Let $k$ be a algebraically closed field, $X = \mathbb{P}^1_k = \mathrm{Proj} k[x,y]$ the projective space and $p = (ax+by) \in X$ a closed point. Let $\mathcal{I}_p = \ker( {\mathcal{O}_X \to p_* \mathcal{O}_{X,p}})$ the ideal sheaf of $p$, I want to prove that $\mathcal{I}_p \simeq \mathcal{O}_X(-1)$.
If I'm understanding it right, $p_* \mathcal{O}_{X,p}$ is the skycraft sheaf at $p$ and $$ \mathcal{I}_p(U) = \begin{cases} \mathcal{O}_X(U) \quad &\text{if} p \notin U\\ \ker (\mathcal{O}_X(U) \to \mathcal{O}_{X,p}) \end{cases} $$ where the last map is the natural insertion. In special $$\mathrm{stalk}_p \mathcal{I}_p = 0$$ Then I really stuck in understand how $\mathcal{I}_p$ can be invertible because a locally free sheaf of rank $n$ has free stalks with the same rank.
Well, returning to original question, let $U_x = D_+(x)$ and $U_y=D_+(y)$ the canonical affine charts of $\mathbb{P}^1$. Let's suppose $p \in U_x$ (i.e., $b \neq 0$), then $$ \mathcal{I}_p(U_x) = \ker (\mathcal{O}_X(U_x) \to p_*\mathcal{O}_p(U_x)) = \ker ( k[x,y]_{(x)} \to k[x,y]_{((ax+by))} ) = 0 $$ Which just doesn't make sense, am I getting it wrong?