The Computational Complexity of Subclasses of Semiperfect Rings

Abstract

This paper studies the computational complexity of subclasses of semiperfect rings from the perspective of computability theory. A ring is semiperfect if the identity can be expressed as a sum of mutually orthogonal local idempotents. Semisimple rings and local rings are typical subclasses of semiperfect rings that play important roles in noncommutative algebra. First, we define a ring to be semisimple if the left regular module can be decomposed as a finite direct sum of simple submodules and prove that the index set of computable semisimple rings is ${\mathsf{\Sigma }}_2^0$-hard within the index set of computable rings. Second, we define local rings by using equivalent properties of non-left invertible elements of rings and show that the index set of computable local rings is ${\mathsf{\Pi }}_2^0$-hard within the index set of computable rings. Finally, based on the ${\mathsf{\Pi }}_2^0$ definition of local rings, computable semiperfect rings can be described by ${\mathsf{\Sigma }}_3^0$ formulas. As a corollary, we find that the index set of computable semiperfect rings can be both ${\mathsf{\Sigma }}_2^0$-hard and ${\mathsf{\Pi }}_2^0$-hard within the index set of computable rings.

1. Introduction

Computable algebra (or effective algebra) is a research area that explores the effective aspects of algebraic objects and algebraic constructions by methods of computability theory [1,2,3], where computability theory is a branch of mathematical logic that uses tools and techniques to examine the computational complexity of problems for computable structures (e.g., [4,5]). For special structures like groups, a lot of fruitful work has been carried out; refer to [6,7,8,9,10]. For instance, Downey and Montalbán proved in [8] that the isomorphism problem for torsion-free abelian groups is ${\mathsf{\Sigma }}_1^1$-complete; Riggs proved in [10] that the index set of computable decomposable torsion-free abelian groups of infinite rank and of finite rank is ${\mathsf{\Sigma }}_1^1$-complete and ${\mathsf{\Sigma }}_3^0$-complete, respectively. For a survey of studying computable abelian groups, we refer to Melnikov [11]. Compared with the research on computable abelian groups, computable rings, or even computable modules, are less studied in the literature. For the work on computable rings, Downey, Lempp, and Mileti constructed a computable commutative ring whose Jacobson radical is ${\mathsf{\Pi }}_2^0$-complete in [12]; Conidis further investigated the complexity of radicals of rings and modules, constructing a computable noncommutative ring whose prime radical is ${\mathsf{\Pi }}_1^1$-complete in [13] and a computable module over a computable ring whose radical is ${\mathsf{\Pi }}_1^1$-complete in [14].

The method of computability theory has also been widely applied to study reverse mathematics [15,16], where reverse mathematics is a branch of mathematical logic that mainly examines the proof-theoretic strength of mathematical problems. We have already studied the complexity of radicals and socles of modules from the viewpoint of reverse mathematics in [17]; for instance, the reverse mathematics result on the socles of modules implies the existence of a computable abelian group such that the socle of it is ${\mathsf{\Sigma }}_1^0$-complete. Motivated by the work on effective aspects of problems on abelian groups and on rings, we are interested in studying the computational complexity of the algebraic properties of rings and modules. For instance, we further studied the complexity of socles of modules by methods of computability theory in [18].

Definition 1.

A ring $(R,{+}_R,{-}_R,{·}_R,0_R,1_R)$ with domain $R\subseteq \mathbb{N}$ is computable if R is a computable set, ${+}_R$, ${-}_R$ and ${·}_R$ are computable functions.

Similar to the study in computable abelian groups and in computable structure theory, we often study the complexity of computable rings possessing a specified property $\mathcal{P}$ by applying the index set method of computability theory. More specifically, we first encode the problem for computable rings satisfying $\mathcal{P}$ by a special subset of natural numbers, namely, the index set of computable rings satisfying $\mathcal{P}$ (to be defined in Section 2), and then examine the computational complexity of the corresponding index set. More recently, in [19], we conducted the study on the complexity of the decomposability problem of computable rings by the index set method; in particular, we proved that the index set of computable decomposable rings is ${\mathsf{\Sigma }}_2^0$-complete within the index set of computable rings.

Semiperfect rings form a significant class of rings in noncommutative algebra, and the theory of such rings is closely connected with homological algebra. For an algebraic background on semiperfect rings, we refer to Chapter 8 of Lam’s book [20]. In this paper, rather than dealing with general rings, we concentrate on the small class of semiperfect rings and investigate the computational complexity of typical subclasses of semiperfect rings, such as semisimple rings and local rings. More precisely, similar to the study of the decomposability problem of computable rings in [19], our aim here is to determine the exact complexity of subclasses of computable semiperfect rings within the whole class of computable rings.

In the following, we only focus on countable rings with identity and mainly consider the index sets of three kinds of rings, namely, semisimple rings, local rings, and semiperfect rings.

1.1. Semisimple Rings

Semisimple rings can be defined in a ring-theoretic way as finite direct products of simple left artinian rings, where a ring R is simple if for any nonzero $r\in R$, the two-sided ideal of R generated by r equals R itself; R is left artinian if for any descending chain $I_0\supseteq I_1\supseteq \cdots \supseteq I_n\supseteq \cdots$ of left ideals of R, there is a number N such that $I_n=I_N$ for all $n\ge N$. Since sequences of left ideals can be encoded by a set, this ring-theoretic definition quantifies over sets, and thus, it is analytic. On the other hand, semisimple rings can be defined in a module-theoretic way. A nonzero R-module S is simple if for any nonzero $x\in S$, S is generated by x, i.e., $S=Rx=\{rx:r\in R\}$.

Definition 2.

A ring R is semisimple if the left regular module is a finite direct sum of simple submodules.

Decomposing a semisimple ring R as a finite direct sum of simple submodules is equivalent to expressing the identity as $1_R=e_1+\cdots +e_n$ with $e_1,\dots ,e_n$ pairwise orthogonal nonzero irreducible idempotents, where an idempotent e is irreducible if $Re$ is a simple R-module. Then, computable semisimple rings can be described by ${\mathsf{\Sigma }}_3^0$ formulas, and the index set of computable semisimple rings is ${\mathsf{\Sigma }}_3^0$ within the index set of computable rings (to be proved in Section 2).

The index set $\mathrm{Fin}$ of finite computably enumerable (c.e. for short) sets is ${\mathsf{\Sigma }}_2^0$-complete. In Section 3, we construct a uniformly computable sequence $\langle R_e:e\in \mathbb{N}\rangle$ of computable rings such that $e\in \mathrm{Fin}$ if and only if $R_e$ is a semisimple ring, obtaining the following lower bound on the complexity of semisimple rings:

• the index set of computable semisimple rings is ${\mathsf{\Sigma }}_2^0$-hard within the index set of computable rings.

Currently, we are unable to prove the following conjecture on the exact complexity of semisimple rings and leave the challenging question open.

Conjecture 1.

The index set of computable semisimple rings is ${\mathsf{\Sigma }}_3^0$-complete within the index set of computable rings.

Since the index set $\mathrm{Cof}$ of cofinite c.e. sets is a typical ${\mathsf{\Sigma }}_3^0$-complete set, to prove Conjecture 1, we need to construct a uniformly computable sequence of computable rings $\langle R_e:e\in \mathbb{N}\rangle$ such that $e\in \mathrm{Cof}$ if and only if $R_e$ is semisimple for all e. Currently, we do not know how to strengthen the ${\mathsf{\Sigma }}_2^0$-completeness construction developed here to the more complicated ${\mathsf{\Sigma }}_3^0$-completeness construction. This is the main obstacle for us to prove the conjecture.

1.2. Local Rings

In algebra texts [20,21], commutative local rings often refer to rings possessing a unique maximal ideal, while noncommutative local rings often refer to rings possessing a unique maximal left ideal. Such definitions for local rings are analytic. However, using equivalent characterizations of local rings via properties of invertible elements, we can define local rings in an arithmetic way as follows.

Definition 3.

A ring R not necessarily commutative is local if for any non-left invertible elements $x,y\in R$, the sum $x+y$ is not left invertible.

Fields are natural examples of commutative local rings, where a commutative ring R is a field if for any nonzero $x\in R$, there is a $y\in R$, such that $xy=1_R$. In general, according to Definition 3, computable local rings can be described by ${\mathsf{\Pi }}_2^0$ formulas, i.e., formulas of the form $\forall x\exists y\phi (x,y)$ with $\phi (x,y)$ containing only bounded number quantifiers (see Proposition 2 in Section 2 for more details).

