Alexandroff Topology of Algebras over an Integral Domain

TL;DR

This paper introduces a topology on the set of $S$-nice subalgebras of an $F$-algebra over an integral domain, enabling the study of their algebraic structure through topological methods.

Contribution

It defines an Alexandroff topology on $S$-nice subalgebras and characterizes irreducible subsets and components within this topological space.

Findings

Irreducible subsets have a supremum in the space.

Characterization of irreducible open sets.

Description of irreducible components.

Abstract

Let $S$ be an integral domain with field of fractions $F$ and let $A$ be an $F$-algebra. An $S$-subalgebra $R$ of $A$ is called $S$-nice if $R$ is lying over $S$ and the localization of $R$ with respect to $S \setminus \{ 0 \}$ is $A$. Let $\mathbb S$ be the set of all $S$-nice subalgebras of $A$. We define a notion of open sets on $\mathbb S$ which makes this set a $T_0$ Alexandroff space. This enables us to study the algebraic structure of $\mathbb S$ from the point of view of topology. We prove that an irreducible subset of $\mathbb S$ has a supremum with respect to the specialization order. We present equivalent conditions for an open set of $\mathbb S$ to be irreducible, and characterize the irreducible components of $\mathbb S$

Equations

$$V(M_1\cup M_2)=V(M_1)\cap V(M_2).$$

$$V({\cup }_{i\in I}{M}_i)={\cap }_{i\in I}V(M_i).$$

$$\bigcap _{i\in I}(\bigcup _{j_i\in J_i}{X}_{i,j_i})=\bigcup _{t\in {\prod }_{i\in I}{J}_i}(\bigcap _{i\in I}{X}_{i,t_i}),$$

$$T\cdot M=\{\sum _{1\le i\le n}{t}_i{m}_i\mid t_i\in T,m_i\in M\}.$$

$$R_{r}=\left(\begin{array}[]{cccccccc}O_{v}&I_{r}\\ J_{1}&O_{v}\\ \end{array}\right),$$

$${\cap }_{R\in U}R={\cap }_{0<r<r_0}{R}_r=R_{r_0}\notin U.$$

$$R^{\prime}=\left(\begin{array}[]{cccccccc}O_{v}&I_{r_{0}}\\ J_{2}&O_{v}\\ \end{array}\right).$$

$$R_{1}=\left(\begin{array}[]{cccccccc}C&C&...&C&I_{1}\\ C&C&...&C&I_{1}\\ .&.&...&.&.\\ .&.&...&.&.\\ .&.&...&.&.\\ C&C&...&C&I_{1}\\ C&C&...&C&C\\ \end{array}\right)\subset R_{2}=\left(\begin{array}[]{cccccccc}C&C&...&C&I_{2}\\ C&C&...&C&I_{2}\\ .&.&...&.&.\\ .&.&...&.&.\\ .&.&...&.&.\\ C&C&...&C&I_{2}\\ C&C&...&C&C\\ \end{array}\right)\subset...$$

$$I={\cup }_{n\in \mathbb{N}}cl\{R_n\};$$

Full text

1 Introduction and some preliminary results

As the title suggests, in this paper we discuss algebras over integral domains from the point of view of Alexandroff topology, which will shortly be defined. In [Sa1] and [Sa2] we studied algebras over valuation domains, concentrating on quasi-valuations that extend a valuation on a field. In [Sa3] we prove several existence theorems of integral domains that may be applied to the study of quasi-valuations. More specifically, let $S$ be an integral domain which is not a field, let $F$ be its field of fractions, and let $A\neq F$ be an $F$-algebra. In [Sa3] we study $S$-subalgebras of $A$, that are lying over $S$ and whose localizations with respect to $S\setminus\{0\}$ is $A$. We call them $S$-nice subalgebras of $A$. Namely, an $S$-subalgebra $R$ of $A$ is called an $S$-nice subalgebra of $A$ if $R\cap F=S$ and $FR=A$; we shall use this notation throughout this paper. We denote by $\mathbb{S}$ the set of all $S$-nice subalgebras of $A$.

We recall now some definitions and results from [Sa3]. The following concept is used quite frequently: let $B$ be a basis of $A$ over $F$. We say that $B$ is $S$-stable if there exists a basis $C$ of $A$ over $F$ such that for all $c\in C$ and $b\in B$, one has $cb\in\sum_{y\in B}Sy$.

We note in [Sa3, Remark 3.4] that if a basis $B$ is closed under multiplication then $B$ is $S$-stable. Thus, for example, every free (noncommutative) $F$-algebra with an arbitrary set of generators has an $S$-stable basis; in particular, every polynomial algebra with an arbitrary set of indeterminates over $F$ has an $S$-stable basis.

We also show in [Sa3, Proposition 3.12] that if $A$ is finite dimensional over $F$, then every basis of $A$ over $F$ is $S$-stable. The first existence theorem is as follows.

Theorem 1.1**.**

(cf. [Sa3, Theorem 3.14]) If there exists an $S$-stable basis of $A$ over $F$, then there exists an $S$-nice subalgebra of $A$.

In particular, if $A$ is finite dimensional over $F$ then there exists an $S$-nice subalgebra of $A$.

The following result is a going-down lemma for $S$-nice subalgebras.

Lemma 1.2**.**

(cf. [Sa3, Lemma 3.20]) Let $S_{1}\subseteq S_{2}$ be integral domains with field of fractions $F$ such that $S_{2}\neq F$. Assume that there exists an $S_{1}$-stable basis of $A$ over $F$. Let $R$ be an $S_{2}$-nice subalgebra of $A$. Then there exists an $S_{1}$-nice subalgebra of $A$, which is contained in $R$.

