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.

  • 1
    $\begingroup$ 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. $\endgroup$
    – hunter
    Feb 21 at 20:04

1 Answer 1


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!


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