In Section 4, we show that the ${\mathsf{\Pi }}_2^0$ complexity for local rings is optimal by proving the following:

• the index set of computable local rings is ${\mathsf{\Pi }}_2^0$-complete within the index set of computable rings.

Since a domain (i.e., a commutative ring with no zero divisors) is semisimple if and only if it is a field, our proof of the result on local rings in Section 4 also implies that

• the index set of computable semisimple domains (=fields) is ${\mathsf{\Pi }}_2^0$-complete within the index set of computable rings.

This reveals that the complexity of computable semisimple domains lies exactly in ${\mathsf{\Pi }}_2^0$ of the arithmetic hierarchy.

1.3. Semiperfect Rings

Similar to the study of semisimple rings and local rings above, we continue to study more generalized semiperfect rings. In algebra texts such as Lam’s book [20], semiperfect rings are often defined based on semisimple rings and the Jacobson radicals of rings. That is, a ring R is semiperfect if the quotient ring $R/rad\left(R\right)$ is semisimple and idempotents of $R/rad\left(R\right)$ can be lifted to R, where $rad\left(R\right)$ is the Jacobson radical of R. On the other hand, based on an equivalent characterization of semiperfect rings in terms of local idempotents (see Theorem 23.6, [20]), we can define semiperfect rings in a much simpler way as follows. An idempotent $e\in R$ is local if $eRe=\{ere:r\in R\}$ is a local ring with identity e.

Definition 4.

A ring R is semiperfect if the identity can be written as a sum of mutually orthogonal local idempotents.

Semisimple rings and local rings are natural examples of semiperfect rings. Based on the ${\mathsf{\Pi }}_2^0$ definition of local rings, semiperfect rings can be described by ${\mathsf{\Sigma }}_3^0$ formulas, and the index set of computable semiperfect rings is ${\mathsf{\Sigma }}_3^0$. Furthermore, our proofs of the results for semisimple rings and local rings also imply corresponding results for semiperfect rings. That is, the index set of computable semiperfect rings is both ${\mathsf{\Sigma }}_2^0$-hard and ${\mathsf{\Pi }}_2^0$-hard within the index set of computable rings. We leave the following conjecture on the exact complexity of semiperfect rings open.

Conjecture 2.

The index set of computable semiperfect rings is ${\mathsf{\Sigma }}_3^0$-complete within the index set of computable rings.

Similar to the study of semisimple rings, to prove Conjecture 2, we may use the index set $\mathrm{Cof}$ of cofinite c.e. sets and construct a uniformly computable sequence of computable rings $\langle R_e:e\in \mathbb{N}\rangle$ such that $e\in \mathrm{Cof}$ if and only if $R_e$ is semiperfect for all e. Then, the ${\mathsf{\Sigma }}_3^0$-complete set $\mathrm{Cof}$ is reducible to the index set of computable semiperfect rings, and then the latter set is ${\mathsf{\Sigma }}_3^0$-complete.

The remaining part is organized as follows. In Section 2, we review necessary notions in computability theory and define the index set of subclasses of computable rings; in particular, we construct an enumeration of computably enumerable sets. In Section 3, we study the complexity of semisimple rings and prove the main complexity result based on the particular enumeration of computably enumerable sets. In Section 4, we prove the complexity result of computable local rings. Finally, we conclude this study of the complexity of subclasses of computable semiperfect rings in Section 5.

2. Preliminaries

In this section, we review necessary notions in computability theory [1,2,3].

A function $f:A\to \mathbb{N}$ with A a subset of $\mathbb{N}$ is partial computable if there is a Turing program that returns $f\left(n\right)$ for any input $n\in A$. Based on the listing of all Turing programs, partial computable functions are often listed as ${\phi }_0,{\phi }_1,\dots ,{\phi }_e,\dots$, where ${\phi }_e$ is the function computed by the e-th Turing program. Partial computable functions with domain $\mathbb{N}$ are total computable (or simply, computable) functions. The characteristic function ${\chi }_A$ of a set $A\subseteq \mathbb{N}$ is defined as ${\chi }_A\left(n\right)=1$ if $n\in A$ and 0 otherwise. A set A is computable if ${\chi }_A$ is a computable function.

A set is arithmetic if it can be defined by a formula with only number quantifiers. According to the appearance of number quantifiers in formulas, arithmetic sets can be divided into subclasses. Let $A\subseteq \mathbb{N}$.

(1)

A is ${\mathsf{\Sigma }}_0^0(={\mathsf{\Pi }}_0^0)$ if it is defined by a formula with bounded number quantifies.

(2)

For each $n\ge 1$, A is ${\mathsf{\Sigma }}_n^0$ if it is defined by arithmetic formulas of the form $\exists x_1\forall x_2\dots Q{x}_n\phi (x_1,x_2,\dots ,x_n)$, where $\phi (x_1,x_2,\dots ,x_n)$ is a ${\mathsf{\Sigma }}_0^0$ formula, Q is ∃ if n is odd and ∀ if n is even.

(3)

For each $n\ge 1$, A is ${\mathsf{\Pi }}_n^0$ if it is defined by arithmetic formulas of the form $\forall x_1\exists x_2\dots Q{x}_n\phi (x_1,x_2,\dots ,x_n)$, where $\phi (x_1,x_2,\dots ,x_n)$ is a ${\mathsf{\Sigma }}_0^0$ formula, Q is ∀ if n is odd and ∃ if n is even.

For example, ${\mathsf{\Sigma }}_3^0$ formulas are formulas of the form $\exists x_1\forall x_2\exists x_3\phi (x_1,x_2,x_3)$ with $\phi (x_1,x_2,x_3)$ only containing bounded number quantifiers. The symbols ${\mathsf{\Sigma }}_n^0$, ${\mathsf{\Pi }}_n^0$ also stand for the class of ${\mathsf{\Sigma }}_n^0$, ${\mathsf{\Pi }}_n^0$ sets, respectively.

For two subsets $A,B$ of $\mathbb{N}$, the set A is many-one reducible (or m-reducible) to B if there is a computable function $f:\mathbb{N}\to \mathbb{N}$ such that $x\in A$ if and only if $f\left(x\right)\in B$, denoted by $A{\le }_{m}B$. For a complexity class $\mathsf{\Gamma }$ such as ${\mathsf{\Pi }}_2^0$ or ${\mathsf{\Sigma }}_3^0$, we view the following m-complete $\mathsf{\Gamma }$ sets as the hardest sets in the class $\mathsf{\Gamma }$ under m-reducibility.

Definition 5.

For a complexity class Γ, a set B is called m-complete Γ (or simply  Γ-complete) if $B\in \mathsf{\Gamma }$, and A is m-reducible to B for any set $A\in \mathsf{\Gamma }$.

Computably enumerable (c.e. for short) sets are domains of partial computable functions. They are often listed as $W_0,W_1,\dots ,W_e,\dots$ with $W_e=dom\left({\phi }_e\right)$, the domain of the e-th partial computable function. We mention the following arithmetic complete sets.

(1)

The index set $\mathrm{Tot}=\{e\in \mathbb{N}:{\phi }_e \mathrm{is} \mathrm{total}\}$ is ${\mathsf{\Pi }}_2^0$-complete.

(2)

The index set $\mathrm{Fin}=\{e\in \mathbb{N}:W_e \mathrm{is} \mathrm{finite}\}$ is ${\mathsf{\Sigma }}_2^0$-complete.

(3)

The index set $\mathrm{Inf}=\{e\in \mathbb{N}:W_e \mathrm{is} \mathrm{infinite}\}$ is ${\mathsf{\Pi }}_2^0$-complete.

(4)

The index set $\mathrm{Cof}=\{e\in \mathbb{N}:W_e \mathrm{is} \mathrm{cofinite}\}$ is ${\mathsf{\Sigma }}_3^0$-complete.

Fix the language ${\mathcal{L}}_r=(+,·,-,0,1)$ of rings, similar to the listing of all partial computable functions, all partial computable rings (i.e., partial computable ${\mathcal{L}}_r$-structures) can be listed as $S_0,S_1,\dots ,S_e,\dots$. This listing includes all computable rings, and we can define the index set of subclasses of computable rings as follows.

Definition 6.

