Zariski neighbourhood of a prime ideal

Question

Consider for the sake of concreteness $X=Spec\ \mathbb{C}[x,y]$, and let $\mathfrak{p}=\langle x\rangle$. Translating this to geometry this is the $y$-axis, in the sense that the $y$-axis is the zero set of $\mathfrak{p}$.

An example of a Zariski neighbourhood of $\mathfrak{p}$ is $$D(y)=\{\mathfrak{q}\mid y\not\in \mathfrak{q}\}$$ translating this into geometry this is the complement of the $x$-axis.

So the geometric picture reads

an open neighbourhood of the $y$-axis is the complement of the $x$-axis

Now I know I shouldn't rely too much on my intuition about the usual Euclidean topology, but this still seems strange to me. This is strange, because the $y$-axis contains the origin, or phrased algebraically $\langle x,y\rangle\in V(\mathfrak{p})$, but the $x$-axis also contains the origin.

So we have an open neighbourhood of the $y$-axis, but it does not contain all points on the $y$-axis. How should I think about this?

It seems like the geometric picture is that an open neighbourhood of a prime ideal (or curve) is a Zariski open subset of $\mathbb{C}[x,y]$ containing a Zariski open subset of the curve

Is that correct? And if so, is the following interpretation also correct:

The local ring $\mathcal{O}_\mathfrak{p}$ of germs of function near the curve $\mathfrak{p}$ are all rational functions that have poles at at most a Zariski closed subset of $\mathfrak{p}$

As you can tell I am very unsure about my intuition about schemes (even the simplest affine ones), so some feedback on this line of reasoning would be appreciated.

Answer 1

I think you should rethink your intuition of what this point represents. The point $\mathfrak p=(x)$ is indeed a point that corresponds to the $y$-axis, yes. But it does not represent the entire $y$-axis as a single piece; rather, you should think of it as a single point that lies "everywhere" on the $y$-axis at once.

That is, your intuition should not be $D(f)$ contains $\mathfrak p$ if and only if it geometrically contains the whole $y$-axis. Rather, $D(f)$ contains $\mathfrak p$ if and only if it contains any (closed) points that lie on the $y$-axis. Indeed, you can check that a distinguished open set $D(f)$ contains $\mathfrak p$ if and only if contains at least one closed point of the form $\mathfrak m=(x,y-b)$ for some $b$, i.e. a closed point lying on the $y$-axis.