I work with toric varieties and took a look at the blowup $\pi:X=\operatorname{Bl}_0(\mathbb{P}^2)\to \mathbb{P}^2$ ($0=(0:0:1)$) with points represented as $(x_0:x_1:x_2),(y_0:y_1)$. Technically we have four principal divisors, the exceptional $E$, two lines $L_1,L_2$ corresponding to the vanishing of $y_0$ and $y_1$ and $H\sim L_1+E$ corresponding to the vanishing of $x_2$. I want to determine basepoint free and ample divisors which worked quite well so far. However, I know want to see how their sheaves look like to verify that sheaves of the divisors with basepoints are not generated by global sections.
For that I found that $D\sim rH+sE$ is only basepoint free for $r\geq0,-r\leq s \leq 0$ and I'm pretty sure that this is correct. What happens if I take $r\geq 0,s > 0$?
My calculation for the sheaf is as follows: $$ \mathcal{O}_X(rH+sE) = \mathcal{O}_X(H)^r\otimes \mathcal{O}_X(E)^s = \mathcal{O}_X(L_1+E)^r\otimes \pi^*\mathcal{O}_{\mathbb{P}^2}(-1)^s $$
However, pulling back a line $L=V(x)\subset\mathbb{P}^2$ yields the divisor $H$, as it's nothing else than adding the exceptional divisor to the strict transform of $L$, so $\pi^*L\sim L_1 + E \sim H$. This means we get $$ \mathcal{O}_X(rH+sE)=\pi^*\mathcal{O}_{\mathbb{P}^2}(1)^r \otimes \pi^*\mathcal{O}_{\mathbb{P}^2}(-1)^s = \pi^*\mathcal{O}_{\mathbb{P}^2}(r)\otimes \pi^*\mathcal{O}_{\mathbb{P}^2}(-s) = \pi^*\mathcal{O}_{\mathbb{P}^2}(r-s) $$
but there is clearly some error as the last sheaf would be generated by global sections for $r>s$ and I'm confident that $rH+sE$ is has basepoints for $r>s$. Where did I go wrong?