Let $\mathcal{P}$ be a property of rings. The index set of computable rings satisfying $\mathcal{P}$ is the set $\{e\in \mathbb{N}:S_e is a computable ring satisfying \mathcal{P}\}.$

Computable rings are partial computable rings with computable domains and computable operations such that the ${\mathsf{\Pi }}_1^0$ axioms of rings (e.g., $\forall x,y,z[x·(y+z)=x·y+x·z]$) are satisfied. Since being a total computable function is a ${\mathsf{\Pi }}_2^0$ property for a partial computable function, the index set of computable rings is ${\mathsf{\Pi }}_2^0$. Furthermore, due to the ${\mathsf{\Pi }}_2^0$-completeness of the index set $\mathrm{Tot}$ of total computable functions, the index set of computable rings is indeed ${\mathsf{\Pi }}_2^0$-complete.

As we are interested in the complexity of subclasses of computable rings, not the complexity of computable rings, it is more appropriate to use the notion of m-reducibility within a given set proposed by Calvert and Knight; see Definition 3.6 [5].

Definition 7

([5]). Let Γ be a complexity class (e.g., ${\mathsf{\Pi }}_2^0,{\mathsf{\Sigma }}_3^0$). For two sets $A\subseteq B$, A is m-complete Γ within B (or simply Γ-complete within B) if

(1) 

A is Γ within B; that is, there is a set $C\in \mathsf{\Gamma }$, such that $A=C\cap B$;

(2) 

For any set $S\in \mathsf{\Gamma }$, S is m-reducible to A within B; that is, there is a computable function $f:\mathbb{N}\to B$ such that $x\in S\iff f\left(x\right)\in A$ for all $x\in \mathbb{N}$.

To measure the lower bound of the complexity of a mathematical problem, we can naturally extend the notion of $\mathsf{\Gamma }$-completeness within a given set to the following $\mathsf{\Gamma }$-hardness within a given set.

Definition 8.

Let ${\mathsf{\Gamma }}_1\subseteq {\mathsf{\Gamma }}_2$ be two complexity classes (e.g., ${\mathsf{\Pi }}_2^0\subseteq {\mathsf{\Sigma }}_3^0$). For two sets $A\subseteq B$, A is m-hard ${\mathsf{\Gamma }}_1$ within B (or simply ${\mathsf{\Gamma }}_1$-hard within B) if

(1) 

A is ${\mathsf{\Gamma }}_2$ within B; that is, there is a set $C\in {\mathsf{\Gamma }}_2$, such that $A=C\cap B$;

(2) 

For any set $S\in {\mathsf{\Gamma }}_1$, S is m-reducible to A within B.

Let $\mathcal{R}$ be the index set of computable rings. To measure the complexity of computable semisimple rings, we will study the complexity of their index set, i.e., $\{e\in \mathbb{N}:S_e \mathrm{is} \mathrm{a} \mathrm{computable} \mathrm{semisimple} \mathrm{ring}\}$, within $\mathcal{R}$. We first prove the following upper bound.

Proposition 1.

The index set of computable semisimple rings is ${\mathsf{\Sigma }}_3^0$ within $\mathcal{R}$.

Proof. 

For a computable ring R and a nonzero $x\in R$, the cyclic R-module $Rx$ is simple if the following ${\mathsf{\Pi }}_2^0$ formula holds:

$$\phi \left(x\right)\equiv \forall r\in R\exists s\in R[rx\ne 0_R\to x=srx].$$

Now R is semisimple if the following ${\mathsf{\Sigma }}_3^0$ formula holds:

$$\begin{array}{cc}\hfill \exists n\ge 1& \exists e_1,\dots ,e_n\in R\forall i,j\in \{1,\dots ,n\}\hfill \\ \hfill & [(e_i=e_i^2\ne 0_R)\wedge \phi \left(e_i\right)\wedge (i\ne j\to e_i{e}_j=0_R)\wedge 1_R=e_1+\cdots +e_n].\hfill \end{array}$$

Then, the index set of computable semisimple rings is ${\mathsf{\Sigma }}_3^0$ within $\mathcal{R}$. □

We have the following upper bound for the complexity of computable local rings.

Proposition 2.

The index set $\{e\in \mathbb{N}:S_e is a computable local ring\}$ is ${\mathsf{\Pi }}_2^0$ within $\mathcal{R}$.

Proof. 

For a computable ring R, it is local if for any non-left invertible elements $x,y\in R$, $x+y$ is not left invertible. This can be described by the ${\mathsf{\Pi }}_2^0$ formula:

$$\forall x,y,z\in R\exists a,b\in R[ax=1_R\vee by=1_R\vee z(x+y)\ne 1_R].$$

So the index set of computable local rings is ${\mathsf{\Pi }}_2^0$ within $\mathcal{R}$. □

Based on the ${\mathsf{\Pi }}_2^0$ property of computable local rings, we can further obtain an upper bound for the complexity of computable semiperfect rings.

Proposition 3.

The index set $\{e\in \mathbb{N}:S_e is a computable semiperfect ring\}$ is ${\mathsf{\Sigma }}_3^0$ within $\mathcal{R}$.

Proof. 

Let R be a computable ring. First, an idempotent $e\in R$ is local (i.e., $eRe$ is a local ring) if the ${\mathsf{\Pi }}_2^0$ formula

$$\psi \left(e\right)\equiv \forall x,y,z\in R\exists a,b\in R[eaeexe=e\vee ebeeye=e\vee ezee(x+y)e\ne e]$$

is true. Second, R is semiperfect if the following ${\mathsf{\Sigma }}_3^0$ formula holds:

$$\begin{array}{cc}\hfill \exists n\ge 1& \exists e_1,\dots ,e_n\in R\forall i,j\in \{1,\dots ,n\}\hfill \\ \hfill & [(e_i=e_i^2\ne 0_R)\wedge \psi \left(e_i\right)\wedge (i\ne j\to e_i{e}_j=0_R)\wedge 1_R=e_1+\cdots +e_n].\hfill \end{array}$$

So the index set of computable semiperfect rings is ${\mathsf{\Sigma }}_3^0$ within $\mathcal{R}$. □

In the rest of this section, we prove a technical Lemma about the computable enumeration of c.e. sets, which will be used to prove our main theorem on the index set result of computable semisimple rings in Section 3.

A sequence $\langle A_e:e\in \mathbb{N}\rangle$ of c.e. subsets of natural numbers is uniformly computably enumerable if there is a computable function f such that $A_e=W_{f\left(e\right)}$ for all e (see Definition 2.5, [2]). Furthermore, a sequence $\langle B_e:e\in \mathbb{N}\rangle$ of computable subsets of natural numbers is uniformly computable if there is a computable function $g:\mathbb{N}\times \mathbb{N}\to \{0,1\}$ such that for each e, the characteristic function of $B_e$ is ${\chi }_{B_e}\left(n\right)=g(e,n)$ for all n.

Definition 9

(Definition 2.8, [2]). A simultaneous computable enumeration (or simply, enumeration) of a uniformly computably enumerable sequence of c.e. sets $\langle A_e:e\in \mathbb{N}\rangle$ is a computable sequence of finite sets $\langle A_{e,s}:e,s\in \mathbb{N}\rangle$ satisfying:

(1) 

$A_{e,s}\subseteq A_{e,s+1}$;

(2) 

$|A_{e,s+1}-A_{e,s}| \le 1$;

(3) 

$A_e={\bigcup }_{s\in \mathbb{N}}{A}_{e,s}$.

The sequence of all c.e. sets $\langle W_e:e\in \mathbb{N}\rangle$ has a standard enumeration $\langle W_{e,s}:e,s\in \mathbb{N}\rangle$, where $W_{e,0}$ is the empty set; for $s\ge 1$, there is at most one $e\le s$, such that at most one $x\le s$ enumerates into $W_{e,s}$. Set constructions in computability theory are often based on the standard enumeration of c.e. sets. Here, to construct proper computable rings, we need a particular enumeration of c.e. sets. We will frequently use the symbol $\langle ·,\cdots ,·\rangle :{\mathbb{N}}^n\to \mathbb{N}$, which stands for a fixed bijection from n-tuples of natural numbers to natural numbers.

Lemma 1.

There is an enumeration $\langle A_{e,s}:e,s\in \mathbb{N}\rangle$ of all c.e. sets such that for all $e\in \mathbb{N}$, $A_e={\bigcup }_{s\in \mathbb{N}}{A}_{e,s}=W_e and s_{\langle e,j\rangle }+(j+1)<s_{\langle e,j+1\rangle }$ if $s_{\langle e,j+1\rangle }$ is defined, where for $j\ge 1$, $s_{\langle e,j\rangle }$ is the j-th stage at which a number goes into $A_e$ if it is defined.