We conclude that a minimal $S$-nice subalgebra of $A$ does not exist. More precisely, we prove

Proposition 1.3**.**

(cf. [Sa3, Proposition 3.21]) Assume that there exists an $S$-stable basis of $A$ over $F$. Let $R$ be an $S$-nice subalgebra of $A$. Then there exists an infinite decreasing chain of $S$-nice subalgebras of $A$ starting from $R$. In particular, a minimal $S$-nice subalgebra of $A$ does not exist.

Definition 1.4**.**

Let $\mathcal{C}$ be a chain of prime ideals of $S$ and let $R$ be a faithful $S$-algebra. Let $\mathcal{D}$ be a chain of prime ideals of $R$. We say that $\mathcal{D}$ covers $\mathcal{C}$ if for every $P\in\mathcal{C}$ there exists $Q\in\mathcal{D}$ lying over $P$; namely, $Q\cap S=P$.

Theorem 1.5**.**

(cf. [Sa3, Theorem 3.24]) Assume that there exists an $S$-stable basis of $A$ over $F$. Let $\mathcal{C}$ be a chain of prime ideals of $S$. If $A$ is commutative then there exists an $S$-nice subalgebra $R$ of $A$ such that there exists a chain of prime ideals $\mathcal{F}$ of $R$ covering $\mathcal{C}$. In fact, there exists an infinite descending chain of such $S$-nice subalgebras of $A$.

Note that by Theorem 1.1 and Proposition 1.3 if $A$ contains an $S$-stable basis then $\mathbb{S}$ is not empty and is, in fact, infinite. In particular, by the above-mentioned note before Theorem 1.1, if $A$ is finite dimensional over $F$ then $\mathbb{S}$ is infinite.

In the second section of this paper, which is its main part, we do not assume that $A$ contains an $S$-stable basis; we merely assume that $\mathbb{S}$ is not empty. We do, however, present an example in which the existence of an $S$-stable basis is assumed. In the third section of this paper we assume that $A$ contains an $S$-stable basis

We present now some of the common definitions we use from order theory. Let $P$ be a set. A relation $\leq$ on $P$ that is reflexive and transitive is called a quasi order or a preorder; if the relation is also antisymmetric then it is called a partial order and $P$ is called a partially ordered set, or a poset. Let $L\subseteq P$. We say that $a\in P$ is a lower (resp. upper) bound of $L$ if $a\leq x$ (resp. $a\geq x$) for all $x\in L$. If the set of lower (resp. upper) bounds of $L$ has a unique greatest (resp. smallest) element, this element is called the greatest lower (resp. least upper) bound of $L$, and is denoted by $\text{inf}L$ (resp. $\text{sup}L$). We say that $L$ is a lower set if $L=\{y\in P\mid y\leq x\text{ for some }x\in L\}$. We say that $L$ is an upper set if $L=\{y\in P\mid y\geq x\text{ for some }x\in L\}$. A subset $\emptyset\neq D\subseteq P$ is called directed if for all $a,b\in D$ there exists $c\in D$ such that $a\leq c$ and $b\leq c$. We say that $P$ is a dcpo (directed complete partial order) if every directed subset of $P$ has a supremum. A subset $I$ of $P$ is called an ideal of $P$ if $I$ is a lower set and directed. $P$ is called an inf semilattice (resp. sup semilattice) if for all $a,b\in P$, $\text{inf}\{a,b\}$ (resp. $\text{sup}\{a,b\}$) exists in $P$. If $P$ is both an inf semilattice and a sup semilattice, we say that $P$ is a lattice. A subset $L$ of $P$ is called a sublattice of $P$ if $L$ is a lattice with respect to the partial order of $P$. Similarly, one defines inf subsemilattice and sup subsemilattice.

We briefly discuss now the notion of an Alexandroff topological space. A topological space whose set of open sets is closed under arbitrary intersections is called an Alexandroff space, after P. alexandroff who first introduced such topological spaces in his paper [Al] from 1937. Equivalently, A topological space is called an Alexandroff space if every element has a minimal open set containing it. A finite topological space is the most important particular case of an Alexandroff space. In fact, Alexandroff spaces share many properties with finite topological spaces; in particular, Alexandroff spaces have all the properties of finite spaces relevant for the theory of digital topology (see [He] and [Kr]). Thus, in the eighties the interest in Alexandroff spaces arose as a consequence of the very important role of finite spaces in digital topology. In 1999 F. G. Arenas studied the topological properties of Alexandroff spaces (see [Ar]).

Let $(T,\tau)$ be a topological space. For $X\subseteq T$ we denote by $clX$ the closure of $X$. It is well known and easy to prove, that if one defines $x\leq_{\tau}y$ whenever $x\in cl\{y\}$, then $\leq_{\tau}$ is a quasi order; i.e., a reflexive and transitive relation. $\leq_{\tau}$ is called the specialization order. Recall that $(T,\tau)$ is called $T_{0}$ if for every two distinct elements in $T$, there exists an open set containing one of them but not the other. It is known that if $(T,\tau)$ is $T_{0}$ then $\leq_{\tau}$ is a partial order. On the other hand, for any quasi order $\leq$ on a set $T$, one can define the topology whose open sets are the upper sets of $T$ with respect to $\leq$, denote it by $A_{\leq}$. So, there are two functors, the specialization order from the class of all topological spaces to the class of all quasi ordered sets, sending $(T,\tau)$ to $(T,\leq_{\tau})$; and the functor in the opposite direction sending $(T,\leq)$ to $(T,A_{\leq})$. If one restrict the class of all topological spaces to the the class of all alexandroff topological spaces, then one has an isomorphism between the categories.

