The fibre of an $S$-scheme $X$ over $s\in S$ is defined as the fibre product $X\times_S \{s\}$ where the one point subscheme $\{s\}$ is given a local ring $k(s)$ equal to the residue field at $s$. This makes $\{s\}$ into an $S$-scheme $Spec(k(s))$.
My question is: is there any special reason for choosing this particular structure $(s,k(s))$? Are there any other local rings we could attach to $\{s\}$ to make it into a scheme and so that the fibre product makes sense?
I get that there is a morphism of schemes $Spec(k(s))\rightarrow S$ which is induced by a homomorphism of rings which is just reduction at the maximal ideal of the local ring $\mathcal{O}_{S,s}\rightarrow \mathcal{O}_{S,s}/\mathcal{M}_{S,s}$, at least in the affine case. In this way I suppose one could think of the resulting $S$-scheme $Spec(k(s))$ as the most natural one.
Best Answer
You absolutely can also consider the local ring $\mathcal{O}_{S,s}$, and produce a different scheme $X \times_S \operatorname{Spec}(\mathcal{O}_{S,s})$.
One nice thing about $X \times_S \operatorname{Spec} k(s)$ is that this is a $k(s)$-scheme -- schemes over fields are nice!
A more important nice thing about $X \times_S \operatorname{Spec} k(s)$ is that the points of this scheme (as a topological space!) are in bijection with the points of $X$ lying over $s$, which is not true of $X \times_S \operatorname{Spec}(\mathcal{O}_{S,s})$ in general.
For example, let $S = \operatorname{Spec} \mathbb{Z}$ and let $X = \operatorname{Spec} \mathbb{Z}[\frac{1}{2}]$. In other words, $X$ is the complement of the point $(2)$ in $S$.
Let's pick $s = (2)$. Note that $X \times_S \operatorname{Spec}(k(s)) = \operatorname{Spec}(\mathbb{Z}[\frac{1}{2}] \otimes \mathbb{Z}/2) = \operatorname{Spec}(0) = \varnothing$, as desired because $X$ is disjoint from $(2)$. On the other hand, $X \times_S \operatorname{Spec}(\mathcal{O}_{S,s}) = \operatorname{Spec}(\mathbb{Z}[\frac{1}{2}] \otimes \mathbb{Z}_{(2)}) = \operatorname{Spec}(\mathbb{Q})$ is nonempty.
It will, however, be true that $X \times_S \operatorname{Spec}(k(s)) \to X$ and $X \times_S \operatorname{Spec}(\mathcal{O}_{S,s}) \to X$ are homeomorphisms onto their image. See https://stacks.math.columbia.edu/tag/01JW for details!