How to imagine the difference between the following schemes?

Question

Consider $A=\operatorname{Spec} k[x]_{(x)}[t]$ and $B=\operatorname{Spec} k[x,t]_{(x)}$ for a field $k$ (Vakil, note 11.3.8). For me, both are infinitesimal neighborhoods of an affine line - the former is the product of an infl nbhd of a point on a line with a line, the latter is simply an infl nbhd of a line on an affine plane. How should I imagine the difference between them, or are they too sophisticated?

Answer 1

This is a very interesting and pedagogical question!

1) The underlying set of the scheme $A=\operatorname{Spec} k[x]_{(x)}[t]$ is the union of
a) The generic point $\eta =(0)$ of the plane $\mathbb A^2_k$
b) The $y$-axis $V(x)\subset \mathbb A^2_k$ .
c) The generic points $\mathfrak p$ of all the irreducible curves $V(\mathfrak p)\subset \mathbb A^2_k$ corresponding to prime ideals $\mathfrak p\subset k[x,y]$ of height $1$.
[Unimportant observation: $(x)$ appears both in b) and c)]
Note carefully that the scheme $B$ is not a subscheme of $\mathbb A^2_k$ because it is not locally closed.
As user oxeimon excellently commented, $A$ has dimension $2$.

2) The scheme $B=\operatorname{Spec} k[x,t]_{(x)}$ is the spectrum of a discrete valuation ring.
It is thus of dimension $1$ and has only two points, namely the generic point $(x)$ of the $y$-axis and the generic point $(0)$ of $\mathbb A^2_k$.
Here also $B$ is not a subscheme of $\mathbb A^2_k$, although its underlying set consists of two points of the plane.

3) It should now be clear that neither of the schemes $A,B$ is an infinitesimal neighbourhood of the $y$-axis $V(x)\subset \mathbb A^2_k$.
The first infinitesimal neighbourhood of $V(x)$ is actually $V(x^2)=\operatorname{Spec} k[x,y]/(x^2)$ .