In this paper the symbol $\subset$ means proper inclusion and the symbol $\subseteq$ means inclusion or equality.

2 The Alexandroff Topology

Inspired by the Zarisky topology on the prime spectrum of a ring, for every $M\subseteq A$ we denote by $V(M)$ the set of all $S$-nice subalgebras of $A$ containing $M$. It is easy to see that $V(\{0\})=\mathbb{S}$, $V(F)=\emptyset$, and for every $M_{1},M_{2}\subseteq A$, we have

[TABLE]

Thus, the set B=$\{V(M)\}_{M\subseteq A}$ is a basis for a topology on $\mathbb{S}$. Namely, every open set in $\mathbb{S}$ is a union of elements of $B$. Moreover, for every set $\{M_{i}\}_{i\in I}$ of subsets of $A$, we have

[TABLE]

Now, an intersection of union of elements of $B$ can be presented as a union of intersection of elements of $B$. Indeed, let $I$ be a set and let $\{J_{i}\}_{i\in I}$ be a set of sets such that for all $i\in I$ and $j_{i}\in J_{i}$ there exists a set $X_{i,j_{i}}$; then

[TABLE]

where $t_{i}$ denotes the $i$-th component of $t$. We apply the above equation to elements of $B$ and deduce that every intersection of open sets of $\mathbb{S}$ is open. Thus, $\mathbb{S}$ is an Alexandroff topological space with respect to the topology defined above.

Let $T$ be a topological space and let $x,y\in T$; then $x\in cl\{y\}$ iff for every open set $U$ containing $x$, we have $y\in U$. If $T$ is alexandroff, this is equivalent to $y\in U_{x}$, where $U_{x}$ denotes the minimal open set containing $x$. Now, let $R\in\mathbb{S}$. It is easy to see that the minimal open set containing $R$ is $V(R)$. Indeed, let $U$ be an open set containing $R$; then $U$ is of the form $\cup_{i\in I}V(M_{i})$ where $M_{i}\subseteq A$ for all $i\in I$. Thus, $R\in V(M_{i})$ for some $i\in I$; hence $M_{i}\subseteq R$ and $V(R)\subseteq V(M_{i})$. On the other hand, $V(R)$ is an open set containing $R$. Thus, the specialization order on $\mathbb{S}$ is the order of inclusion; i.e., for $R_{1},R_{2}\in\mathbb{S}$, $R_{1}\leq R_{2}$ iff $R_{1}\subseteq R_{2}$. Moreover, as in any alexandroff topological space, $U\subseteq\mathbb{S}$ is open iff $U$ is an upper set with respect to the specialization order; dually, $C\subseteq\mathbb{S}$ is closed iff $C$ is a lower set.

We will frequently use the following four basic lemmas.

Lemma 2.1**.**

1. Let $R_{1}$ and $R_{2}$ be two elements of $\mathbb{S}$ and let $R$ be an $S$-algebra satisfying $R_{1}\subseteq R\subseteq R_{2}$; then $R$ is an $S$-nice subalgebra of $A$. 2. Let $\{R_{i}\}$ be a finite subset of $\mathbb{S}$; then $\cap_{1\leq i\leq n}R_{i}\in\mathbb{S}$.

Proof.

Straightforward.

∎

For subsets $M\subseteq A$ and $T\subseteq F$ we define

[TABLE]

Lemma 2.2**.**

Let $K=\{R_{i}\}_{i\in I}$ be a nonempty subset of $\mathbb{S}$ and denote $R_{0}=\cup_{i\in I}R_{i}$. Then the following three properties are valid:

(a) $R_{0}\cap F=S$;

(b) $F\cdot R_{0}=A$; and

(c) If $R_{0}$ is closed under addition then $S\cdot R_{0}\subseteq R_{0}$.

In particular, if $R_{0}$ is a ring then it is an $S$-nice subalgebra of $A$.

Proof.

(a) Clearly, $S\subseteq R_{0}\cap F$. Let $\alpha\in R_{0}\cap F$; then $\alpha\in R_{1}$ for some $R_{1}\in K$. Since $R_{1}\cap F=S$, we have $\alpha\in S$. (b) Since $K$ is not empty there exists $R_{1}\in K$ such that $FR_{1}=A$; thus, $A=FR_{1}\subseteq FR_{0}\subseteq A$. (c) Let $\{r_{j}\}_{1\leq j\leq n}\subseteq R_{0}$ and let $\{s_{j}\}_{1\leq j\leq n}\subseteq S$; then for all $1\leq j\leq n$, $r_{j}\in R_{i_{j}}$ for appropriate $R_{i_{j}}\in K$. Thus, for all $1\leq j\leq n$, $s_{j}r_{j}\in R_{i_{j}}\subseteq R_{0}$; hence, $S\cdot R_{0}\subseteq R_{0}$. ∎

Note that $R_{0}$ is not necessarily an $S$-algebra since it is not necessarily a ring.

Lemma 2.3**.**

Let $C=\{R_{i}\}_{i\in I}$ be a nonempty chain in $\mathbb{S}$. Then the supremum of $C$ exists in $\mathbb{S}$.

