线性代数 | excellent algebraic space

注：本文为 “excellent algebraic space” 相关讨论。
英文引文，机翻未校。
内嵌索引较多，未全部引入翻译。
如有内容异常，请看原文。

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What is an excellent algebraic space?

什么是优秀代数空间？

What does it mean to say that an algebraic space $S$ is excellent? One knows that excellence of a Noetherian ring is not a property that is etale local (that is, excellence cannot be checked over an etale cover; there are counterexamples in EGA). Thus I wonder: what does excellence actually mean in the context of algebraic spaces?
当我们说一个代数空间 $S$ 是“优秀的（excellent）”时，其含义是什么？我们知道，诺特环（Noetherian ring）的优秀性并非平展局部（etale local）性质——也就是说，无法通过平展覆盖（etale cover）来验证优秀性，在《代数几何基础》（EGA）中存在相关反例。因此我想知道：在代数空间的背景下，优秀性的实际含义究竟是什么？

The question comes from looking at this paper http://imrn.oxfordjournals.org/content/2006/75273 of Max Lieblich. He uses the phrase “excellent algebraic space” five times in the paper without discussing its meaning (as far as I can tell), so I presume the notion is standard. I would be very grateful if someone could explain what it means.
这个问题源于阅读马克斯·利布利希（Max Lieblich）的一篇论文。据我观察，他在论文中五次使用了“优秀代数空间（excellent algebraic space）”这一表述，却未对其含义进行说明，因此我推测这是一个标准概念。若有人能解释其含义，我将不胜感激。

asked May 1, 2015 at 16:12
O-Ren Ishii

A variant of your question would be: Is being excellent an etale local property of schemes? An affirmative answer to this would give an unambiguous meaning to ‘excellent algebraic space’. Excellent is made of three conditions - G-ring + J2 + universally catenary. (see stacks.math.columbia.edu/tag/07QS) Unfortunately, the property of being universally catenary does not seem to be an etale local property (see stacks.math.columbia.edu/tag/0355).
你的问题可衍生出另一个版本：优秀性是否为概形（schemes）的平展局部性质？若该问题能得到肯定回答，“优秀代数空间”的含义便会清晰明确。优秀性由三个条件构成，即 G-环（G-ring）、J2 条件（J2 condition）与泛链条件（universally catenary）。遗憾的是，泛链性质似乎并非平展局部性质。

Amit H
Commented May 2, 2015 at 10:45

Let $X$ be a Noetherian algebraic space.
设 $X$ 为一个诺特代数空间（Noetherian algebraic space）。

We say $X$ is quasi-excellent if the following equivalent conditions hold: (1) for every scheme $U$ and etale morphism $U\to X$ the scheme $U$ is quasi-excellent, and (2) for some scheme $U$ and surjective etale morphism $U\to X$ the scheme $U$ is quasi-excellent.
若满足以下两个等价条件，则称 $X$ 为“拟优秀的（quasi-excellent）”：（1）对于任意概形 $U$ 及平展态射 $U\to X$，概形 $U$ 是拟优秀的；（2）存在某个概形 $U$ 及满平展态射 $U\to X$，使得概形 $U$ 是拟优秀的。

We say $X$ is universally catenary if for every quasi-separated morphism $Y\to X$ locally of finite type, the topological space $|Y|$ of $Y$ is catenary. (Note that the space $|Y|$ is a sober locally Noetherian topological space.)
若对于任意拟分离（quasi-separated）且局部有限型（locally of finite type）的态射 $Y\to X$，$Y$ 的拓扑空间 $|Y|$ 是链状的（catenary），则称 $X$ 满足“泛链条件（universally catenary）”。（注：空间 $|Y|$ 是一个清醒的（sober）局部诺特拓扑空间。）

We say $X$ is excellent if it is quasi-excellent and universally catenary.
若 $X$ 既为拟优秀的，又满足泛链条件，则称 $X$ 是“优秀的（excellent）”。

