Grothendieck vanishing on a Noetherian space

I prove the assertion by induction on $d\ge 0$. The case $d=-1$, needed as the input for $d=0$, is the assertion that every Noetherian space $Y$ with $\dim Y\le -1$ satisfies $H^q(Y,\mathcal{G})=0$ for every sheaf $\mathcal{G}$ on $Y$ and every $q>-1$. If $Y\mathrm{\ne }\mathrm{\varnothing }$ and $y\in Y$, the singleton $\{y\}$ is irreducible [F25], so by [F8] its closure in $Y$ is a nonempty irreducible closed subset of $Y$, which is a strict chain of length $0$; hence $0\le \dim Y$ by [F2], and $\dim Y\le -1$ forces $Y=\mathrm{\varnothing }$. For $Y=\mathrm{\varnothing }$ and any sheaf $\mathcal{G}$ the group of global sections is $\mathrm{\Gamma }(\mathrm{\varnothing },\mathcal{G})=\mathcal{G}(\mathrm{\varnothing })$, a singleton by [F18], that is the zero group; the functor $\mathrm{\Gamma }(\mathrm{\varnothing },-)$ is therefore the zero functor, which is exact, and by [F19] its positive derived functors vanish, so $H^q(\mathrm{\varnothing },\mathcal{G})=0$ for every $q\ge 0$ and the case $d=-1$ holds. Now fix $d\ge 0$ and assume the assertion for $d-1$, i.e. for all Noetherian spaces of dimension at most $d-1$ and all sheaves on them. Since a Noetherian space has only finitely many irreducible components [F9], it is enough to prove, by induction on $t\ge 0$, the auxiliary statement $Q(t)$: every Noetherian space $X$ with $\dim X\le d$ having at most $t$ irreducible components satisfies $H^q(X,\mathcal{F})=0$ for every sheaf of abelian groups $\mathcal{F}$ on $X$ and every $q>d$.

Assume $t\ge 2$ and $Q(t-1)$, and let $X$ be Noetherian with $\dim X\le d$ and at most $t$ irreducible components; the spaces with fewer than $t$ components are covered by $Q(t-1)$, so let $X$ have exactly $t$ components. Choose one component $C$ and let $W$ be the union of the remaining components of $X$, a finite union by [F9], hence a closed subset of $X$ by [F3] and [F23]. Since $X$ is the union of its irreducible components by [F4], $X=C\cup W$. Moreover $C\subseteq ̸W$: otherwise $C=C\cap W={\bigcup }_{C^{\mathrm{\prime }}\mathrm{\ne }C}(C\cap C^{\mathrm{\prime }})$, a finite union of subsets closed in $C$ by [F3], and $C$ is irreducible and nonempty [F25], so $C=C\cap C^{\mathrm{\prime }}$ for some component $C^{\mathrm{\prime }}\mathrm{\ne }C$, that is $C\subseteq C^{\mathrm{\prime }}$, contradicting maximality of $C$ [F24]. The irreducible components of the space $W$ are exactly the components $C^{\mathrm{\prime }}\mathrm{\ne }C$ of $X$: each such $C^{\mathrm{\prime }}$ is an irreducible closed subset of $W$ by [F3], no $C^{\mathrm{\prime }}$ is contained in the union of the others by the same maximality argument just given, and $W$ is that finite union, so [F6] applies; in particular $W$ has exactly $t-1$ irreducible components, and $W$ is Noetherian by [F16] with $\dim W\le d$ by [F2]. Finally let $Z^{\mathrm{\prime }}:=\overline{X\setminus W}$, the closure in $X$ of $X\setminus W$; by [F7], whose hypotheses $X=C\cup W$, $C$ irreducible and $C\subseteq ̸W$ have just been verified, this $Z^{\mathrm{\prime }}$ is irreducible, and $Z^{\mathrm{\prime }}$ is nonempty because $X\setminus W=C\setminus W\mathrm{\ne }\mathrm{\varnothing }$. By [F16] and [F2] the space $Z^{\mathrm{\prime }}$ is Noetherian of dimension at most $d$, and by [F5] it has exactly one irreducible component; since $t\ge 2$ we have $1\le t-1$, so $Q(t-1)$ applies to $Z^{\mathrm{\prime }}$ as well as to $W$.