Proof.

Let $R_{0}=\cup_{i\in I}R_{i}$; since $C$ is a chain, $R_{0}$ is ring. By Lemma 2.2, $R_{0}\in\mathbb{S}$. It is clear that $R_{0}$ is the smallest $S$-nice subalgebra of $A$ containing every element of $C$.

∎

Lemma 2.4**.**

Let $U$ be a nonempty open set of $\mathbb{S}$. Then there exists a maximal element in $U$, which is also a maximal element in $\mathbb{S}$.

Proof.

Consider $U$ with the partial order of containment. Let $C$ be a nonempty chain in $U$. By Lemma 2.3, $\text{sup}C$ exists in $\mathbb{S}$. Now, $\text{sup}C\supseteq R_{1}$ for some $R_{1}\in U$; thus, by the definition of an open set in $\mathbb{S}$, $\text{sup}C\in U$. Therefore, by Zorn’s Lemma, there exists a maximal element in $U$.

∎

Remark 2.5*.*

In view of Lemma 2.4, one can be more precise. In fact, for every maximal chain $C$ in a nonempty open set $U$ of $\mathbb{S}$ (there exists such a chain by Zorn’s Lemma), $\text{sup}C=\cup_{R\in C}R$ is a maximal element of $C$, $U$ and $\mathbb{S}$. Moreover, for the same reason, for any $R\in U$ there exists a maximal element $R^{\prime}\in U$ containing $R$; indeed, take any maximal chain in $U$ containing $R$, and use the reasoning above.

Let $U\subseteq\mathbb{S}$ be an open set. We consider the following properties:

(a) $U$ is of the form $V(R_{0})$ for some $R_{0}\in\mathbb{S}$.

(b) $U$ is closed under arbitrary nonempty intersection.

(c) $V(\cap_{R\in U}R)=U$.

(d) $V(\cap_{R\in U}R)\subseteq U$.

(e) $U$ is closed under finite nonempty intersection; i.e., $U$ is an inf subsemilattice of $\mathbb{S}$.

(f) $U$ contains no more than one minimal element.

Proposition 2.6**.**

Notation as above; the following implications hold:

$(a)\Leftrightarrow(b)\Rightarrow(c)\Leftrightarrow(d)\Rightarrow(e)\Rightarrow(f)$.

Proof.

$(a)\Rightarrow(b)$. By definition, every element of $U$ contains $R_{0}$. Let $R^{\prime}=\cap_{R\in W}R$ where $\emptyset\neq W\subseteq U$. Then $R_{0}\subseteq R^{\prime}\subseteq R^{\prime\prime}$ for some $R^{\prime\prime}\in W$; hence by Lemma 2.1, $R^{\prime}\in\mathbb{S}$ and thus, by the definition of $U$, $R^{\prime}\in U$. $(b)\Rightarrow(a)$ and $(b)\Rightarrow(c)$. By assumption $\cap_{R\in U}R\in U\subseteq\mathbb{S}$, which is clearly the smallest element of $U$; since every $R\in U$ contains $\cap_{R\in U}R$ and $U$ is an open set, we get $U=V(\cap_{R\in U}R)$. $(c)\Rightarrow(d)$ is trivial. $(d)\Rightarrow(c)$. As above, every $R\in U$ contains $\cap_{R\in U}R$; thus, $V(\cap_{R\in U}R)\supseteq U$. $(d)\Rightarrow(e)$. Let $R_{1},R_{2}\in U$; by Lemma 2.1, $R_{1}\cap R_{2}\in\mathbb{S}$; clearly, $R_{1}\cap R_{2}\supseteq\cap_{R\in U}R$, and thus $R_{1}\cap R_{2}\in U$. $(e)\Rightarrow(f)$. Assume to the contrary that there exists two different minimal elements $R_{1},R_{2}$ in $U$. Then, by assumption $R_{1}\cap R_{2}\in U$, but it is clearly strictly contained in both $R_{1}$ and $R_{2}$, a contradiction.

∎

We present now examples which demonstrate that the left to right implications in the previous proposition cannot be reversed.

Example 2.7*.*

To show that the implication $(c)\Rightarrow(b)$ may not hold, we can consider any case in which $\mathbb{S}$ is taken as the open set and $\cap_{R\in\mathbb{S}}R\notin\mathbb{S}$; it is clear that $V(\cap_{R\in\mathbb{S}}R)=\mathbb{S}$. As an explicit example, let $S=\mathbb{Z}$, $A=M_{2}(\mathbb{Q})$ and $U=\mathbb{S}$. More generally, in [Sa3, discussion after Proposition 3.21] we showed that whenever $A$ contains an $S$-stable basis, we have $\cap_{R\in\mathbb{S}}R\notin\mathbb{S}$.