Proof. 

Fix an effective bijection $\langle ·,·\rangle :{\mathbb{N}}^2\to \mathbb{N}$. We build the enumeration $\langle A_{e,s}:e,s\in \mathbb{N}\rangle$ from $\langle W_{e,s}:e,s\in \mathbb{N}\rangle$ by stages.

Construction

Stage 0. For all e, let $A_{e,0}:=\varnothing$, the empty set.

Stage $s=\langle e,t\rangle \ge 1$. For all $e^{\prime }$, if $e^{\prime }\ne e$, let $A_{e^{\prime },s}:=A_{e^{\prime },s-1}$; otherwise, $e^{\prime }=e$, define $A_{e,s}$ as follows:

(1)

If $W_{e,s}=\varnothing$, set $A_{e,s}:=\varnothing$.

(2)

If $W_{e,s}\ne \varnothing$, there are two subcases.

(2.1)

If $A_{e,s-1}=\varnothing$, let $x_1$ be the least number in $W_{e,s}$. Enumerate $x_1$ into $A_{e,s}$, set $s_{\langle e,1\rangle }:=s$.

(2.2)

Otherwise, $A_{e,s-1}\ne \varnothing$, let $A_{e,s-1}=\{x_1,\dots ,x_n\}$, where for $1\le j\le n$, $x_j$ is the j-th number enumerating into $A_e$. Let $s_{\langle e,j\rangle }$ be the stage at which $x_j$ enumerates into $A_e$.

*

If $s>s_{\langle e,n\rangle }+n+1$ and $|W_{e,s}-A_{e,s-1}|\ne \varnothing$, let $x_{n+1}$ be the least number in $W_{e,s}-A_{e,s-1}$. Enumerate $x_{n+1}$ into $A_{e,s}$, set $s_{\langle e,n+1\rangle }:=s$. In this case, $A_{e,s}=A_{e,s-1}\cup \left\{x_{n+1}\right\}$.

*

Otherwise, do nothing. Set $A_{e,s}:=A_{e,s-1}$.

This ends the construction.

For all $s\ge 1$, $A_{e,s-1}\subseteq A_{e,s}$. For all $j\ge 1$, the desired property $s_{\langle e,j\rangle }+(j+1)<s_{\langle e,j+1\rangle }$ holds. Let $A_e={\bigcup }_{s\in \mathbb{N}}{A}_{e,s}$. We show that $A_e=W_e$. Since elements of $A_e$ are taken from $W_e$, we have $A_e\subseteq W_e$.

Suppose otherwise that $A_e\ne W_e$; then, there is a least number $x\in W_e-A_e$. Let u be the stage with $x\in W_{e,u}-W_{e,u-1}$. Then, for all $s\ge u$, we have $x\in W_{e,s}-A_{e,s-1}$. Let $t_1>u$ be the least stage, such that all numbers of $W_e-A_e$ less than x (if any) have been enumerated into $A_e$ by stage $t_1$. Then, for all $s\ge t_1$, x is the least number in $W_{e,s}-A_{e,s-1}$. After stage $t_1$, there are infinitely many number of stages of the form $\langle e,t\rangle$, and we have to enumerate x into $A_e$. This contradicts the assumption that $x\notin A_e$. So, we have $A_e=W_e$. □

3. Semisimple Rings

In this section, based on the enumeration $\langle A_{e,s}:e,s\in \mathbb{N}\rangle$ of all c.e. sets with $A_e={\bigcup }_{s\in \mathbb{N}}{A}_{e,s}=W_e$ in Lemma 1 above, we prove the following main theorem.

Theorem 1.

The index set of computable semisimple rings is ${\mathsf{\Sigma }}_2^0$-hard within the index set of computable rings.

Proof. 

To prove the theorem, we need to build an effective reduction from the ${\mathsf{\Sigma }}_2^0$-complete set $\mathrm{Fin}=\{e\in \mathbb{N}:W_e \mathrm{is} \mathrm{finite}\}$ to the index set of computable semisimple rings. Let $S=\mathbb{Q}[x_i:i\in \mathbb{N}]$ be the commutative polynomial ring with indeterminates from $\{x_i:i\in \mathbb{N}\}$. S is computable under a fixed computable presentation, i.e., an effective coding of S into natural numbers. Consider the quotient ring $R=S/(x_i-x_i^2:i\in \mathbb{N})$ of the S modulo the ideal generated by polynomials $\{x_i-x_i^2:i\in \mathbb{N}\}$. For all $i\in \mathbb{N}$, $x_i$ is an idempotent of the quotient ring R. Furthermore, monomials of R are of the form $x_{i_1}\cdots x_{i_n}$ with $0\le i_1<\cdots <i_n$ and $n\ge 1$. For each polynomial $f\in S$, we can find a reduced form of it in R. For instance, the reduced form of $f=1+3x_3^4+2x_1{x}_4^2$ equals $f^{\ast }=1+3x_3+2x_1{x}_4$. Therefore, R is a computable ring. As a $\mathbb{Q}$-vector space, R can be written as

$$R=\mathbb{Q}\oplus \underset{n=1}{\overset{\infty }{⨁}}\underset{i_1<\cdots <i_n}{⨁}\mathbb{Q}{x}_{i_1}\cdots x_{i_n}.$$

Using the ${\mathsf{\Sigma }}_2^0$-complete set $\mathrm{Fin}=\{e\in \mathbb{N}:W_e \mathrm{is} \mathrm{finite}\}$, we build a uniformly computable sequence of computable ideals $\langle I_e:e\in \mathbb{N}\rangle$ of R, such that the quotient ring $R_e=R/I_e$ meets the requirements:

${\mathcal{P}}_e$:

If $e\in \mathrm{Fin}$, then $R_e=R/I_e$ is a semisimple ring.

${\mathcal{Q}}_e$:

If $e\notin \mathrm{Fin}$, then $R_e=R/I_e$ is not a semisimple ring.

Let $A_{e,s}(e,s\in \mathbb{N})$ be the effective approximation of all c.e. sets with $A_e={\bigcup }_{s\in \mathbb{N}}{A}_{e,s}=W_e$ in Lemma 1.

Definition 10.

Let $s\ge 1$. For any $e\le s$, say that ${\mathcal{P}}_e$ requires attention at stage s if $A_{e,s}\ne A_{e,s-1}$, i.e., some number goes into $A_e=W_e$.

We build the sequence $\langle I_e={\bigcup }_{s\in \mathbb{N}}{I}_{e,s}:e\in \mathbb{N}\rangle$ of ideals of R by stages, where $I_{e,s}$ is the ideal of R generated by polynomials enumerated into $I_e$ by stage s.

Construction

Stage 0. For all e, the ${\mathcal{P}}_e$-requirement does not require attention; add $x_0$ into $I_{e,0}$.

Stage $s\ge 1$. For all $e^{\prime }\ge s+1$, ${\mathcal{P}}_{e^{\prime }}$ has not yet required attention; add $x_s$ into $I_{e^{\prime },s}$.

For each $e\le s$, add generators into $I_{e,s}$ depending on whether ${\mathcal{P}}_e$ requires attention at stage s or not.

Case 1. If ${\mathcal{P}}_e$ does not require attention at the current stage s, act as follows:

(1)

If ${\mathcal{P}}_e$ has not required attention before, add $x_s$ into $I_{e,s}$.

(2)

Otherwise, there are stages $<s$ at which ${\mathcal{P}}_e$ required attention, let $s_1<\cdots <s_m$ be all such stages $<s$. For $1\le j\le m$, $s_j$ is the j-th stage at which ${\mathcal{P}}_e$ requires attention, then $s_j$ is also the j-th stage at which a number enumerates into $A_e$. According to the choice of $A_e$ in Lemma 1, we have $s_{j+1}>s_j+j+1\ge s_j+2$ for all $1\le j\le m-1$. Act as follows:

(a)

add $x_s-x_{s-1}$ into $I_{e,s}$ if $s>s_m+2$;

(b)

add $x_s-x_{s-2}$ if $s=s_m+2$, i.e., ${\mathcal{P}}_e$ required attention at stage $s-2$;

(c)

do nothing if $s=s_m+1$.

