{Corollary 2.2}"} @@ -36,7 +36,7 @@ For $f$ to be smooth at $x$, it is necessary and sufficient that $X$ be flat (or ::: ::: {.proof} -Only the sufficiency needs proving, and this follows from [(2.1)](#II.2.1), combined with the flatness criterion [(I 5.9)](#I.5.9). +Only the sufficiency needs proving, and this follows from [2.1](#II.2.1), combined with the flatness criterion [I 5.9](#I.5.9). ::: To state the following result, "recall" that a morphism $f\colon X\to Y$ locally of finite type is said to be *equidimensional* at the point $x\in X$ if (setting $y=f(x)$) we can find an open neighbourhood $U$ of $x$ of which every component dominates a component of $Y$ such that, for all $y'\in Y$, the irreducible components of $f^{-1}(y')\cap U$ all have the same dimension, independent of $y$. diff --git a/sga-6/index.Rmd b/sga-6/index.Rmd index d7129bf..30d446e 100644 --- a/sga-6/index.Rmd +++ b/sga-6/index.Rmd @@ -16,6 +16,10 @@ description: "Séminaire de Géométrie Algébrique du Bois Marie: Book 6." \providecommand{\coh}{\mathrm{coh}} \providecommand{\perf}{\mathrm{perf}} \providecommand{\ob}{\operatorname{ob}} +\providecommand{\cl}{\operatorname{cl}} +\providecommand{\Kb}{\operatorname{K}^\mathrm{b}} +\providecommand{\Db}{\operatorname{D}^\mathrm{b}} +\providecommand{\HH}{\operatorname{H}} # Introduction {-} diff --git a/sga-6/sga-6-iv/sga-6-iv-1.Rmd b/sga-6/sga-6-iv/sga-6-iv-1.Rmd index 956df27..0c8cdec 100644 --- a/sga-6/sga-6-iv/sga-6-iv-1.Rmd +++ b/sga-6/sga-6-iv/sga-6-iv-1.Rmd @@ -1,5 +1,72 @@ ## 1. Reminders and generalities on Grothendieck groups {#IV.1} -::: {.rmenv #IV.1.1 title="1.1"} +::: {.rmenv #IV.1.1 title="1.1" latex="{1.1}"} +\oldpage{1 (274)}Let $\cal{C}$ be a *triangulated* category. +Recall ([I 6.3](#I.6.3) and [SGA 5 VIII 2]) that a map $f$ from $\ob\cal{C}$ to an abelian group $G$ is said to be *additive* if we have $f(E)=f(E')+f(E'')$ for every distinguished triangle $E'\to E\to E''\to E'[1]$. +The additive maps from $\ob\cal{C}$ to $G$ form an abelian group, which depends functorially on $G$. +The functor thus obtained is represented by an abelian group $k(\cal{C})$ and a universal additive map $\cl\colon\ob\cal{C}\to k(\cal{C})$ (denoted by $\cl$ when there is no risk of confusion). + +The group $k(\cal{C})$ depends functorially on $\cal{C}$ with respect to exact functors. +If $\cal{C}$ and $\cal{C}'$ are triangulated categories, then any two isomorphic exact functors from $\cal{C}$ to $\cal{C}'$ induce the same homomorphism $k(\cal{C})\to k(\cal{C}')$; +in particular, if $u\colon\cal{C}\to\cal{C}'$ is an exact functor that is an equivalence of categories, then $k(u)\colon k(\cal{C})\to k(\cal{C}')$ is an isomorphism. +::: + +::: {.itenv #IV.1.2 title="Lemma 1.2" latex="{Lemma 1.2}"} +Let $\cal{C}$ be a triangulated category, and $L\in\ob\cal{C}$. + +a. For every $n\in\mathbb{Z}$, we have + $$ + \cl(L[n]) + = (-1)^n\cl(L). + $$ + If $L'$ and $L''$ are objects of $\cal{C}$ such that we have $L\cong L'\oplus L''$, then we have + $$ + \cl(L) + = \cl(L') + \cl(L''). + $$ +b. Suppose that there exists $n\in\mathbb{Z}$ such that $\coprod_{i\geq0}L[2ni]$ is representable. + Then $\cl(L)=0$. +::: + +::: {.proof} +Claim (a) follows directly from the definitions (cf. [SGA 5 VIII 2]). +For (b), it suffices to remark that we have +$$ + \coprod_{i\geq0} L[2ni] + \cong L\oplus\left(\coprod_{i\geq0} L[2ni]\right)[2n] +$$ +\oldpage{2 (275)}and to apply (a). +::: + +::: {.itenv #IV.1.3 title="Lemma 1.3" latex="{Lemma 1.3}"} +Let $\cal{C}$ be a triangulated category, and $\cal{A}$ a thick subcategory ([@IV-V, I 2.1]). +Then the inclusion and passage to the quotient functors define an exact sequence +$$ + k(\cal{A}) + \to k(\cal{C}) + \to k(\cal{C}/\cal{A}) + \to 0. +$$ +::: + +::: {.proof} +See [SGA 5 VIII 3.1]. +::: + +::: {.itenv #IV.1.4 title="Lemma 1.4" latex="{Lemma 1.4}"} +Let $\cal{A}$ be an additive (resp. abelian) category, and $f$ an additive function ([1.1](#IV.1.1)) on $\ob\Kb(A)$ (resp. $\Db(A)$). +For $E\in\ob\Kb(A)$ (resp. $\Db(A)$) we have +$$ + f(E) + = \sum (-1)^i f(E^i) +$$ +(resp. $f(E)=\sum(-1)^i f(\HH^i(E))$). +::: + +::: {.proof} +Left as an exercise to the reader. +::: + +::: {.rmenv #IV.1.5 title="1.5" latex="{1.5}"} Let :::