To show that the implication $(e)\Rightarrow(d)$ may not hold, one can consider an infinite chain of $S$-nice subalgebras of $A$ such that their intersection is an $S$-nice subalgebra of $A$ that is strictly contained in each of them. As an explicit example, let $S=O_{v}$ be a valuation domain with value group $\mathbb{R}$, and let $A=M_{2}(F)$, where $F$ is the field of fractions of $O_{v}$. Take $0<r_{0}\in\mathbb{R}$ and for all $0<r\leq r_{0}$ let $I_{r}=\{x\in O_{v}\mid v(x)\geq r\}$. For all $0<r\leq r_{0}$ denote

[TABLE]

where $J_{1}$ is any nonzero ideal of $O_{v}$. Let $U=\cup_{0<r<r_{0}}V(R_{r})$. Since $\{I_{r}\}_{0<r<r_{0}}$ is a chain, $U$ is closed under finite nonempty intersection; however,

[TABLE]

Finally we demonstrate that $(f)$ does not necessarily imply $(e)$. With the notation presented above, let $J_{2}$ be an ideal of $O_{v}$ strictly containing $J_{1}$; and let

[TABLE]

Let $U=\cup_{0<r<r_{0}}V(R_{r})\cup V(R^{\prime})$. Then the unique minimal element of $U$ is $R^{\prime}$, but $U$ in not closed under finite nonempty intersection since for all $0<r<r_{0}$, we have $R^{\prime}\supset R^{\prime}\cap R_{r}$; thus, $R^{\prime}\cap R_{r}\notin U$

Proposition 2.8**.**

Let $H$ be a nonempty subset of $\mathbb{S}$; then

1. There exists a lower bound for $H$ iff the infimum of $H$ exists.

2. There exists an upper bound for $H$ iff the supremum of $H$ exists.

Proof.

Clearly, the right to left implication in both statements is trivial. We prove the left to right implication of the first statement. Let $R_{0}\in\mathbb{S}$ be a lower bound of $H$, let $R_{1}=\cap_{R\in H}R$ and let $R_{2}$ be any element of $H$ (note that $H$ is not empty). Then by Lemma 2.1, $R_{1}\in\mathbb{S}$ and it is clearly the infimum of $H$. We prove now the left to right implication of the second statement. Let $K$ denote the set of all upper bounds of $H$; by assumption, $K$ is not empty. Since $H$ is not empty, $K$ has a lower bound; indeed, any element of $H$ is a lower bound for $K$. Thus, by the first statement, the infimum of $K$, $R_{3}=\cap_{R\in K}R$, exists in $\mathbb{S}$. Now, let $R\in H$, then for every $R^{\prime}\in K$, we have $R\subseteq R^{\prime}$; thus, $R\subseteq R_{3}$. Therefore, $R_{3}\in K$ and $R_{3}$ is the supremum of $H$.

∎

We note that the assumption that $H\neq\emptyset$ is crucial. Indeed, the empty set clearly has an upper bound but the supremum of $\emptyset$ does not exist, since $\mathbb{S}$ may not contain a smallest element, as shown in Example 2.7. Also, the empty set clearly has a lower bound but the infimum of $\emptyset$ exists iff $\mathbb{S}$ is irreducible, as we shall see in Theorem 2.11.

Dually to Proposition 2.6, let $C\subseteq\mathbb{S}$ be a closed set. We consider the following properties:

(a) $C=cl\{R\}$ for some $R\in\mathbb{S}$.

(b) Every nonempty subset of $C$ has a supremum, which belongs to $C$.

(c) $\text{sup}C\in\mathbb{S}$ and $C=cl\{\text{sup}C\}$.

(d) $\text{sup}C\in\mathbb{S}$ and $cl\{\text{sup}C\}\subseteq C$.

(e) $C$ is a sup subsemilattice of $\mathbb{S}$.

(f) $C$ contains no more than one maximal element.

The proof of the following proposition is quite similar to the proof of Proposition 2.6. The implications $(a)\Rightarrow(b)$, $(c)\Rightarrow(b)$, and $(d)\Rightarrow(e)$ rely on Proposition 2.8; we shall not prove it here.

Proposition 2.9**.**

Notation as above; the following implications hold:

$(a)\Leftrightarrow(b)\Leftrightarrow(c)\Leftrightarrow(d)\Rightarrow(e)\Rightarrow(f)$.

Note the small difference in the implication $(c)\Rightarrow(b)$, which is valid in Proposition 2.9 but not in Proposition 2.6. The reason for this difference is the fact that $V(M)$ is defined for all $M\subseteq A$ whereas $cl\{R\}$ is defined only for $R\in\mathbb{S}$. While in condition (c) of Proposition 2.6, $\cap_{R\in U}R$ is not necessarily in $\mathbb{S}$, in conditions (c) and (d) of Proposition 2.9 we require that $\text{sup}C$ would be in $\mathbb{S}$.

We also note that condition (e) of Proposition 2.9 implies that $C$ is an ideal of $\mathbb{S}$, in the sense of order theory defined in the introduction. Indeed, by assumption $C$ is closed and thus it is a lower set. By the assumption in (e), $C$ is a sup subsemilattice of $\mathbb{S}$; in particular, $C$ is directed. In fact, in this case $C$ is actually a sublattice of $\mathbb{S}$. To show this, let $R_{1},R_{2}\in C$, the infimum of $R_{1}$ and $R_{2}$ is their intersection which is in $\mathbb{S}$; by assumption $C$ is a closed set and thus it is a lower set, so $\text{inf}\{R_{1},R_{2}\}\in C$; and by the assumption in condition (e), $C$ is a sup subsemilattice of $\mathbb{S}$.

The following theorem is important to our study.

Theorem 2.10**.**

Let $I$ be an irreducible subset of $\mathbb{S}$. Then $\cup_{R\in I}R$ is an $S$-nice subalgebra of $A$; in particular, $\text{sup}I=\cup_{R\in D}R$.

Proof.