Case 2. If ${\mathcal{P}}_e$ requires attention at stage s, act as follows:

(1)

If s is the first stage at which ${\mathcal{P}}_e$ requires attention, add $1_R-(x_s+x_{s+1})$ into $I_{e,s}$.

(2)

Otherwise, let $s_1<\cdots <s_m$ be all stages $<s$ at which ${\mathcal{P}}_e$-requirement required attention before, then $s=s_{m+1}$, the $(m+1)$-th stage at which ${\mathcal{P}}_e$ requires attention. Again, by the choice of the enumeration $\langle A_{e,s}:e,s\in \mathbb{N}\rangle$, we have $s=s_{m+1}>s_m+m+1\ge s_m+2$. Add $x_{s_m}-(x_s+x_{s+1})$ into $I_{e,s}$.

This ends the construction.

An ideal I of a computable ring R is computable if there is an algorithm that determines the membership of I, that is, whether $r\in I$ or not for an element $r\in R$.

Lemma 2. 

For all $e,s\ge 0$, $I_{e,s}$ is a computable ideal of R.

Proof. 

Fix $e,s\ge 0$. To show that $I_{e,s}$ is computable, we need to provide an algorithm to decide whether $f\in I_{e,s}$ or not for a polynomial $f\in R$. First of all, each polynomial $f\in R$ can be written as

$$f=a+\sum _{k\in K}\sum _{(j_1,\dots ,j_k)\in J}{b}_{j_1\dots j_k}{x}_{j_1}\cdots x_{j_k}$$

with $a,b_{j_1\cdots j_k}\in \mathbb{Q}$ for $(j_1,\dots ,j_k)\in J$, $k\in K$, and finite sets $K\subseteq \mathbb{N}$, $J\subseteq {\bigcup }_{k=1}^{\infty }\{(j_1,\dots ,j_k)\in {\mathbb{N}}^k:j_1<\cdots <j_k\}$.

If $s=0$, $I_{e,0}$ is the ideal of R generated by $x_0$. Given $f\in R$, calculate a polynomial $f^{\ast }\in R$ by replacing each appearance of $x_0$ in f (if any) with $0_R$. Then $f\in I_{e,0}\iff f^{\ast }=0_R$. So $I_{e,0}$ is computable.

If $s\ge 1$ and ${\mathcal{P}}_e$ has not yet required attention by the end of stage s, then $I_{e,s}$ is the ideal of R generated by $\{x_t:0\le t\le s\}$. Given $f\in R$, calculate a polynomial $f^{\ast }\in R$ by replacing each appearance of $x_t(0\le t\le s)$ in f (if any) with $0_R$. Then, $f\in I_{e,s}\iff f^{\ast }=0_R$, and $I_{e,s}$ is computable.

If $s\ge 1$ and there are stages $\le s$ at which ${\mathcal{P}}_e$ requires attention, let $s_1<\cdots <s_m$ be all such stages.

Case 1. $m=1$ with $s_1=s$. $I_{e,s}$ is the ideal of R generated by $\{x_t:0\le t\le s-1\}\cup \{1_R-(x_s+x_{s+1})\}.$ Elements of $I_{e,s}$ are of the form

$$f=\sum _{t=0}^{s-1}{g}_t{x}_t+h[1_R-(x_s+x_{s+1})]$$

with $g_t,h\in R$. By replacing indeterminates of the form $x_t(0\le t<s)$ in f with $0_R$, we obtain $f^{\prime }=h^{\prime }[1_R-(x_s+x_{s+1})]$, where $h^{\prime }$ is obtained from h by the same actions. Now, elements of $I_{e,s}$ are of the form $f^{\prime }=h^{\prime }[1_R-(x_s+x_{s+1})]$.

Let $f\in R$. Determine whether $f\in I_{e,s}$ or not as follows.

(1)

Substitute each $x_t(t<s)$ by $0_R$ in f to obtain a polynomial $f^{\prime }\in R$.

(2)

As $x_s{x}_{s+1}=x_s[(x_s+x_{s+1})-1_R]\in I_{e,s}$, we can assume that $f^{\prime }=q+g{x}_s+h{x}_{s+1}$, where $q,g,h\in R$ do not contain any $x_i$ with $i\le s+1$.That is, $q,g,h$ are polynomials of the form

$$a+\sum _{k\in K}\sum _{(j_1,\dots ,j_k)\in J}{b}_{j_1\dots j_k}{x}_{j_1}\cdots x_{j_k}$$

with $a,b_{j_1\cdots j_k}\in \mathbb{Q}$ for $k\in K$, $(j_1,\dots ,j_k)\in J$, and finite sets $K\subseteq \mathbb{N}$ and $J\subseteq {\bigcup }_{k=1}^{\infty }\{(j_1,\dots ,j_k)\in {\mathbb{N}}^k:s+1<j_1<\cdots <j_k\}$.

Now, we have

$$\begin{array}{cc}\hfill f\in I_{e,s}\iff f^{\prime }& =q+g{x}_s+h{x}_{s+1}\in I_{e,s}\hfill \\ \hfill \iff f^{\prime }& =h^{\prime }[1_R-(x_s+x_{s+1})]\hfill \end{array}$$

for some $h^{\prime }\in R$. Suppose that $f^{\prime }\in I_{e,s}$.

(i)

Take $x_s=1_R$ and $x_{s+1}=0_R$; we have $q+g=0$.

(ii)

Take $x_{s+1}=1_R$ and $x_s=0_R$; we have $q+h=0$.

Furthermore, we can check that

$$\begin{array}{c}\hfill f^{\prime }\in I_{e,s}\iff g=h=-q.\end{array}$$

In this case, $f^{\prime }=q[1_R-(x_s+x_{s+1})]$ with $q\in R$.

So, $I_{e,s}$ is a computable ideal of R.

Case 2. $m=1$ with $s_1<s$. $I_{e,s}$ is the ideal of R generated by $\{x_{t_0}:0\le t_0\le s_1-1\}\cup \{1_R-(x_{s_1}+x_{s_1+1})\}\cup \{x_{s_1+2}-x_{s_1}\}\cup \{x_{t_1}-x_{t_1-1}:s_1+3\le t_1\le s\}.$ Elements of $I_{e,s}$ can be expressed as

$$f=\sum _{t_0=0}^{s_1-1}{g}_{t_0}{x}_{t_0}+h_1[1_R-(x_{s_1}+x_{s_1+1})]+h_2(x_{s_1+2}-x_{s_1})+\sum _{t_1=s_1+3}^s{g}_{t_1}(x_{t_1}-x_{t_1-1})$$

with $g_{t_0},g_{t_1},h_1,h_2\in R$ for $0\le t_0<s_1$, $s_1+3\le t_1\le s$. By replacing each $x_{t_0}(0\le t_0<s_1)$ in f with $0_R$ and then replacing each $x_{t_1}(s_1+2\le t_1\le s)$ with $x_{s_1}$, we obtain $f^{\prime }=h^{\prime }[1_R-(x_{s_1}+x_{s_1+1})]$. Now, elements of $I_{e,s}$ are of the form $f^{\prime }=h^{\prime }[1_R-(x_{s_1}+x_{s_1+1})]$ with neither $x_{t_0}(0\le t_0<s_1)$ nor $x_{t_1}(s_1+2\le t_1\le s)$ occurring in $h^{\prime }$. Similar to Case 1, one can determine whether a polynomial $f\in R$ is of the form $h^{\prime }[1_R-(x_{s_1}+x_{s_1+1})]$ for some $h^{\prime }\in R$. So, $I_{e,s}$ is computable.

Case 3. $m=2$ with $s_1<s_2=s$. $I_{e,s}$ is the ideal of R generated by

$$\begin{array}{cc}\hfill & \{x_{t_0}:0\le t_0\le s_1-1\}\cup \{1_R-(x_{s_1}+x_{s_1+1})\}\cup \{x_{s_1+2}-x_{s_1}\}\hfill \\ \hfill & \cup \{x_{t_1}-x_{t_1-1}:s_1+3\le t_1\le s-1\}\cup \{x_{s_1}-(x_s+x_{s+1})\}.\hfill \end{array}$$

Similar to Case 2, one can reduce elements of $I_{e,s}$ to polynomials of the form

$$f^{\prime }=h_1^{\prime }[1_R-(x_{s_1}+x_{s_1+1})]+h_2^{\prime }[x_{s_1}-(x_s+x_{s+1})],$$

such that $h_1^{\prime },h_2^{\prime }$ do not contain $x_{t_0}(0\le t_0<s_1)$ or $x_{t_1}(s_1+2\le t_1\le s-1)$. Then, by replacing $x_{s_1}$ with $x_s+x_{s+1}$ in $f^{\prime }$, we obtain a polynomial $f^{″}=h_1^{″}[1_R-(x_s+x_{s+1}+x_{s_1+1})]$ with $h_1^{″}$ the corresponding polynomial obtained from $h_1^{\prime }$.