Discussion: The problem with this definition is that it is hard to check (so it is not clear that there exist excellent algebraic spaces, besides the ones we know about, namely algebraic spaces of finite type over an excellent base scheme). On the other hand, everywhere in Max’s paper you can replace excellent by quasi-excellent. In fact, in order to use Artin’s results on representability all you need is to work over a base where the local rings are G-rings. Moreover, to establish a stack is algebraic, you may work etale locally on the base (if I understand correctly, this is how the algebraic space you are talking about arises in his paper), hence you can always assume the base is a scheme.
讨论：该定义存在一个问题，即难以验证——除了我们已知的、在优秀基概形（excellent base scheme）上有限型的代数空间外，目前尚不清楚是否存在其他优秀代数空间。不过，在马克斯的论文中，所有“优秀的”表述实际上都可替换为“拟优秀的”。事实上，若要应用阿廷（Artin）关于可表性的结论，只需在局部环为 G-环的基上进行研究即可。此外，要证明一个层（stack）是代数的，可在基上进行平展局部研究（若我理解无误，你所提及的代数空间在其论文中正是通过这种方式构造的），因此始终可假设基为一个概形。

answered May 8, 2015 at 12:08
answered May 8, 2015 at 12:08
Dracula

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15.52 Excellent rings

15.52 优秀环（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.
本节我们讨论格罗滕迪克（Grothendieck）提出的“优秀环”概念。关于 G-环、J-2 环和泛链环（universally catenary rings）的定义，可参考定义 15.50.1、定义 15.47.1 以及《代数学》（Algebra）中的定义 10.105.3。

Definition 10.105.3. A Noetherian ring R is said to be universally catenary if every R-algebra of finite type is catenary.
若一个诺特环（Noetherian ring）$R$ 上的所有有限型 $R$-代数（$R$-algebra of finite type）均为链环（catenary ring），则称 $R$ 为泛链环（universally catenary ring）。

注：“链环（catenary ring）”是指环中任意两个素理想之间的所有极大素理想链长度相等，而“泛链环”进一步要求其所有有限型扩张代数均满足这一链长相等性，是交换代数中刻画环维度性质的重要概念。

Definition 15.52.1

定义 15.52.1

Let $R$ be a ring.
设 $R$ 为一个环。

• We say $R$ is quasi-excellent if $R$ is Noetherian, a G-ring, and J-2.
  若 $R$ 是诺特环（Noetherian ring）、G-环且满足 J-2 条件，则称 $R$ 为拟优秀环（quasi-excellent ring）。
• We say $R$ is excellent if $R$ is quasi-excellent and universally catenary.
  若 $R$ 是拟优秀环且为泛链环，则称 $R$ 为优秀环（excellent ring）。

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 15.110) 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 $R$.
因此，一个诺特环若满足以下两个条件，则为拟优秀环：1）其形式纤维（formal fibres）是几何正则的（geometrically regular）；2）其上任意有限型代数（finite type algebra）的奇异集（singular set）是闭集。而该环要成为优秀环，还需额外满足（局部）存在一个“良好的维数函数（good dimension function）”。后续我们将在 15.110 节看到：“泛链环”这一性质可等价表述为：对 $R$ 的任意极大理想 $\mathfrak{m}$，局部环同态 $R_{\mathfrak{m}}\to R_{\mathfrak{m}}^{\wedge }$（$R_{\mathfrak{m}}^{\wedge }$ 为 $R_{\mathfrak{m}}$ 的完备化）满足特定条件。

Lemma 15.52.2

引理 15.52.2

Any localization of a finite type ring over a (quasi-)excellent ring is (quasi-)excellent.
（拟）优秀环上有限型环的任意局部化（localization）仍是（拟）优秀环。

Proof

证明

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. $\square$
对于有限型代数，其满足 J-2 条件和泛链性的结论可由定义直接推出；关于 G-环的相关性质，参见命题 15.50.10。局部化保持（拟）优秀性的证明此处略去。$\square$

Proposition 15.52.3

命题 15.52.3

The following types of rings are excellent:
以下类型的环均为优秀环：

• fields,
  域；

• Noetherian complete local rings,
  诺特完备局部环（Noetherian complete local rings）；

• $\mathbf{Z}$ (the ring of integers),
  整数环 $\mathbf{Z}$；

• Dedekind domains with fraction field of characteristic zero,
  分式域（fraction field）特征为 0 的戴德金整环（Dedekind domains）；

