
 \subsektion{Coherence of higher direct images}

 Last time I said that you can think of $H^1$ as extensions. That's true, but it is
 better to start the induction at $q=0$, where the statement is vacuous.

 $(*)_0\Rightarrow (*)_1'$. Choose $\alpha\in H^1(X,F)$. By the same reason as last time,
 there exists an affine \'etale morphism $p:U\to X$ such that $\alpha\mapsto 0$ in
 $H^1(U,p^*F)$. Now consider
 \[
    0\to F\to p_*p^*F\to Q\to 0
 \]
 which gives us
 \[
    H^0(X,p_*p^*F)\twoheadrightarrow H^0(X,Q)\to H^1(X,F)\xrightarrow{\alpha\mapsto 0}
    \underbrace{H^1(X,p_*p^*F)}_{\hookrightarrow H^1(U,p^*F)}
 \]
 The descent is hidden in the (obvious) statement that global sections are the same in
 the Zariski and \'etale topologies.

 \begin{theorem}
   Let $f:X\to Y$ be a proper morphism of locally noetherian algebraic spaces and $F$ a
   coherent sheaf on $X$. Then $R^q\!f_*F$ is a coherent $\O_Y$-module for every $q\ge
   0$.
 \end{theorem}
 We use the working definition of properness.
 \begin{proof}
   Reduce to the case $Y$ is affine.

   Devissage is on the homework: $X$ a noetherian locally separated algebraic space, and
   let $K=\coh \O_X$. Let $K'\subseteq K$ be a subcategory such that
   \begin{enumerate}
     \item $0\in K'$
     \item for every exact sequence $0\to A'\to A\to A''\to 0$ in $K$, if two out of
     three are in $K'$, then so is the third.
     \item If $A_1\oplus A_2\in K'$, then $A_1,A_2\in K'$.
     \item for every reduced and irreducible closed subspace $Z\subseteq X$, there exists a $G\in K'$
     sith $\supp G=Z$.
   \end{enumerate}
   Then $K'=K$. Integral means reduced and irreducible \anton{Ogus: well, we had that weird
   example where we have two tangent directions on the line ... that is integral, but
   doesn't feel like it should be}

   Take $K'\subseteq \coh X$ to be the subcategory of coherent sheaves for which the
   theorem holds. (1) zero is in there, (2) use the long exact sequence, (3) cohomology
   is the direct sum, and we must check (4).
   \begin{remark}
     \anton{strengthening of the theorem as follows} $X$ separated of finite type
     algebraic space over a noetherian affine scheme $Y$ \anton{let's change it to over a
     field}, then there exists $d$ such that for \emph{every sheaf} $F$ on $X$,
     $H^q(X,F)=0$ for $q\ge 0$. Do this by taking a dense open which is a scheme and do
     some kind of devissage. The $d$ is not the dimension in general.
   \end{remark}

   Check (4). Let $Z\subseteq X$. We can choose $\pi$ birational proper surjective by
   Chow's lemma.
   \[\xymatrix{
    Z'_U\ar[dr]_{\pi_U}& Z'\ar@{^(->}[r]^{closed} \ar[dr] & \PP^n_Z\ar[d]\\
    &U\ar[r] & Z \ar@{^(->}[r]\ar[d] & X \ar[dl]\\
    & &Y
   }\]
    Let $L=\O_{Z'}(1)$. For $m \gg 0$, $\pi^*\pi_*L^{\otimes m}\to L^{\otimes m}$ is
   surjective \anton{you don't need this surjectivity} and $T^q\pi_*L^{\otimes m}=0$ for $q>0$. To check these, it is enough to
   find $U\to Z$, with $\pi_U:Z'_U\to U$ \anton{use Serre's vanishing theorem}. This implies that
   $H^p(Z,\pi_*L^{\otimes m})=H^p(Z',L^{\otimes m})$ because we have the spectral
   sequence $E^{p,q}_2=H^p(Z,R^q\pi_*L^{\otimes m})\Rightarrow H^{p+q}(Z',L^{\otimes
   m})$, and we have that $E^{p,q}_2=0$ for $q>0$. This proves that
   $H^p(Z,\pi_*L^{\otimes m})$ are finite type $\Ga(Y,\O_Y)$-modules. So $\pi_*L^{\otimes
   m}\in K'$. We still need to show that $\supp \pi_*L^{\otimes m}=Z$.

   The support is defined by the ideal of sections which annihilate that module. Suppose
   $f\in \O_Z(U)$ annihilates $\pi_*L^{\otimes m}$. Then we claim that $f=0$. To see
   this, choose a dense open $U^0\subseteq U$ such that $Z_{U^0}'\to U^0$ is an
   isomorphism. The restriction of $\pi_*L^{\otimes m}$ to $U^0$ is a locally free sheaf
   of non-zero rank, so $f\mapsto 0\in \O_Z(U^0)$. But then $f=0$ because $U$ is reduced.

   Note that we produced for any reduced subspace a sheaf with that support. We
   didn't need irreducible.
 \end{proof}
 Next week, we'll define stacks. Now we'll define a proper morphism.