A space with no irreducible component is empty, because every point of a space lies in an irreducible component by [F4]; thus $Q(0)$ says that $H^q(\mathrm{\varnothing },\mathcal{F})=0$ for all $q>d$, which is the computation of [step 1.1] for the empty space, the classes of sheaves on $\mathrm{\varnothing }$ being trivial and $H^0$ also vanishing there.

Keep the notation of [step 1.2] and let $\mathcal{F}$ be a sheaf of abelian groups on $X$; write $U:=X\setminus W$ and let $j:U\hookrightarrow X$ and $i:W\hookrightarrow X$ be the inclusions. By [F13] there is a short exact sequence $0\to \mathcal{G}\to \mathcal{F}\to i_{\ast }(\mathcal{F}{\mathrm{\mid }}_W)\to 0,\mathcal{G}:=j_{!}(\mathcal{F}{\mathrm{\mid }}_U)\mathrm{.}$ I claim ${\mathcal{G}}_x=0$ for every $x\mathrm{\notin }{Z}^{\mathrm{\prime }}$. Indeed, if $x\mathrm{\notin }{Z}^{\mathrm{\prime }}=\bar{U}$, then by [F26] there is an open neighbourhood $V$ of $x$ with $V\cap U=\mathrm{\varnothing }$; every section of $\mathcal{G}$ over an open $V^{\mathrm{\prime }}\subseteq V$ has $V^{\mathrm{\prime }}\cap U=\mathrm{\varnothing }$, so $\mathcal{G}(V^{\mathrm{\prime }})=0$ by [F28], and these $V^{\mathrm{\prime }}$ are cofinal among the open neighbourhoods of $x$, so the colimit of [F27] computing ${\mathcal{G}}_x$ is a colimit of zero groups, that is ${\mathcal{G}}_x=0$. Therefore [F14] applies to $\mathcal{G}$ and the closed subset $Z^{\mathrm{\prime }}$, and the unit $\mathcal{G}\to i_{\ast }^{\mathrm{\prime }}(i^{\mathrm{\prime }-1}\mathcal{G})$, $i^{\mathrm{\prime }}:Z^{\mathrm{\prime }}\hookrightarrow X$ the inclusion, is an isomorphism of sheaves; by [F17] it induces an isomorphism $H^p(X,\mathcal{G})\cong H^p(X,i_{\ast }^{\mathrm{\prime }}(i^{\mathrm{\prime }-1}\mathcal{G}))$ for every $p$, and by [F11] the latter group is $H^p(Z^{\mathrm{\prime }},i^{\mathrm{\prime }-1}\mathcal{G})$; for $p>d$ this vanishes by $Q(t-1)$ applied to $Z^{\mathrm{\prime }}$ in [step 1.2]. Likewise $H^p(X,i_{\ast }(\mathcal{F}{\mathrm{\mid }}_W))\cong H^p(W,\mathcal{F}{\mathrm{\mid }}_W)=0$ for $p>d$ by [F11] and $Q(t-1)$ applied to $W$.

