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Show that $H^1(\mathbf{A}^1_k, \mathbf{Z}_U) \neq 0$ for $U = \mathbf{A}^1_k \setminus \{P,Q\}$, $k$ infinite field.

Is it really neccessary that $P \neq Q$?

My proof is as follows: Take the long exact sequence of $0 \to j_!j*\mathbf{Z} \to \mathbf{Z} \to i^*i_*\mathbf{Z} \to 0$ and get ($H^1(\mathbf{A}^1_k, \mathbf{Z}) = 0$ since $k$ is infinite, so the space is irreducible)

$0 \to \mathbf{Z} \to \mathbf{Z} \to \mathbf{Z}^{|X \setminus U|} \to H^1(\mathbf{A}^1_k, \mathbf{Z}_U) \to 0$. Now tensor with $\mathbf{Q}$ and count the ranks: $1 - 1 + |X \setminus U| - rk H^1(\mathbf{A}^1_k, \mathbf{Z}_U) = 0$, so $H^1(\mathbf{A}^1_k, \mathbf{Z}_U) \neq 0$ for $|X \setminus U| > 0$.

Am I missing something here?

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I haven't checked if you missed something, but if that exact sequence is correct, notice the map $Z\to Z$ has to be injective, for its cokernel is a subgroup of a torsion free group. –  Mariano Suárez-Alvarez Oct 21 '11 at 20:57
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up vote 1 down vote accepted

Yes, it is really necessary that $P\neq Q$.
Indeed, if $V=\mathbb A^1_k\setminus \lbrace P\rbrace$ , then $H^1(\mathbb A^1_k,\mathbb Z_V) =0$ because the sheaf $\mathbb Z_V$ on $\mathbb A^1_k$ is flasque and flasque sheaves are acyclic.
This follows from the explicit description of its groups of sections, which is easy if you remember that $\mathbb Z_V \subset \mathbb Z$:

The sections of $\mathbb Z_V$
-For an open subset $W\subset \mathbb A^1_k$ not containing $P$, that is $W\subset V$, we have $\Gamma (W,\mathbb Z_V)=\mathbb Z$ , since $\mathbb Z_V|V=\mathbb Z $ as sheaves on V .

-For an open subset $W\subset \mathbb A^1_k$ containing $P$ we have $\Gamma (W,\mathbb Z_V)=\mathbb 0$, since $\Gamma (\mathbb A^1_k,\mathbb Z_V)\subset \Gamma (\mathbb A^1_k,\mathbb Z)$ and the stalk of $\mathbb Z_V$ at $P$ is zero: $(\mathbb Z_V)_P=0$

A variant One could also use the long exact sequence associated to $$0\to \mathbb Z_V \to \mathbb Z \to j_\ast (\mathbb Z|\lbrace P\rbrace) \to 0 \quad ( \ast)$$ and get: $$\quad 0\to \text {don't care}\to \Gamma (\mathbb A^1_k,\mathbb Z)=\mathbb Z \stackrel {=}{\to}\Gamma (\mathbb A^1_k,j_\ast (\mathbb Z|\lbrace P\rbrace) =\mathbb Z\to H^1(\mathbb A^1_k,\mathbb Z_V)\to H^1(\mathbb A^1_k,\mathbb Z)=0 \to \cdots $$ The notable points are that $H^1(\mathbb A^1_k,\mathbb Z)=0$ because constant sheaves are flasque on irreducible spaces and above all that the morphism $\Gamma (\mathbb A^1_k,\mathbb Z)=\mathbb Z \to \Gamma (\mathbb A^1_k,j_\ast (\mathbb Z|\lbrace P\rbrace) =\mathbb Z$ is equality, which results from the definition of the maps in $(\ast)$.
(I think this variant is what you wanted to write in your analysis, but your first "$\mathbb Z$" [where I wrote "don't care" !] should actually be zero, and you shouldn't tensor by $\mathbb Q$)

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