TITLE: Showing Grothendieck's Vanishing Theorem provides a strict bound
QUESTION [10 upvotes]: The following result is due to Grothendieck:

If $X$ is a noetherian topological space of dimension $n$, then for all $i>n$ and all sheaves of abelian groups $\mathscr{F}$ on $X$, we have $H^i(X,\mathscr{F})=0$.

Exercise III.2.1 in Hartshorne is designed to show that the bound is strict.  I will restate the exercise here.

(a) Let $X=\mathbb{A}_k^1$ be the affine line over an infinite field $k$.  Let $P,Q$ be distinct closed points of $X$, and let $U=X-\{P,Q\}$.  Show that $H^1(X,\mathbb{Z}_U)\ne 0$.
(b) More generally, let $Y\subseteq X=\mathbb{A}_k^n$ be the union of $n+1$ hyperplanes in general position, and let $U=X- Y$.  Show that $H^n(X,\mathbb{Z}_U)\ne 0$.

Here, $\mathbb{Z}_U:=j_!(\mathbb{Z}|_{U})$ is the sheaf obtained by extending $\mathbb{Z}|_U$ by zero outside of $U$, where $j:U\to X$ is inclusion.  Similarly, for the inclusion $i:Y\to X$ where $Y := \{P, Q\}$, define $\mathbb{Z}_Y:=i_*(\mathbb{Z}|_Y)$.  I've been able to solve $(a)$ by making use of the exact sequence of sheaves on $X$
$$0\to\mathbb{Z}_U\to\mathbb{Z}\to\mathbb{Z}_Y\to 0$$
This gives rise to a long exact sequence in cohomology, from which we find $H^1(X,\mathbb{Z}_U)\ne 0$.
I'm having trouble with part $(b)$ and would appreciate some help.  I'm trying to proceed by induction, with the base case given by part $(a)$.  It seems to me that the inductive step should make use of the fact that if $Y=H_1\cup\ldots\cup H_{n+1}$ is the union of $n+1$ hyperplanes in general position in $\mathbb{A}_k^n$, then $Y-(H_2\cup\ldots\cup H_{n+1})$ is the complement of $n$ hyperplanes in general position in $H_1=\mathbb{A}_k^{n-1}$, but I can't seem to work it out.  Thanks in advance for any help.

REPLY [2 votes]: The simplest example showing that the bound in Grothendieck's theorem is sharp is given by projective space of dimension $n$ over a field $k$:$$H^n(\mathbb P^n_k,\mathcal O_{\mathbb P^n_k}(-n-1))\simeq k$$ (Hartshorne, III Theorem 5.1 (c), page 225)