Elements in projective space

Question

Consider the projective line for simplicity. I was wondering was it means to look at the stalk at $0$. What is $0$ even referring to? The ideal generated by $0$ in one of the copies of the affine line? And if that is the case which one? Thanks for the help, even though this should be trivial.

Answer 1

You can view the projective line $\mathbb P^1$ as a one point compactification of $\mathbb A^1$. You take one of the standard open subspaces of $\mathbb P^1$, for example the space $D_+(x_0) = \{(x_0:x_1):\mathbb P^1|x_0\neq 0\}$ where the first coordinate does not vanish. That subspace is isomorphic to $\mathbb A^1$ via the map which sends $t$ to $(1:t)$. The inverse sends $(x_0:x_1)$ to $x_1/x_0$. Now the $0$ in $\mathbb P^1$ is (relative to the arbitrary choice of $D_+(x_0)$ the image of $0\in\mathbb A^1$ under the inclusion $\mathbb A^1 \hookrightarrow \mathbb P^1$. You get the point $(1:0)$. You can view the one point $(0:1)$ which isn't contained in $D_+(x_0)$ as the point $\infty =(0:1)$ at infinity. The stalk of $\mathbb P^1$ at $0$ can now be computed in the normal way, as a colimit over the rings $\mathcal O_{\mathbb P}(U)$ where $U$ ranges over the (Zariski, when you do scheme theory) open subspaces of $\mathbb P^1$ which contain $p$. It will be the case that $$\mathcal O_{\mathbb P^1,(1:0)}=\mathcal O_{\mathbb A^1,0} = k[x]_{(x-0)}$$ It is the ring $k[x]$ with inverses to all polynomials which do not vanish at $0$ freely adjoint.