Denote $R_{0}=\cup_{R\in I}R$. Let $a,b\in R_{0}$; we prove that there exists $R_{1}\in I$ such that $a,b\in R_{1}$. Assume to the contrary that there exists no such $R_{1}$. Let $G_{1}$ denote the set of all elements in $\mathbb{S}$ not containing $a$, and let $G_{2}$ denote the set of all elements in $\mathbb{S}$ not containing $b$. It is clear that $G_{1}$ and $G_{2}$ are closed in $\mathbb{S}$. However, by our assumption $I\subseteq G_{1}\cup G_{2}$, while $I\nsubseteq G_{1}$ and $I\nsubseteq G_{2}$, a contradiction. Thus, $R_{0}$ is a ring. By Lemma 2.2, $R_{0}\in\mathbb{S}$.

∎

Theorem 2.11**.**

Let $U\subseteq\mathbb{S}$ be a nonempty open set. The following statements are equivalent:

(a) $U$ has a greatest element.

(b) There exists $R_{0}\in\mathbb{S}$ such that $clU=cl\{R_{0}\}$.

(c) $U$ has a unique maximal element.

(d) $U$ is irreducible.

(e) $U$ is a sup subsemilattice of $\mathbb{S}$.

(f) Every nonempty subset $H\subseteq U$ has a supremum.

Proof.

We prove $(a)\Rightarrow(b)\Rightarrow(c)\Rightarrow(d)\Rightarrow(e)\Rightarrow(f)\Rightarrow(a)$. To show that $(a)\Rightarrow(b)$ we denote by $R_{0}$ the greatest element of $U$. It is clear that every closed set containing $U$ also contains $R_{0}$; on the other hand, since $R_{0}$ is the greatest element of $U$, by the definition of the topology on $\mathbb{S}$, every closed set containing $R_{0}$ also contains $U$. $(b)\Rightarrow(c)$. We prove that $R_{0}$ is the unique maximal element of $U$. It is clear that $R_{0}$ is a maximal element of $U$, since otherwise $U\setminus cl\{R_{0}\}\neq\emptyset$ and thus $clU\setminus cl\{R_{0}\}\neq\emptyset$. Similarly, Assuming there exists another maximal element $R_{1}\in U$, we get $R_{1}\in U\setminus cl\{R_{0}\}$. $(c)\Rightarrow(d)$. Let $R_{0}$ denote the unique maximal element of $U$. Assume to the contrary that $U\subseteq G_{1}\cup G_{2}$ where $G_{1},G_{2}$ are closed in $\mathbb{S}$ while $U\nsubseteq G_{1}$ and $U\nsubseteq G_{2}$. Let $R_{1}\in U\setminus G_{1}$ and $R_{2}\in U\setminus G_{2}$. By Remark 2.5, there exist maximal elements $R_{1}^{\prime},R_{2}^{\prime}\in U$ containing $R_{1}$ and $R_{2}$, respectively. Since $R_{0}$ is the unique maximal element of $U$, we have $R_{1}^{\prime}=R_{2}^{\prime}=R_{0}$. Thus, $R_{0}\notin G_{1}$ and $R_{0}\notin G_{2}$, a contradiction. We prove $(d)\Rightarrow(e)$. Let $R_{1},R_{2}\in U$; by assumption $U$ is irreducible and thus by Theorem 2.10, $R_{0}=\cup_{R\in U}R$ is an $S$-nice subalgebra of $A$. Clearly, $R_{0}$ contains both $R_{1}$ and $R_{2}$. Thus, by Proposition 2.8, $\text{sup}\{R_{1},R_{2}\}$ exists. Hence, by the definition of an open set, $\text{sup}\{R_{1},R_{2}\}\in U$. We prove now $(e)\Rightarrow(f)$. Let $\emptyset\neq H\subseteq U$. By Remark 2.5, for every $R\in H$ there exists a maximal element $T_{R}\in U$ containing $R$. By assumption, every two elements of $U$ have a supremum, thus these $T_{R}$ must all be equal. So, $H$ is bounded from above and therefore, by Proposition 2.8, the supremum of $H$ exists. Finally, we show $(f)\Rightarrow(a)$. By assumption $U$ has a supremum, and since $U$ is an open set, its supremum is its greatest element.

∎

In view of Theorem 2.11, we characterize now the irreducible components of $\mathbb{S}$.

Proposition 2.12**.**

$I\subseteq\mathbb{S}$* is an irreducible component of $\mathbb{S}$ iff $I=cl\{R\}$ for some maximal $R\in\mathbb{S}$.*

Proof.

Let $R$ be a maximal element of $\mathbb{S}$. It is clear that $cl\{R\}$ is irreducible. Assume to the contrary that there exists an irreducible set $G\supset cl\{R\}$. Let $R^{\prime}\in G\setminus cl\{R\}$. Then $R\nsubseteq R^{\prime}$ and $R^{\prime}\nsubseteq R$. By Theorem 2.10, $\cup_{T\in G}T\in\mathbb{S}$. However, $\cup_{T\in G}T$ strictly contains $R$, a contradiction. On the other hand, let $I\subseteq\mathbb{S}$ be an irreducible component of $\mathbb{S}$. Let $R_{0}=\cup_{T\in I}T$; by Theorem 2.10, $R_{0}\in\mathbb{S}$. Thus, $I\subseteq cl\{R_{0}\}$. Since $cl\{R_{0}\}$ is irreducible and $I$ is an irreducible component of $\mathbb{S}$, one has $I=cl\{R_{0}\}$. Now, $R_{0}$ is maximal in $\mathbb{S}$, since otherwise there exists $R_{0}\subset R_{1}\in\mathbb{S}$, but then $cl\{R_{0}\}\subset cl\{R_{1}\}$, a contradiction.