• finite type ring extensions of any of the above.
  上述任意一类环的有限型环扩张（finite type ring extensions）。

Proof

证明

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. $\square$
参见命题 15.50.12 和 15.48.7，可知上述环均为 G-环且满足 J-2 条件。任意科恩 - 麦考利环（Cohen-Macaulay ring）都是泛链环（参见《代数学》引理 10.105.9）；特别地，域、戴德金环，以及更一般的正则环（regular rings）均为泛链环。借助科恩结构定理（Cohen structure theorem）可证：完备局部环是泛链环（参见《代数学》注记 10.160.9）。$\square$

The material developed above has some consequences for Nagata rings.
上述内容对 Nagata 环（永田环）有若干推论。

Lemma 15.52.4

引理 15.52.4

Let $(A,\mathfrak{m})$ be a Noetherian local ring. The following are equivalent:
设 $(A,\mathfrak{m})$ 为诺特局部环，则以下两个条件等价：

1. $A$ is Nagata, and
   $A$ 是 Nagata 环；

2. the formal fibres of $A$ are geometrically reduced.
   $A$ 的形式纤维是几何既约的（geometrically reduced）。

Proof

证明

Assume (2). By Algebra, Lemma 10.162.14 we have to show that if $A\to B$ is finite, $B$ is a domain, and ${\mathfrak{m}}^{\prime }\subset B$ is a maximal ideal, then $B_{{\mathfrak{m}}^{\prime }}$ is analytically unramified. Combining Lemmas 15.51.9 and 15.51.4 and Proposition 15.51.5 we see that the formal fibres of $B_{{\mathfrak{m}}^{\prime }}$ are geometrically reduced. In particular $B_{{\mathfrak{m}}^{\prime }}^{\wedge }{\otimes }_{B}L$ is reduced where $L$ is the fraction field of $B$. It follows that $B_{{\mathfrak{m}}^{\prime }}^{\wedge }$ is reduced, i.e., $B_{{\mathfrak{m}}^{\prime }}$ is analytically unramified.
假设条件（2）成立。由《代数学》引理 10.162.14，需证明：若 $A\to B$ 是有限同态、$B$ 是整环（domain），且 ${\mathfrak{m}}^{\prime }\subset B$ 是极大理想，则局部环 $B_{{\mathfrak{m}}^{\prime }}$ 是解析无分歧的（analytically unramified）。结合引理 15.51.9、15.51.4 及命题 15.51.5 可知，$B_{{\mathfrak{m}}^{\prime }}$ 的形式纤维是几何既约的；特别地，若 $L$ 是 $B$ 的分式域，则张量积 $B_{{\mathfrak{m}}^{\prime }}^{\wedge }{\otimes }_{B}L$ 是既约环（reduced ring）。由此可推出 $B_{{\mathfrak{m}}^{\prime }}^{\wedge }$ 是既约环，即 $B_{{\mathfrak{m}}^{\prime }}$ 是解析无分歧的。

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 $x_1,\dots ,x_n\in K$ which generate $K$ over $\kappa (\mathfrak{q})$ and which satisfy monic polynomials $P_i(T)=T^{d_i}+a_{i,1}{T}^{d_i-1}+\dots +a_{i,d_i}=0$ with $a_{i,j}\in \mathfrak{m}$. Then let $B$ be the $A$-subalgebra of $K$ generated by $x_1,\dots ,x_n$. (For more details see the proof of Algebra, Lemma 10.162.14.) Then
假设条件（1）成立。设 $\mathfrak{q}\subset A$ 为素理想，且 $K/\kappa (\mathfrak{q})$ 为有限扩张（$\kappa (\mathfrak{q})=A/\mathfrak{q}$ 为剩余域），需证明 $A^{\wedge }{\otimes }_{A}K$ 是既约环。取局部子环 $B$ 满足 $A/\mathfrak{q}\subset B\subset K$，其中 $B$ 在 $A$ 上有限，且 $B$ 的分式域为 $K$。$B$ 的构造方式如下：选取 $x_1,\dots ,x_n\in K$，使得 $x_1,\dots ,x_n$ 生成扩张 $K/\kappa (\mathfrak{q})$，且每个 $x_i$ 满足首一多项式 $P_i(T)=T^{d_i}+a_{i,1}{T}^{d_i-1}+\dots +a_{i,d_i}=0$（其中 $a_{i,j}\in \mathfrak{m}$）；令 $B$ 为由 $x_1,\dots ,x_n$ 生成的 $K$ 的 $A$-子代数（细节参见《代数学》引理 10.162.14 的证明）。此时有：

