As far as I can tell, the topological spaces associated to the schemes in the title are both sets with two elements, with the discrete topology since both have prime ideals $(x)$ and $(x-1)$ which are maximal and thus closed.

In regards to the sheaf structure, obviously the global functions are different. However, as far as I can tell, the local rings of $\mathbb{C}[x]/(x^2-x)$ at $(x)$ and $(x-1)$ are isomorphic to $\mathbb{C}$, as is the local ring of $\mathbb{C}[x]/(x^3-x^2)$ at $(x-1)$, but the local ring of $\mathbb{C}[x]/(x^3-x^2)$ at $(x)$ is a bit larger in some sense (since for example $x$ is not identified with some element of $\mathbb{C}$).

What I would like to know is: how does one translate this algebraic difference (which feels minor to me, because in my mind, both schemes are pairs of points, and if we think about the local rings are local functions, despite the second having more elements in the local ring at $(x)$, there is still only one prime ideal in said local ring to evaluate the functions on, so all functions are still in some sense constant) into a geometrically satisfying picture?


So to simplify the situation, I would suggest throwing out the point $\text{Spec}\, \mathbb{C}[x]/(x-1)$, because it is common to both of the schemes under consideration. You're left with $\text{Spec}\, \mathbb{C}[x]/(x) = \text{Spec}\, \mathbb{C}$ and $\text{Spec}\, \mathbb{C}[x]/(x^2)$. Both of these schemes are points, topologically speaking, as you rightly observe. However, $\text{Spec}\, \mathbb{C}$ is reduced, because its ring of global sections, namely $\mathbb{C}$, has no nonzero nilpotent elements. On the other hand, $\text{Spec}\, \mathbb{C}[x]/(x^2)$ is nonreduced, because the element $x \in \mathbb{C}[x]/(x^2)$ is a nonzero nilpotent: its square is $0$.

How do you view the distinction between $\text{Spec}\, \mathbb{C}[x]/(x) = \text{Spec}\, \mathbb{C}$ and $\text{Spec}\, \mathbb{C}[x]/(x^2)$ geometrically? You can view $\text{Spec}\, \mathbb{C}[x]/(x) = \text{Spec}\, \mathbb{C}$ as a point and $\text{Spec}\, \mathbb{C}[x]/(x^2)$ as a point with a fuzzy ball around it. The fuzz is meant to indicate the nonreducedness of the point.

Alternatively, consider a polynomial $f(x) \in \mathbb{C}[x]$. Then the value of $f(x)$ at $x = 0$ is the constant term of $f(x)$; i.e., it is the image of $f(x)$ under the evaluation map $\mathbb{C}[x] \to \mathbb{C}$ sending $x \to 0$. The value of $f(x)$ under the reduction map $\mathbb{C}[x] \to \mathbb{C}[x]/(x^2)$ tells you not only the constant term of $f(x)$, but also the coefficient of the linear term. In this way, you can think of the nonreduced point as also specifying some linear data, which you can picture geometrically as a little vector based at the point.


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