Let $f\in R$. Determine whether $f\in I_{e,s}$ or not as follows.

(1)

In f, by substituting each $x_{t_0}(0\le t_0<s_1)$ with $0_R$, and each $x_{t_1}(s_1+2\le t_1\le s-1)$ with $x_{s_1}$, we obtain $f^{\prime }\in R$; then substituting each $x_{s_1}$ with $x_s+x_{s+1}$ in $f^{\prime }$, we obtain $f^{″}\in R$.

(2)

Since $1_R-(x_{s_1}+x_{s_1+1}),x_{s_1}-(x_s+x_{s+1})\in I_{e,s}$, and

$$\begin{array}{cc}\hfill x_s{x}_{s+1}& =x_s{x}_{s+1}+x_s{x}_{s+1}-x_s{x}_{s+1}\hfill \\ \hfill & =x_s{x}_{s+1}{x}_s+x_s{x}_{s+1}{x}_{s+1}-x_s{x}_{s+1}[x_{s_1}+x_{s_1+1}+1_R-(x_{s_1}+x_{s_1+1})]\hfill \\ \hfill & =x_s{x}_{s+1}(x_s+x_{s+1}-x_{s_1})-x_s{x}_{s+1}[1_R-(x_{s_1}+x_{s_1+1})]-x_s{x}_{s+1}{x}_{s_1+1},\hfill \end{array}$$

we have $x_s{x}_{s+1}(1_R+x_{s_1+1})\in I_{e,s}$. Together with

$$\begin{array}{cc}\hfill x_s{x}_{s+1}(1_R-x_{s_1+1})& =x_s{x}_{s+1}(1_R-x_{s_1+1}^2)\hfill \\ \hfill & =x_s{x}_{s+1}(1_R+x_{s_1+1})(1_R-x_{s_1+1})\in I_{e,s},\hfill \end{array}$$

we have

$$2x_s{x}_{s+1}=x_s{x}_{s+1}(1_R+x_{s_1+1})+x_s{x}_{s+1}(1_R-x_{s_1+1})\in I_{e,s},$$

and thus $x_s{x}_{s+1}\in I_{e,s}$. Then,

$$\begin{array}{cc}\hfill x_s{x}_{s_1+1}& =x_s[(x_{s_1}+x_{s_1+1})-1_R]-(x_s{x}_{s_1}-x_s)\hfill \\ \hfill & =x_s[(x_{s_1}+x_{s_1+1})-1_R]-[x_s(x_{s_1}-x_s-x_{s+1})+x_s(x_s+x_{s+1})-x_s^2]\hfill \\ \hfill & =x_s[(x_{s_1}+x_{s_1+1})-1_R]-x_s(x_{s_1}-x_s-x_{s+1})-x_s{x}_{s+1}\in I_{e,s}.\hfill \end{array}$$

By a similar argument, $x_{s+1}{x}_{s_1+1}\in I_{e,s}$. We can further assume that $f^{″}$ in (1) is of the form $f^{″}=q+g{x}_s+h_1{x}_{s+1}+h_2{x}_{s_1+1}\in R,$ where $q,g,h_1,h_2\in R$ do not contain any $x_t$ for $0\le t\le s+1$. Now,

$$\begin{array}{cc}\hfill f\in I_{e,s}\iff f^{″}& =q+g{x}_s+h_1{x}_{s+1}+h_2{x}_{s_1+1}\in I_{e,s}\hfill \\ \hfill \iff f^{″}& =h^{\prime }[1_R-(x_s+x_{s+1}+x_{s_1+1})]\hfill \end{array}$$

for some $h^{\prime }\in R$. Suppose that $f^{″}\in I_{e,s}$.

(i)

Take $x_s=1_R$ and $x_{s+1}=x_{s_1+1}=0_R$; we have $q+g=0$.

(ii)

Take $x_{s+1}=1_R$ and $x_s=x_{s_1+1}=0_R$; we have $q+h_1=0$.

(iii)

Take $x_{s_1+1}=1_R$ and $x_s=x_{s+1}=0_R$; we have $q+h_2=0$.

Then, $f^{″}\in I_{e,s}$ if and only if $g=h_1=h_2=-q.$ In this case, we have

$$f^{″}=q[1_R-(x_s+x_{s+1}+x_{s_1+1})].$$

So, $I_{e,s}$ is a computable ideal of R.

In general, when $m\ge 2$ with $s_1<\cdots <s_m\le s$, as in Case 3 above, elements of $I_{e,s}$ are finite R-linear sums of generators added by stage s of the construction. For any polynomial $f\in R$, by reducing generators of $I_{e,s}$, we obtain a polynomial $f^{\prime }$ with $f-f^{\prime }\in I_{e,s}$ satisfying:

$$f^{\prime }\in I_{e,s}\iff f^{\prime }=h[1_R-(x_{s_m}+x_{s_m+1}+\cdots +x_{s_1+1})]$$

for some $h\in R$. Furthermore, we can prove by induction on m with $m\ge 1$ that $x_{s_m}{x}_{s_i+1}\in I_{e,s}$ for all $1\le i\le m$ and $x_{s_i+1}{x}_{s_j+1}\in I_{e,s}$ for all $1\le i<j\le m$ (refer to ($\ast 1$), ($\ast 2$) in the proof of Lemma 3 below). Now, we can obtain a polynomial $f^{″}$ with $f^{\prime }-f^{″}\in I_{e,s}$ such that

$$\begin{array}{c}\hfill f^{″}=q+g{x}_{s_m}+\sum _{k=1}^m{h}_k{x}_{s_k+1},\end{array}$$

with $q,g,h_k\in R(1\le k\le m)$ containing no $x_t$ for $0\le t\le s_m+1$. Similar to Case 3 above, we see that $f^{″}\in I_{e,s}\iff g=h_k=-q$ for $1\le k\le m$, in which case we have

$$f^{″}=q[1_R-(x_{s_m}+x_{s_m+1}+\cdots +x_{s_1+1})].$$

Therefore, $I_{e,s}$ is a computable ideal of R.

This completes the proof of Lemma 2. □

Note that a sequence $\langle I_e:e\in \mathbb{N}\rangle$ of computable ideals of a computable ring R is uniformly computable if and only if the coding set $X=\{\langle e,r\rangle \in \mathbb{N}\times R:r\in I_e\}$ is a computable subset of $\mathbb{N}\times R$. Based on Lemma 2, we explain that the sequence $\langle I_e={\bigcup }_{s\in \mathbb{N}}{I}_{e,s}:e\in \mathbb{N}\rangle$ of ideals of R is uniformly computable. Let

$$f=a+\sum _{k\in K}\sum _{(j_1,\dots ,j_k)\in J}{b}_{j_1\dots j_k}{x}_{j_1}\cdots x_{j_k}\in R$$

with $a,b_{j_1\cdots j_k}\in \mathbb{Q}$ for $(j_1,\dots ,j_k)\in J$, $k\in K$, and finite sets $K\subseteq \mathbb{N}$, $J\subseteq {\bigcup }_{k=1}^{\infty }\{(j_1,\dots ,j_k)\in {\mathbb{N}}^k:j_1<\cdots <j_k\}$. Let $N_f$ be the largest index i such that $x_i$ occurs in the expression of f above. We can check that $f\in I_e\iff f\in I_{e,N_f}$. Indeed, suppose that ${\mathcal{P}}_e$ has required attention for exactly m many times by stage $N_f$ for some $m\ge 1$. Let $s_1<\cdots <s_m\le N_f$ be all stages at which ${\mathcal{P}}_e$ required attention. Then, the reduced form of f at stage $N_f$ is

$$f^{\ast }=a+b{x}_{s_m}+c_1{x}_{s_m+1}+c_2{x}_{s_{m-1}+1}+\cdots +c_m{x}_{s_1+1}$$

with $a,b,c_i\in \mathbb{Q}$ for $1\le i\le m$. Then, $f\in I_e$ if and only if $b=c_i=-a$ for all $1\le i\le m$ if and only if

$$f^{\ast }=a[1_R-(x_{s_m}+x_{s_m+1}+\cdots +x_{s_1+1})]$$

if and only if $f\in I_{e,N_f}$.

