Lemma 29.6.5 (02JQ)—The Stacks project

Lemma 29.6.5. Let $f : X \to Y$ be a quasi-compact morphism. Let $Z$ be the scheme theoretic image of $f$ . Let $z \in Z$ ¹. There exists a valuation ring $A$ with fraction field $K$ and a commutative diagram

$\xymatrix{ \mathop{\mathrm{Spec}}(K) \ar[rr] \ar[d] & & X \ar[d] \ar[ld] \\ \mathop{\mathrm{Spec}}(A) \ar[r] & Z \ar[r] & Y }$

such that the closed point of $\mathop{\mathrm{Spec}}(A)$ maps to $z$ . In particular any point of $Z$ is the specialization of a point of $f(X)$ .

Proof. Let $z \in \mathop{\mathrm{Spec}}(R) = V \subset Y$ be an affine open neighbourhood of $z$ . By Lemma 29.6.3 the intersection $Z \cap V$ is the scheme theoretic image of $f^{-1}(V) \to V$ . Hence we may replace $Y$ by $V$ and assume $Y = \mathop{\mathrm{Spec}}(R)$ is affine. In this case $X$ is quasi-compact as $f$ is quasi-compact. Say $X = U_1 \cup \ldots \cup U_ n$ is a finite affine open covering. Write $U_ i = \mathop{\mathrm{Spec}}(A_ i)$ . Let $I = \mathop{\mathrm{Ker}}(R \to A_1 \times \ldots \times A_ n)$ . By Lemma 29.6.3 again we see that $Z$ corresponds to the closed subscheme $\mathop{\mathrm{Spec}}(R/I)$ of $Y$ . If $\mathfrak p \subset R$ is the prime corresponding to $z$ , then we see that $I_{\mathfrak p} \subset R_{\mathfrak p}$ is not an equality. Hence (as localization is exact, see Algebra, Proposition 10.9.12) we see that $R_{\mathfrak p} \to (A_1)_{\mathfrak p} \times \ldots \times (A_ n)_{\mathfrak p}$ is not zero. Hence one of the rings $(A_ i)_{\mathfrak p}$ is not zero. Hence there exists an $i$ and a prime $\mathfrak q_ i \subset A_ i$ lying over a prime $\mathfrak p_ i \subset \mathfrak p$ . By Algebra, Lemma 10.50.2 we can choose a valuation ring $A \subset K = \kappa (\mathfrak q_ i)$ dominating the local ring $R_{\mathfrak p}/\mathfrak p_ iR_{\mathfrak p} \subset \kappa (\mathfrak q_ i)$ . This gives the desired diagram. Some details omitted. $\square$

[1] By Lemma 29.6.3 set-theoretically

$Z$ agrees with the closure of

$f(X)$ in

$Y$ .

Comments (5)

Comment #2762 by BCnrd on August 10, 2017 at 16:35

The notation $f.f.$ to denote "fraction field" (twice) near the end of the proof seems kind of awful (not only because in the argument there is a map called $f$ ). Is this really the standard notation in this document? Why not ${\rm{Frac}}$ instead?

Comment #2765 by Dario Weißmann on August 10, 2017 at 20:58

Why not use the notation $Q(-)$ introduced in Example 10.9.8 (3)? Or - as in this case we are talking about the residue field at a prime - use the notation $\kappa(-)$ introduced in 25.2.1 (2)?

Comment #2766 by BCnrd on August 11, 2017 at 00:18

Dario's suggestion of $\kappa(\mathfrak{p})$ for the fraction field of $A/\mathfrak{p}$ seems fine since such notation has been standard from EGA, and I think the meaning of ${\rm{Frac}}$ is also immediately recognized by anyone, but I am not a fan of $Q(\cdot)$ to denote the fraction field of a domain since it is not sufficiently universally known and so potentially a bit obscure (though admittedly anyone who has actually understood the proof up to there can infer what it must mean). In a work this massive, which essentially nobody reads linearly, it is best not to use slightly non-standard notation unless there is an easily-identified and systematically maintained list of all such notation, but alas there seems to be no such list; the Notation section in the "Preliminaries" Part is a bit brief!)

Comment #2767 by sdf on August 11, 2017 at 17:03

There is a fairly standard notation of $\mathrm{ff}$ or $\mathrm{FF}$ or $\mathrm{FoF}$ for field of fractions at least in my part of the world. But having a full-stop as part of the notation is aesthetically less than ideal, this has been pointed out before I think.

Comment #2875 by Johan on October 05, 2017 at 20:04

OK already! I have now changed this by not having any notation for the fraction field of a domain. It turns out it isn't that frequently used and often it is used (as in the current situation) when we are taking the residue field at a prime. See here to see why this kind of change is painful.

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