$Spec(k[x]_{(x)})$ as a shred of curve.

Question

Let $k$ be a field and consider the ring $k[x]$. x is a prime ideal so we can take the localization $k[x]_{(x)}$ in which there are all inverses except for $x$. So we have that $Spec(k[x]_{(x)})=\{(0),(x) \}$. In what sense can I think of it as a shred of a particular nice smooth curve?

Answer 1

If $X$ is a scheme and $x \in X$, then there is a canonical morphism of schemes $\mathrm{Spec}(\mathcal{O}_{X,x}) \to X$. It is a topological embedding, and the image consists of all $y \in X$ such that $x \prec y$, i.e. of all generizations (often also called generalizations) of $x$. In your case, we look at the generizations of $0 \in \mathbb{P}^1$ (or just $0 \in \mathbb{A}^1$). In the case of a curve, a closed point has only two generizations, namely itsself and the generic point. So it's a quite small "shred" of the curve.

Answer 2

Of course $\text{Spec}(k[x]_{(x)})$ is a locally ringed space, and carries more information than just the underlying topological space. In particular, the stalk over the point $x$ (i.e., the prime ideal $(x)$) is $k[x]_{(x)}$; intuitively, this is the ring of germs of functions at the point $x$. That is to say, equivalence classes of pairs $(f, U)$ where $U$ is an open neighborhood of $x$ in the Zariski topology on $\text{Spec}(k[x])$, and $f \in k(x)$ has no poles in $U$, with $(f, U) \sim (g, V)$ if $f$ and $g$ agree on the intersection. So while the underlying space is not too interesting, the whole package is interesting in that it remembers local information of the scheme near $x$.