As an example, consider the simple case of $m=1$; that is, ${\mathcal{P}}_e$ has required attention exactly once by stage $N_f$. Let $s_1\le N_f$ be the first stage at which ${\mathcal{P}}_e$ required attention. We have

$$1_R-(x_{s_1}+x_{s_1+1}),x_{s_1+2}-x_{s_1},x_t-x_{t-1}\in I_{e,N_f}$$

for $s_1+2<t\le N_f$ (if any). The reduced form of each $x_i$ with $i<s_1$ in f is $0_R$, the reduced form of each $x_t(s_1+2\le t\le N_f)$ is $x_{s_1}$. Also, note that $x_{s_1}{x}_{s_1+1}\in I_{e,s}$, then the reduced form of f at stage $N_f$ is

$$f^{\ast }=a+b{x}_{s_1}+c{x}_{s_1+1}$$

for some $a,b,c\in \mathbb{Q}$. Although $f^{\ast }$ may be assigned new expressions by replacing $x_{s_1}$ with $x_{s_2}+x_{s_2+1}$ at next stage $s_2>N_f$ at which ${\mathcal{P}}_e$ requires attention and then by replacing $x_{s_2}$ with $x_{s_3}+x_{s_3+1}$ at next stage $s_3>s_2>N_f$ at which ${\mathcal{P}}_e$ requires attention, and so on, we see that $f\in I_e$ if and only if $b=c=-a$ if and only if $f^{\ast }=a[1_R-(x_{s_1}+x_{s_1+1})]$ if and only if $f\in I_{e,N_f}$.

For a number e and a given polynomial $f\in R$, we can first find the largest index $N_f$ such that $x_{N_f}$ occurs in the expression of f as a finite $\mathbb{Q}$-linear sum of monomials of R, and then proceed as in Lemma 2 to see whether $f\in I_{e,N_f}$ or not. This shows that the sequence $\langle R_e=R/I_e:e\in \mathbb{N}\rangle$ of quotient rings of R is uniformly computable.

For any $e\in \mathbb{N}$ and $f,g\in R$, if $f-g\in I_e$, we write $f{=}_{R_e}g$, meaning that $f=g$ in the quotient ring $R_e=R/I_e$.

Lemma 3. 

For all e, ${\mathcal{P}}_e$ is satisfied.

Proof. 

Assume that $e\in \mathrm{Fin}$; then, $A_e=W_e$ is finite and there are only finitely many stages at which ${\mathcal{P}}_e$ requires attention. Let $s_1<s_2<\cdots <s_n$ be all the consecutive stages at which ${\mathcal{P}}_e$ requires attention.

If $n=0$, $W_e=\varnothing$. ${\mathcal{P}}_e$ required no attention during the construction, so $x_t{= }_{R_e} 0_R$ for all $t\in \mathbb{N}$. $R_e=\mathbb{Q}$ is the field of rational numbers, which is also a semisimple ring.

If $n=1$, then the following equalities hold in the quotient ring $R_e= R/I_e$.

(1)

$x_0 {= }_{R_e} x_1 {= }_{R_e}\cdots {= }_{R_e} x_{s_1-1} {= }_{R_e} 0_R$.

(2)

$1_R {= }_{R_e} x_{s_1}+x_{s_1+1}$ and $x_{s_1}{x}_{s_1+1} {= }_{R_e} 0_R$.

(3)

For all $t>s_1+1$, $x_t {= }_{R_e} x_{s_1}$.

In this case, $R_e$ can be decomposed as $R_e= R_e{x}_{s_1}\oplus R_e{x}_{s_1+1}.$ Furthermore, $R_e{x}_{s_1}= \mathbb{Q}{x}_{s_1}$ and $R_e{x}_{s_1+1}=\mathbb{Q}{x}_{s_1+1}$ are simple $R_e$-modules. So $R_e$ is a semisimple ring.

If $n\ge 2$, then the following equalities hold in the quotient ring $R_e$.

(1)

$x_0 {= }_{R_e} x_1{= }_{R_e}\cdots {= }_{R_e} x_{s_1-1} {= }_{R_e} 0_R$.

(2)

$1_R {= }_{R_e} x_{s_1}+x_{s_1+1}$ and $x_{s_1}{x}_{s_1+1} {= }_{R_e} 0_R$.

(3)

$x_{s_1} {= }_{R_e} x_{s_1+2} {= }_{R_e} \cdots {= }_{R_e} x_{s_2-1}$ and $x_{s_1} {= }_{R_e} x_{s_2}+x_{s_2+1}$. Then, we have $x_{s_2}{x}_{s_2+1} {= }_{R_e} 0_R$. Indeed,

$$\begin{array}{cc}\hfill x_{s_2}{x}_{s_2+1}& {= }_{R_e} x_{s_2}{x}_{s_2+1}+x_{s_2}{x}_{s_2+1}-x_{s_2}{x}_{s_2+1}\hfill \\ \hfill & {= }_{R_e} x_{s_2}{x}_{s_2+1}{x}_{s_2}+x_{s_2}{x}_{s_2+1}{x}_{s_2+1}-x_{s_2}{x}_{s_2+1}(x_{s_1}+x_{s_1+1})\hfill \\ \hfill & {= }_{R_e} x_{s_2}{x}_{s_2+1}(x_{s_2}+x_{s_2+1}-x_{s_1})-x_{s_2}{x}_{s_2+1}{x}_{s_1+1}\hfill \\ \hfill & {= }_{R_e} -x_{s_2}{x}_{s_2+1}{x}_{s_1+1}\hfill \\ \hfill & {= }_{R_e} -(-x_{s_2}{x}_{s_2+1}{x}_{s_1+1})x_{s_1+1}\hfill \\ \hfill & {= }_{R_e} x_{s_2}{x}_{s_2+1}{x}_{s_1+1} {= }_{R_e} 0_R.\hfill \end{array}$$

(4)

For $1\le k<n$, $x_{s_k} {= }_{R_e} x_{s_k+2} {= }_{R_e} x_{s_k+3} {= }_{R_e} \cdots {= }_{R_e} x_{s_{k+1}-1}$ and $x_{s_k} {= }_{R_e} x_{s_{k+1}}+x_{s_{k+1}+1}$.

(5)

For all $t>s_n+1$, $x_t {= }_{R_e} x_{s_n}$.

In $R_e$, we have $1_R {= }_{R_e} x_{s_n}+x_{s_n+1}+\cdots +x_{s_2+1}+x_{s_1+1}.$ Furthermore, the following equalities hold:

($\ast 1$)

for all $1\le i\le n$, $x_{s_n}{x}_{s_i+1} {= }_{R_e} 0_R$;

($\ast 2$)

for all $1\le i<j\le n$, $x_{s_i+1}{x}_{s_j+1} {= }_{R_e} 0_R$.

We provide the proof of $(\ast 1)$ and $(\ast 2)$ by induction on n with $n\ge 2$.

Basic Step. As $x_{s_1}{x}_{s_1+1} {= }_{R_e} x_{s_2}{x}_{s_2+1} {= }_{R_e} 0_R$, $x_{s_1} {= }_{R_e} x_{s_2}+x_{s_2+1}$, we have

$$\begin{array}{cc}\hfill x_{s_2+1}{x}_{s_1+1}& {= }_{R_e} (x_{s_1}-x_{s_2})x_{s_1+1}\hfill \\ \hfill & {= }_{R_e} 0_R-x_{s_2}{x}_{s_1+1}\hfill \\ \hfill & {= }_{R_e} (-x_{s_2}{x}_{s_1+1})(-x_{s_2}{x}_{s_1+1}) {= }_{R_e} x_{s_2}{x}_{s_1+1} {= }_{R_e} 0_R.\hfill \end{array}$$

$(\ast 1)$ and $(\ast 2)$ hold for $n=2$.

