# Definition of a one point subscheme and fibre at a point

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.

• It might help to think of the motivating example of varieties over an algebrically closed fields $k$ -- let's even restrict to affine varieties (the question is local on the target... it's not local on the source, but we're just looking for motivation). The fiber of a map $X \to Y$ of affine varieties over a point $p$ of $Y$ ought to be, as a set, the pre-image. It's worth checking that tensoring with the residue field gives you this, whereas tensoring with some big local ring does not. Feb 21 at 20:04

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!