a few pages of 6.iv

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 $y$.

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 {-}

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
 :::