Hartshorne Exercise III.2.1(a) 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?

Yes, it is really necessary that $P\neq Q$.
Indeed,  if $V=\mathbb A^1_k\setminus \lbrace P\rbrace$ , then $H^1(\mathbb A^1_k,\mathbb Z_V) =0$ 
Proof
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 $(\ast)$.   