∎

Recall from [GHKLMS, Def. O-5.6.] that a topological space $T$ is called sober if for every irreducible closed subset $C$ of $T$, there exists a unique $t\in T$ such that $C$ is the closure of $t$; i.e., $C$ has a unique generic point. Also recall that a poset $P$ is called dcpo (directed complete partial order) if every directed subset of $P$ has a supremum. It is known (see, for example [GHKLMS, Ex. O-5.15.]) that every sober space is a dcpo, under the specialization order.

Note that by Lemma 2.4 there exists a maximal $S$-nice subalgebra of $A$. In [Sa3, discussion after Corollary 3.21] we noted that in case $S$ is a valuation domain of $F$ and $A$ is a field, then the maximal $S$-nice subalgebras of $A$ are precisely the valuation domains (whose valuations extend $v$) of $A$. So, by Proposition 2.12, the closures of these valuation domains are precisely the irreducible components of $\mathbb{S}$.

We also showed in [Sa3, Example 3.26] that even in the case of a central simple $F$-algebra, one can have an infinite ascending chain of $S$-nice subalgebras of $A$ (even when $S$ is a valuation domain). For the reader’s convenience we present here the example.

Example 2.13*.*

(cf. [Sa3, Example 3.26]) Let $C$ be a non-Noetherian integral domain with field of fractions $F$. Let $\{0\}\neq I_{1}\subset I_{2}\subset I_{3}\subset...$ be an infinite ascending chain of ideals of $C$ and let $A=M_{n}(F)$. Then

[TABLE]

is an infinite accending chain of $C$-nice subalgebras of $A$. So, let

[TABLE]

then $I$ is closed and irreducible with no generic point.

In particular, we have an example in which $\mathbb{S}$ is not sober. Nevertheless, in a subsequent paper we will show that $\mathbb{S}$ is indeed a dcpo and has some interesting properties from the point of view of domain theory.

Remark 2.14*.*

Note that by Lemma 2.1, $\mathbb{S}$ is an inf semilattice, where the infimum is actually an intersection of sets. In particular, taking $U=\mathbb{S}$ in Theorem 2.11, the conditions of the theorem are also equivalent to the condition that $\mathbb{S}$ is a lattice. In fact we can say even more. By the definition of an irreducible space, every finite intersection of nonempty open sets is nonempty. In our case, whenever $\mathbb{S}$ satisfies the equivalent conditions of Theorem 2.11, the intersection of all nonempty open sets of $\mathbb{S}$ is the singleton $\{R_{0}\}$, where $R_{0}$ denotes the greatest element of $\mathbb{S}$. Moreover, this property is also equivalent to the equivalent conditions presented in Theorem 2.11.

Recall (cf. [En, Theorem 16.4]) that a valuation $v$ on a field $K$ is called henselian if $v$ extends uniquely to every algebraic field extension of $K$; in this case, one also says that the corresponding valuation domain is henselian. Thus, in view of the previous remark and the discussion before Example 2.13, we have,

Example 2.15*.*

If $S$ is an henselian valuation domain, $F$ is its field of fractions and $A$ is an algebraic field extension of $F$, then $\mathbb{S}$ is a lattice (viewed from the point of view of order theory), and an irreducible topological space (viewed as an Alexandroff topological space).

3 Prime ideals

In this short section we discuss the prime spectra of $S$-nice subalgebras of $A$ and the subsets of $\text{Spec}(S)$ covered by them. We assume that $A$ is commutative (in this case whenever $R_{1}\subset R_{2}$ are $S$-subalgebras of $A$ and $Q$ is a prime ideal of $R_{2}$, then $Q\cap R_{1}$ is a prime ideal of $R_{1}$). We also assume that $A$ contains an $S$-stable basis, in order to be able to use the going down lemma for $S$-nice subalgebras of $A$ (Lemma 1.2) and Theorem 1.5.

For a ring $T$, we denote by $\text{Spec}(T)$ the prime spectrum of $T$; i.e., the set of all prime ideals of $T$. Recall the following definition from [Sa3]: if for every $P\in\text{Spec}(S)$ there exists $Q\in\text{Spec}(R)$ lying over $P$, we say that $R$ satisfies “Lying Over” (LO, in short) over $S$. We denote by $\text{Spec}_{R}(S)$ the set of all prime ideals of $S$ having a prime ideal of $R$ lying over them; namely,

$\text{Spec}_{R}(S)=\{P\in\text{Spec}(S)\mid$ there exists $Q\in\text{Spec}(R)$ lying over $P\}$. Note that, by definition $\text{Spec}_{R}(S)\subset\text{Spec}(S)$ iff $R$ does not satisfy LO over $S$.

As usual, we use the term “almost all” to mean that a property is satisfied to all but finitely many elements.

In the following lemma we show that whenever $R$ does not satisfy LO over $S$, there exists an $S$-nice subalgebra of $A$ whose prime spectrum lies over a larger set of primes of $S$ than the prime spectrum of $R$.

Lemma 3.1**.**

Let $R_{1}$ be an $S$-nice subalgebra of $A$ such that $\text{Spec}_{R_{1}}(S)\neq\text{Spec}(S)$, and let $P\in\text{Spec}(S)\setminus\text{Spec}_{R_{1}}(S)$. Then there exists $R_{1}\supset R\in\mathbb{S}$ such that $\text{Spec}_{R_{1}}(S)\subset\text{Spec}_{R}(S)$ and $P\in\text{Spec}_{R}(S)$.

