The Stacks project

\begin{equation*} \DeclareMathOperator\Coim{Coim} \DeclareMathOperator\Coker{Coker} \DeclareMathOperator\Ext{Ext} \DeclareMathOperator\Hom{Hom} \DeclareMathOperator\Im{Im} \DeclareMathOperator\Ker{Ker} \DeclareMathOperator\Mor{Mor} \DeclareMathOperator\Ob{Ob} \DeclareMathOperator\Sh{Sh} \DeclareMathOperator\SheafExt{\mathcal{E}\mathit{xt}} \DeclareMathOperator\SheafHom{\mathcal{H}\mathit{om}} \DeclareMathOperator\Spec{Spec} \newcommand\colim{\mathop{\mathrm{colim}}\nolimits} \newcommand\lim{\mathop{\mathrm{lim}}\nolimits} \newcommand\Qcoh{\mathit{Qcoh}} \newcommand\Sch{\mathit{Sch}} \newcommand\QCohstack{\mathcal{QC}\!\mathit{oh}} \newcommand\Cohstack{\mathcal{C}\!\mathit{oh}} \newcommand\Spacesstack{\mathcal{S}\!\mathit{paces}} \newcommand\Quotfunctor{\mathrm{Quot}} \newcommand\Hilbfunctor{\mathrm{Hilb}} \newcommand\Curvesstack{\mathcal{C}\!\mathit{urves}} \newcommand\Polarizedstack{\mathcal{P}\!\mathit{olarized}} \newcommand\Complexesstack{\mathcal{C}\!\mathit{omplexes}} \newcommand\Pic{\mathop{\mathrm{Pic}}\nolimits} \newcommand\Picardstack{\mathcal{P}\!\mathit{ic}} \newcommand\Picardfunctor{\mathrm{Pic}} \newcommand\Deformationcategory{\mathcal{D}\!\mathit{ef}} \end{equation*}

Lemma 28.33.11. Fibres of unramified morphisms.

  1. Let $X$ be a scheme over a field $k$. The structure morphism $X \to \mathop{\mathrm{Spec}}(k)$ is unramified if and only if $X$ is a disjoint union of spectra of finite separable field extensions of $k$.

  2. If $f : X \to S$ is an unramified morphism then for every $s \in S$ the fibre $X_ s$ is a disjoint union of spectra of finite separable field extensions of $\kappa (s)$.

Proof. Part (2) follows from part (1) and Lemma 28.33.5. Let us prove part (1). We first use Algebra, Lemma 10.147.7. This lemma implies that if $X$ is a disjoint union of spectra of finite separable field extensions of $k$ then $X \to \mathop{\mathrm{Spec}}(k)$ is unramified. Conversely, suppose that $X \to \mathop{\mathrm{Spec}}(k)$ is unramified. By Algebra, Lemma 10.147.5 for every $x \in X$ the residue field extension $k \subset \kappa (x)$ is finite separable. Since $X \to \mathop{\mathrm{Spec}}(k)$ is locally quasi-finite (Lemma 28.33.10) we see that all points of $X$ are isolated closed points, see Lemma 28.19.6. Thus $X$ is a discrete space, in particular the disjoint union of the spectra of its local rings. By Algebra, Lemma 10.147.5 again these local rings are fields, and we win. $\square$


Comments (3)

Comment #2245 by comment on

I do not quite follow how one concludes in the proof that X is discrete. Does one apply that from Tag 00PJ by using the fact that the local ring at is a field (as follows from in Tag 00UW )?

Comment #2247 by JuanPablo on

I didn't see immediately how one concludes that is discrete, either. I think that one can argue as follows:

is unramified so locally of finite type, so is locally Noetherian. So the affine opens of are spectra of Noetherian rings, and by Lemma 10.36.2 (tag 00FR) they're finite discrete. So points are open.

Maybe state that locally Noetherian zero dimensional squemes are discrete as a separate lemma in section 27.10? (it's similar to lemma 27.10.6 tag 0CKV).

Comment #2280 by on

Dear comment and JuanPablo, instead of your suggestions (which are fine too), I've fixed it by using that an unramified morphism is locally quasi-finite and that unramified morphisms have discrete fibres. Here is the commit.


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