$A^{\wedge }{\otimes }_{A}K=(A^{\wedge }{\otimes }_{A}B{)}_{\mathfrak{q}}=B_{\mathfrak{q}}^{\wedge }$

Since $B^{\wedge }$ is reduced by Algebra, Lemma 10.162.14 the proof is complete. $\square$
由《代数学》引理 10.162.14 可知 $B^{\wedge }$ 是既约环，故证明完成。$\square$

Lemma 15.52.5

引理 15.52.5

A quasi-excellent ring is Nagata.
拟优秀环是 Nagata 环。

Proof

证明

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. $\square$
设 $R$ 为拟优秀环。由引理 15.52.2 可知，$R$ 上的有限型代数仍是拟优秀环；结合《代数学》引理 10.162.3，只需证明“任意拟优秀整环是 N-1 环”即可。应用《代数学》引理 10.161.15（并利用“拟优秀环满足 J-2 条件”），可进一步将问题简化为“证明拟优秀局部整环 $R$ 是 N-1 环”。由于同态 $R\to R^{\wedge }$ 是正则的（regular），由引理 15.42.1 可知 $R^{\wedge }$ 是既约环，即 $R$ 是解析无分歧的。再由《代数学》引理 10.162.10 可得 $R$ 是 N-1 环。故证明完成。$\square$

Lemma 15.52.6

引理 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.
设 $(A,\mathfrak{m})$ 为诺特局部环。若 $A$ 是正规环（normal ring），且 $A$ 的形式纤维是正规的（例如 $A$ 是优秀环或拟优秀环时），则 $A$ 的完备化 $A^{\wedge }$ 是正规环。

Proof

证明

Follows immediately from Algebra, Lemma 10.163.8. $\square$
直接由《代数学》引理 10.163.8 可得。$\square$

Lemma 10.163.8 (0C22)—The Stacks project

引理 10.163.8（0C22）

Lemma 10.163.8. Let $\phi :R\to S$ be a ring map. Assume
引理 10.163.8. 设 $\phi :R\to S$ 为一环同态。若满足以下条件：

1. $R$ is Noetherian,
   $R$ 是诺特环（Noetherian ring）；

2. $S$ is Noetherian,
   $S$ 是诺特环；

3. $\phi$ is flat,
   $\phi$ 是平坦同态（flat homomorphism）；

4. the fibre rings $S{\otimes }_R\kappa (\mathfrak{p})$ are normal, and
   纤维环（fibre ring）$S{\otimes }_R\kappa (\mathfrak{p})$ 是正规环（normal ring）（其中 $\kappa (\mathfrak{p})=R/\mathfrak{p}$ 为 $R$ 在素理想 $\mathfrak{p}$ 处的剩余域）；

5. $R$ is normal.
   $R$ 是正规环。

Then $S$ is normal.
则 $S$ 是正规环。

Proof

证明

For a Noetherian ring being normal is the same as having properties $(S_2)$ and $(R_1)$, see Lemma 10.157.4.
对诺特环而言，“是正规环”等价于“同时满足 $(S_2)$ 性质和 $(R_1)$ 性质”，参见引理 10.157.4（注：$(S_2)$ 为塞尔深度条件，$(R_1)$ 为正则性条件，二者是判断诺特环正规性的核心准则）。

Thus we know $R$ and the fibre rings have these properties.
由此可知，$R$（条件 5）和纤维环 $S{\otimes }_R\kappa (\mathfrak{p})$（条件 4）均满足 $(S_2)$ 和 $(R_1)$ 性质。