Proof.

Consider the chain $\{P\}$. By Theorem 1.5, there exists $R_{2}\in\mathbb{S}$ having a prime ideal lying over $P$. By Lemma 2.1, $R=R_{1}\cap R_{2}$ is an $S$-nice subalgebra of $A$. Now, since $A$ is commutative, every prime ideal of $R_{i}$ ($i=1,2$) intersect to a prime ideal of $R$. Thus, $\text{Spec}_{R_{1}}(S)\subset\text{Spec}_{R}(S)$ with $P\in\text{Spec}_{R}(S)$.

∎

Proposition 3.2**.**

There exists $R\in\mathbb{S}$ satisfying LO over $S$ iff there exists $R_{1}\in\mathbb{S}$ whose prime spectrum is lying over almost all prime ideals of $S$; namely, $\text{Spec}(S)\setminus\text{Spec}_{R_{1}}(S)$ is finite.

Proof.

$(\Rightarrow)$ is trivial; just take $R_{1}=R$. $(\Leftarrow)$ Let $R_{1}$ be an $S$-nice subalgebra of $A$ whose prime spectrum is lying over almost all prime ideals of $S$. If $R_{1}$ satisfies LO over $S$ then we are done. Otherwise, by Lemma 3.1 there exists $R_{1}\supset R_{2}\in\mathbb{S}$ with $\text{Spec}_{R_{1}}(S)\subset\text{Spec}_{R_{2}}(S)$. After finitely many such steps we get $R_{n}\in\mathbb{S}$ that satisfies LO over $S$.

∎

Remark 3.3*.*

Note that if there exists $R\in\mathbb{S}$ that satisfies LO over $S$, then for every nonempty closed set $C\subseteq\mathbb{S}$ there exists $R_{1}\in C$ that satisfies LO over $S$; indeed, take any $R_{2}\in C$ and denote $R_{1}=R\cap R_{2}$.

In view of the previous remark, in the following proposition we present equivalent conditions for the non-existence of an $S$-nice subalgebra of $A$ that satisfies LO over $S$.

Proposition 3.4**.**

The following conditions are equivalent:

(a) There exists no $R\in\mathbb{S}$ that satisfies LO over $S$.

(b) There is no maximal subset $Y\subseteq\text{Spec}(S)$ (with respect to inclusion) such that there exists $R\in\mathbb{S}$ with $\text{Spec}_{R}(S)=Y$.

(c) For every $R\in\mathbb{S}$ there exists an infinite descending chain $\{R_{i}\}_{i\in I}$ of $S$-nice subalgebras of $A$ such that $R_{i}\subseteq R$ for all $i\in I$, and $\text{Spec}_{R_{i}}(S)\subset\text{Spec}_{R_{j}}(S)$, whenever $R_{j}\subset R_{i}$.

(d) Every nonempty closed subset of $\mathbb{S}$ contains an infinite descending chain $\{R_{i}\}_{i\in I}$ of $S$-nice subalgebras of $A$ such that $\text{Spec}_{R_{i}}(S)\subset\text{Spec}_{R_{j}}(S)$, whenever $R_{j}\subset R_{i}$.

Proof.

We prove $(a)\Rightarrow(b)\Rightarrow(c)\Rightarrow(d)\Rightarrow(a)$. To show $(a)\Rightarrow(b)$, assume to the contrary that there exists $R_{1}\in\mathbb{S}$ such that $Y=\text{Spec}_{R_{1}}(S)$ is a maximal subset of $\text{Spec}(S)$. By assumption $\text{Spec}_{R_{1}}(S)\neq\text{Spec}(S)$. By Lemma 3.1 there exists $R_{2}\in\mathbb{S}$ such that $\text{Spec}_{R_{1}}(S)\subset\text{Spec}_{R_{2}}(S)$, a contradiction to the maximality of $Y$. We prove now $(b)\Rightarrow(c)$. Let $R_{1}\in\mathbb{S}$. By assumption $\text{Spec}_{R_{1}}(S)\neq\text{Spec}(S)$, since otherwise $\text{Spec}_{R_{1}}(S)$ is a maximal subset of $\text{Spec}(S)$. By Lemma 3.1 there exists $R_{1}\supset R_{2}\in\mathbb{S}$ such that $\text{Spec}_{R_{1}}(S)\subset\text{Spec}_{R_{2}}(S)$. By assumption $\text{Spec}_{R_{2}}(S)\neq\text{Spec}(S)$; so again by Lemma 3.1, there exists $R_{2}\supset R_{3}\in\mathbb{S}$ such that $\text{Spec}_{R_{2}}(S)\subset\text{Spec}_{R_{3}}(S)$. We continue this way to get an infinite descending chain $\{R_{k}\}_{k\in\mathbb{N}}$ of $S$-nice subalgebras of $A$ such that $\text{Spec}_{R_{i}}(S)\subset\text{Spec}_{R_{j}}(S)$, whenever $i<j$. To prove that $(c)\Rightarrow(d)$, let $C$ be a nonempty closed subset of $\mathbb{S}$ and let $R_{1}\in C$. The result follows by applying the assumption on $R_{1}$ and recalling the definition of the topology on $\mathbb{S}$. Finally, we show $(d)\Rightarrow(a)$. Assume to the contrary that there exists $R_{1}\in\mathbb{S}$ satisfying LO over $S$. Then the closed subset $cl\{R_{1}\}$ contains only elements that satisfy LO over $S$, a contradiction.

∎