$Q(1)$: let $X$ be Noetherian with $\dim X\le d$ and at most one irreducible component, and let $U\subseteq X$ be open; I show $H^p(X,j_{U!}{\mathbb{Z}}_U)=0$ for every $p>d$. If $X=\mathrm{\varnothing }$ this is [step 2.1], so assume $X\mathrm{\ne }\mathrm{\varnothing }$; then $X$ is irreducible, because [F4] puts every point in its sole irreducible component. If $U=\mathrm{\varnothing }$ then ${\mathbb{Z}}_U$ is a sheaf on the empty space, so its only section group is zero by [F18], and ${\mathbb{Z}}_U=0$ and $j_{U!}{\mathbb{Z}}_U=0$, whence $H^p(X,j_{U!}{\mathbb{Z}}_U)=0$ in every degree [F19]. Assume now $U\mathrm{\ne }\mathrm{\varnothing }$, put $Z:=X\setminus U$, and note that $Z$ is closed in $X$ and $Z\mathrm{\ne }X$. I claim $\dim Z\le d-1$. Let $Z_0⊊\cdots ⊊Z_s$ be a strict chain of nonempty irreducible closed subsets of the subspace $Z$; each $Z_i$ is also closed in $X$ by [F29] since $Z$ is closed in $X$, and irreducibility of $Z_i$ is the same whether tested in $Z$ or in $X$, while $X$ itself is a nonempty irreducible closed subset of $X$ strictly containing $Z_s$; so $Z_0⊊\cdots ⊊Z_s⊊X$ is a strict chain of nonempty irreducible closed subsets of $X$, of length $s+1$, and $s+1\le \dim X\le d$ by [F2], that is $s\le d-1$; hence $\dim Z\le d-1$ by [F2]. Apply [F13] to $\mathcal{F}={\mathbb{Z}}_X$: there is a short exact sequence $0\to j_{!}({\mathbb{Z}}_U)\to {\mathbb{Z}}_X\to i_{\ast }({\mathbb{Z}}_X{\mathrm{\mid }}_Z)\to 0,$ the first term being identified with $j_{!}({\mathbb{Z}}_U)$ by clause 2 of [F13]. In the long exact sequence of [F15], the exact portion $H^{p-1}(X,i_{\ast }({\mathbb{Z}}_X{\mathrm{\mid }}_Z))\to H^p(X,j_{!}({\mathbb{Z}}_U))\to H^p(X,{\mathbb{Z}}_X)$ has vanishing outer terms for $p>d$: the first is $H^{p-1}(Z,{\mathbb{Z}}_X{\mathrm{\mid }}_Z)$ by [F11], and $Z$ is Noetherian by [F16] with $\dim Z\le d-1$, so it vanishes by the induction hypothesis for $d-1$ because $p-1>d-1$; the second vanishes by [F12] because $X$ is irreducible and $p>d\ge 0$ gives $p>0$. Therefore $H^p(X,j_{U!}{\mathbb{Z}}_U)=0$ for every $p>d$ and every open $U\subseteq X$. Since $X$ is Noetherian with $\dim X\le d$, [F10] now gives $H^p(X,\mathcal{F})=0$ for every sheaf of abelian groups $\mathcal{F}$ on $X$ and every $p>d$, which is $Q(1)$.

Let $p>d$. The long exact sequence of [F15] attached to the short exact sequence of [step 2.2] contains the exact portion $H^p(X,\mathcal{G})\to H^p(X,\mathcal{F})\to H^p(X,i_{\ast }(\mathcal{F}{\mathrm{\mid }}_W)),$ and both outer groups vanish for $p>d$ by [step 2.2]; hence $H^p(X,\mathcal{F})=0$. As $\mathcal{F}$ was arbitrary this is $Q(t)$ for the space $X$ with exactly $t$ components, and spaces with fewer components are $Q(t-1)$, so $Q(t)$ holds for all spaces of dimension at most $d$ with at most $t$ components; by induction on $t$, using that all these spaces have finitely many components [F9], the assertion holds for every Noetherian space of dimension at most $d$, which completes the induction on $d$ started in [step 1.1] and proves the theorem. The final clause follows from [F20], which identifies the condition $\mathrm{cd}\le d$ with vanishing above $d$ for every sheaf in the class of all abelian sheaves. The Axiom of Choice [F21] is used for the supplied injective resolutions and the reduction [F10], and supplies DC for resolution comparison through [F22]: the cohomology groups used throughout are the derived functors of global sections computed from injective resolutions, as in [F15], [F11], [F12], [F17], [F19] and [F10], and the existence of those resolutions is what requires the choice principle; all topological arguments in [step 1.1], [step 3.1], [step 1.2] and [step 2.2] are choice-free. ∎