Hence we may apply Lemmas 10.163.4 and 10.163.5 and we see that $S$ is $(S_2)$ and $(R_1)$, in other words normal by Lemma 10.157.4 again. $\square$
因此可应用引理 10.163.4 和 10.163.5（注：这两个引理分别给出“平坦同态下 $(S_2)$ 性质的传递性”和“平坦同态下 $(R_1)$ 性质的传递性”），推出 $S$ 满足 $(S_2)$ 和 $(R_1)$ 性质；再由引理 10.157.4 可知，$S$ 是正规环。$\square$

Remark 10.164.8 (0355)—The Stacks project

注记 10.164.8（0355）

The property of being “universally catenary” does not descend; not even along étale ring maps.
“泛链性（universally catenary）”这一性质不具有下降性（descend）；即便沿着平展环同态（étale ring map），该性质也无法下降。

In Examples, Section 110.19 there is a construction of a finite ring map $A\to B$ with $A$ local Noetherian and not universally catenary, $B$ semi-local with two maximal ideals $\mathfrak{m}$, $\mathfrak{n}$ with $B_{\mathfrak{m}}$ and $B_{\mathfrak{n}}$ regular of dimension 2 and 1 respectively, and the same residue fields as that of $A$.
在“例子（Examples）”部分的 110.19 节中，构造了一个有限环同态 $A\to B$，满足以下条件：$A$ 是诺特局部环（local Noetherian ring）且非泛链环；$B$ 是半局部环（semi-local ring），有两个极大理想 $\mathfrak{m}$ 和 $\mathfrak{n}$，其中局部环 $B_{\mathfrak{m}}$ 是维数为 2 的正则环（regular ring），$B_{\mathfrak{n}}$ 是维数为 1 的正则环，且二者与 $A$ 具有相同的剩余域（residue field）。

Moreover, ${\mathfrak{m}}_A$ generates the maximal ideal in both $B_{\mathfrak{m}}$ and $B_{\mathfrak{n}}$ (so $A\to B$ is unramified as well as finite).
此外，$A$ 的极大理想 ${\mathfrak{m}}_A$ 在 $B_{\mathfrak{m}}$ 和 $B_{\mathfrak{n}}$ 中均生成其极大理想（因此环同态 $A\to B$ 既是有限的（finite），也是无分歧的（unramified））。

By Lemma 10.152.3 there exists a local étale ring map $A\to A^{\prime }$ such that $B{\otimes }_A{A}^{\prime }=B_1\times B_2$ decomposes with $A^{\prime }\to B_i$ surjective.
由引理 10.152.3 可知，存在局部平展环同态 $A\to A^{\prime }$，使得张量积 $B{\otimes }_A{A}^{\prime }$ 分解为直积 $B_1\times B_2$，且同态 $A^{\prime }\to B_i$（$i=1,2$）是满射。

This shows that $A^{\prime }$ has two minimal primes ${\mathfrak{q}}_i$ with $A^{\prime }/{\mathfrak{q}}_i\cong B_i$.
这表明 $A^{\prime }$ 有两个极小素理想 ${\mathfrak{q}}_i$（$i=1,2$），且商环 $A^{\prime }/{\mathfrak{q}}_i\cong B_i$。

Since $B_i$ is regular local (since it is étale over either $B_{\mathfrak{m}}$ or $B_{\mathfrak{n}}$) we conclude that $A^{\prime }$ is universally catenary.
由于 $B_i$ 是正则局部环（因为 $B_i$ 是 $B_{\mathfrak{m}}$ 或 $B_{\mathfrak{n}}$ 上的平展环），故可推出 $A^{\prime }$ 是泛链环。

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via：

• ag.algebraic geometry - What is an excellent algebraic space? - MathOverflow
  https://mathoverflow.net/questions/204458/what-is-an-excellent-algebraic-space
• Section 15.52 (07QS): Excellent rings—The Stacks project
  https://stacks.math.columbia.edu/tag/07QS
• Remark 10.164.8 (0355)—The Stacks project
  https://stacks.math.columbia.edu/tag/0355