Let $k$ be a field. The morphism $\mathbb{P}^1_k\to \operatorname{Spec}k$ is a separated morphism of schemes. I am pretty confident about the following statements:
a. The diagonal morphism $\mathbb{P}^1_k\to \mathbb{P}^1_k\times_k \mathbb{P}^1_k$ is a closed immersion and defines a closed subscheme $\Delta$ of codimension $1$ (this is the separatedness assumption).
b. There is a Weil divisor associated to $\Delta$ and hence an invertible sheaf $\mathcal{O}(\Delta)$ defined by a Cartier divisor.
c. The corresponding Cartier divisor is defined as: $$\left\{\left(\operatorname{Spec}k[x,y],x-y \right), ~\left(\operatorname{Spec}k\left[x,y^{-1}\right],xy^{-1}-1 \right),~\left(\operatorname{Spec}k\left[x^{-1},y\right],1-x^{-1}y \right),~\left(\operatorname{Spec}k\left[x^{-1},y^{-1}\right],x^{-1}-y^{-1} \right)\right\}.$$ where $x$ and $y$ corresponds to the coordinates on $\textbf{A}^1_k\times_k \textbf{A}^1_k=\operatorname{Spec}k[x,y]$.
I have several elementary questions in mind to improve my understanding of the situation. Those are surely silly and easy but I feel the need to enlighten some of my thoughts on the diagonal morphism.
Can I use a.-b.-c. backward to prove that $\mathbb{P}^1_k\to \operatorname{Spec}k$ is separated?
Is $\mathcal{O}(\Delta)$ non principal? In particular, different from $\mathcal{O}(x-y)$.
If I replace $\mathbb{P}^1_k$ by any curve $X$ such that $X\to \operatorname{Spec}k$ is separated, is the following true : Let $\operatorname{Spec} R$ be an affine subscheme of $X$. Then, $\mathcal{O}(\Delta)(\operatorname{Spec} R\otimes_k R)$ corresponds to the kernel of $R\otimes_k R\to R,~r\otimes r'\mapsto rr'$ ?
Many thanks!