Definition 15.52.1. Let $R$ be a ring.
We say $R$ is quasi-excellent if $R$ is Noetherian, a G-ring, and J-2.
We say $R$ is excellent if $R$ is quasi-excellent and universally catenary.
15.52 Excellent rings
In this section we discuss Grothendieck's notion of excellent rings. For the definitions of G-rings, J-2 rings, and universally catenary rings we refer to Definition 15.50.1, Definition 15.47.1, and Algebra, Definition 10.105.3.
Thus a Noetherian ring is quasi-excellent if it has geometrically regular formal fibres and if any finite type algebra over it has closed singular set. For such a ring to be excellent we require in addition that there exists (locally) a good dimension function. We will see later (Section 15.109) that to be universally catenary can be formulated as a condition on the maps $R_\mathfrak m \to R_\mathfrak m^\wedge $ for maximal ideals $\mathfrak m$ of $R$.
Lemma 15.52.2. Any localization of a finite type ring over a (quasi-)excellent ring is (quasi-)excellent.
Proof. For finite type algebras this follows from the definitions for the properties J-2 and universally catenary. For G-rings, see Proposition 15.50.10. We omit the proof that localization preserves (quasi-)excellency. $\square$
Proposition 15.52.3. The following types of rings are excellent:
fields,
Noetherian complete local rings,
$\mathbf{Z}$,
Dedekind domains with fraction field of characteristic zero,
finite type ring extensions of any of the above.
Proof. See Propositions 15.50.12 and 15.48.7 to see that these rings are G-rings and have J-2. Any Cohen-Macaulay ring is universally catenary, see Algebra, Lemma 10.105.9. In particular fields, Dedekind rings, and more generally regular rings are universally catenary. Via the Cohen structure theorem we see that complete local rings are universally catenary, see Algebra, Remark 10.160.9. $\square$
The material developed above has some consequences for Nagata rings.
Lemma 15.52.4. Let $(A, \mathfrak m)$ be a Noetherian local ring. The following are equivalent
$A$ is Nagata, and
the formal fibres of $A$ are geometrically reduced.
Proof. Assume (2). By Algebra, Lemma 10.162.14 we have to show that if $A \to B$ is finite, $B$ is a domain, and $\mathfrak m' \subset B$ is a maximal ideal, then $B_{\mathfrak m'}$ is analytically unramified. Combining Lemmas 15.51.9 and 15.51.4 and Proposition 15.51.5 we see that the formal fibres of $B_{\mathfrak m'}$ are geometrically reduced. In particular $B_{\mathfrak m'}^\wedge \otimes _ B L$ is reduced where $L$ is the fraction field of $B$. It follows that $B_{\mathfrak m'}^\wedge $ is reduced, i.e., $B_{\mathfrak m'}$ is analytically unramified.
Assume (1). Let $\mathfrak q \subset A$ be a prime ideal and let $K/\kappa (\mathfrak q)$ be a finite extension. We have to show that $A^\wedge \otimes _ A K$ is reduced. Let $A/\mathfrak q \subset B \subset K$ be a local subring finite over $A$ whose fraction field is $K$. To construct $B$ choose $x_1, \ldots , x_ n \in K$ which generate $K$ over $\kappa (\mathfrak q)$ and which satisfy monic polynomials $P_ i(T) = T^{d_ i} + a_{i, 1} T^{d_ i - 1} + \ldots + a_{i, d_ i} = 0$ with $a_{i, j} \in \mathfrak m$. Then let $B$ be the $A$-subalgebra of $K$ generated by $x_1, \ldots , x_ n$. (For more details see the proof of Algebra, Lemma 10.162.14.) Then
\[ A^\wedge \otimes _ A K = (A^\wedge \otimes _ A B)_\mathfrak q = B^\wedge _\mathfrak q \]
Since $B^\wedge $ is reduced by Algebra, Lemma 10.162.14 the proof is complete. $\square$
Lemma 15.52.5. A quasi-excellent ring is Nagata.
Proof. Let $R$ be quasi-excellent. Using that a finite type algebra over $R$ is quasi-excellent (Lemma 15.52.2) we see that it suffices to show that any quasi-excellent domain is N-1, see Algebra, Lemma 10.162.3. Applying Algebra, Lemma 10.161.15 (and using that a quasi-excellent ring is J-2) we reduce to showing that a quasi-excellent local domain $R$ is N-1. As $R \to R^\wedge $ is regular we see that $R^\wedge $ is reduced by Lemma 15.42.1. In other words, $R$ is analytically unramified. Hence $R$ is N-1 by Algebra, Lemma 10.162.10. $\square$
Lemma 15.52.6. Let $(A, \mathfrak m)$ be a Noetherian local ring. If $A$ is normal and the formal fibres of $A$ are normal (for example if $A$ is excellent or quasi-excellent), then $A^\wedge $ is normal.
Proof. Follows immediately from Algebra, Lemma 10.163.8. $\square$
Post a comment
Your email address will not be published. Required fields are marked.
In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the toolbar).
All contributions are licensed under the GNU Free Documentation License.
Comments (2)
Comment #8664 by Owen B on
Comment #9394 by Stacks project on