Let consider the projective line $\mathbb{P}_k^1$ as scheme over algebraically closed field $k$ and a closed point $x \in \mathbb{P}^1$ consideres as closed sub scheme. This gives arise for exact sequence
$$0 \to J_x \to \mathcal{O}_{\mathcal{\mathbb{P}^1}} \to \mathcal{O}_x \to 0$$
with corresponding ideal sheaf $J_x$ to $x$.
I often heard that $J_x$ equals $\mathcal{O}_{\mathcal{\mathbb{P}^1}}(-1)$ but the argument seems not clear to me.
My following thoughts lead me to another conclusion that $J_x=J_x$ equals $\mathcal{O}_{\mathcal{\mathbb{P}^1}}(-1)$$ but I don't find an error in my reasonings:
I considered as following:
Since $k$ is algebraically closed every closed point corresponds to a maximal ideal which is here always given by a linear polynomial.
Since $ \mathcal{O}_x =\mathcal{O}_{\mathcal{\mathbb{P}^1}}/J_x $ and $\mathcal{O}_{\mathcal{\mathbb{P}^1}}(1)$ consists locally of polynomials of degree one (therefore linear polynomials) I would would suppose that $J_x=\mathcal{O}_{\mathcal{\mathbb{P}^1}}(1)$ what contradicts the fact that $J_x=\mathcal{O}_{\mathcal{\mathbb{P}^1}}(-1)$.
Can anybody tell we where I make here an error in my reasoning? And if there exist a way to see intuitively why $J_x=\mathcal{O}_{\mathcal{\mathbb{P}^1}}(-1)$ holds.
But why $J_x = \mathcal{O}_{\mathcal{\mathbb{P}^1}}(-1)$?
There are two steps in the argument:
1) As shown in this thread (see the other answer Georges links in the comments), given any point $p\in\mathbb{P}^1$ we have the an isomorphism $O_{\mathbb{P}^1}(p)\cong O_{\mathbb{P}^1}(1)$.
2) The ideal sheaf corresponding to a divisor $D$ is precisely $O(-D)$.