Inductive Step. Assume that $(\ast 1)$ and $(\ast 2)$ hold for $n-1$ with $n-1\ge 1$. We prove that the equalities hold for n. For $(\ast 1)$. First of all, we have

$$\begin{array}{cc}\hfill x_{s_n}{x}_{s_n+1}& {= }_{R_e} x_{s_n}{x}_{s_n+1}+x_{s_n}{x}_{s_n+1}-x_{s_n}{x}_{s_n+1}\hfill \\ \hfill & {= }_{R_e} x_{s_n}{x}_{s_n+1}{x}_{s_n}+x_{s_n}{x}_{s_n+1}{x}_{s_n+1}-x_{s_n}{x}_{s_n+1}{1}_R\hfill \\ \hfill & {= }_{R_e} x_{s_n}{x}_{s_n+1}(x_{s_n}+x_{s_n+1})-x_{s_n}{x}_{s_n+1}(x_{s_n}+x_{s_n+1}+\sum _{k=1}^{n-1}{x}_{s_k+1})\hfill \\ \hfill & {= }_{R_e} -\sum _{k=1}^{n-1}{x}_{s_n}{x}_{s_n+1}{x}_{s_k+1}.\hfill \end{array}$$

By induction, $x_{s_{n-1}}{x}_{s_i+1} {= }_{R_e} (x_{s_n}+x_{s_n+1})x_{s_i+1} {= }_{R_e} 0_R$ for all $1\le i\le n-1$, and thus, we have $x_{s_n}{x}_{s_i+1} {= }_{R_e} -x_{s_n+1}{x}_{s_i+1}$. Then,

$$\begin{array}{cc}\hfill x_{s_n}{x}_{s_i+1} {= }_{R_e} x_{s_n}{x}_{s_i+1}{x}_{s_n}{x}_{s_i+1}& {= }_{R_e} x_{s_n}{x}_{s_i+1}(-x_{s_n+1}{x}_{s_i+1})\hfill \\ \hfill & {= }_{R_e} -x_{s_n}{x}_{s_n+1}{x}_{s_i+1}\hfill \\ \hfill & {= }_{R_e} (\sum _{k=1}^{n-1}{x}_{s_n}{x}_{s_n+1}{x}_{s_k+1})x_{s_i+1} {= }_{R_e} x_{s_n}{x}_{s_n+1}{x}_{s_i+1}.\hfill \end{array}$$

Now, $x_{s_n}{x}_{s_i+1} {= }_{R_e} x_{s_n}{x}_{s_n+1}{x}_{s_i+1} {= }_{R_e} -x_{s_n}{x}_{s_n+1}{x}_{s_i+1} {= }_{R_e} 0_R$ for all $1\le i\le n-1$. Then, we have

$$x_{s_n}{x}_{s_n+1} {= }_{R_e} -\sum _{k=1}^{n-1}{x}_{s_n}{x}_{s_n+1}{x}_{s_k+1} {= }_{R_e} 0_R.$$

This shows that $(\ast 1)$ holds for n.

For $(\ast 2)$. By induction, we only need to show that $x_{s_n+1}{x}_{s_i+1} {= }_{R_e} 0_R$ for all $1\le i<n$. Indeed, we have already seen that $x_{s_{n-1}}{x}_{s_i+1} {= }_{R_e} 0_R$ and $x_{s_n}{x}_{s_i+1} {= }_{R_e} 0_R$, together with $x_{s_{n-1}}=x_{s_n}+x_{s_n+1}$, we have

$$\begin{array}{cc}\hfill x_{s_n+1}{x}_{s_i+1}& {= }_{R_e} (x_{s_n}+x_{s_n+1})x_{s_i+1}-x_{s_n}{x}_{s_i+1}\hfill \\ \hfill & {= }_{R_e} x_{s_{n-1}}{x}_{s_i+1}-x_{s_n}{x}_{s_i+1} {= }_{R_e} 0_R.\hfill \end{array}$$

This shows that $(\ast 2)$ holds for n.

This ends the proof of $(\ast 1),(\ast 2)$.

The ring $R_e$ can be decomposed as

$$R_e=R_e{x}_{s_n}\oplus R_e{x}_{s_n+1}\oplus \cdots \oplus R_e{x}_{s_1+1}.$$

We show that $R_e{x}_{s_n}$ and $R_e{x}_{s_j+1}$ are simple $R_e$-modules for $1\le j\le n$. Take a polynomial

$$f=a+\sum _{k\in K}\sum _{(j_1,\dots ,j_k)\in J}{b}_{j_1\dots j_k}{x}_{j_1}\cdots x_{j_k}\in R,$$

where $a,b_{j_1\cdots j_k}\in \mathbb{Q}$ for $(j_1,\dots ,j_k)\in J$, $k\in K$ with numbers $j_1<\cdots <j_k$, $k\ge 1$ and finite sets $J,K$. In $R_e$, f can be written as follows:

$$\begin{array}{cc}\hfill & f {= }_{R_e} f(x_{s_n}+x_{s_n+1}+\cdots +x_{s_1+1})\hfill \\ \hfill & {= }_{R_e} (a+\sum _{k\in K}\sum _{(j_1,\dots ,j_k)\in J}{b}_{j_1\dots j_k}{x}_{j_1}\cdots x_{j_k})(x_{s_n}+\sum _{j=1}^n{x}_{s_j+1})\hfill \\ \hfill & {= }_{R_e} a{x}_{s_n}+\sum _{j=1}^{n}a{x}_{s_j+1}+\sum _{k\in K}\sum _{(j_1,\dots ,j_k)\in J}{b}_{j_1\dots j_k}{x}_{j_1}\cdots x_{j_k}{x}_{s_n}+\sum _{k\in K}\sum _{(j_1,\dots ,j_k)\in J}\sum _{j=1}^n{b}_{j_1\dots j_k}{x}_{j_1}\cdots x_{j_k}{x}_{s_j+1}\hfill \end{array}$$

For any $k\ge 1$ and $j_1<\cdots <j_k$, suppose that $x_{j_1} \cdots x_{j_k}{\ne }_{R_e}{0}_R$. As $x_i {= }_{R_e} 0_R$ for $0\le i<s_1$ and $x_j {= }_{R_e} x_{s_n}$ for $j\ge s_n+2$, it suffices to prove the case that $s_1\le j_1<\cdots <j_k\le s_n+1$. Note that $x_{s_i+1}{x}_{s_j} {= }_{R_e} 0_R$ for $1\le i\le j\le n$ and $x_{s_i+1}{x}_{s_k+1} {= }_{R_e} 0_R$ for $1\le i<k\le n$. Also, note that each $x_j$ with $s_1\le j\le s_n+1$ is equal to some $x_{s_i}$ or some $x_{s_i+1}$ with $1\le i\le n$. There are two cases in the expression of $x_{j_1}\cdots x_{j_k}$ in $R_e$.

(1)

For each $1\le t\le k$, $x_{j_t}$ is equal to some $x_{s_i}$ with $1\le i\le n$. Without loss of generality, assume that $x_{j_1}\cdots x_{j_k} {= }_{R_e} x_{s_{i_1}}\cdots x_{s_{i_l}}$ for some $1\le i_1<\cdots <i_l\le n$. We first consider the expression of $x_{s_i}{x}_{s_j}$ with $1\le i<j\le n$.

$$\begin{array}{cc}\hfill x_{s_i}{x}_{s_j}& {= }_{R_e} (x_{s_{i+1}}+x_{s_{i+1}+1})x_{s_j}\hfill \\ \hfill & {= }_{R_e} \cdots {= }_{R_e} (x_{s_j}+x_{s_j+1}+\sum _{k=i+1}^{j-1}{x}_{s_k+1})x_{s_j} {= }_{R_e} x_{s_j}.\hfill \end{array}$$

Now, we have $x_{j_1}\cdots x_{j_k} {= }_{R_e} x_{s_{i_1}}\cdots x_{s_{i_l}} {= }_{R_e} x_{s_{i_l}}$.

(2)

There is at least one $1\le t\le k$ such that $x_{j_t}$ is equal to some $x_{s_j+1}$ with $1\le j\le n$. Since we assume that $x_{j_1}\cdots x_{j_k}{\ne }_{R_e}{0}_R$, there is exactly one such $x_{s_j+1}$ in the expression of $x_{j_1}\cdots x_{j_k}$. As $x_{s_i}{x}_{s_j+1} {= }_{R_e} 0_R$ if $1\le j\le i\le n$, assume that $x_{j_1}\cdots x_{j_k} {= }_{R_e} x_{s_{i_1}}\cdots x_{s_{i_l}}{x}_{s_j+1}$ for some $1\le i_1<\cdots <i_l<j\le n$. Again, we first consider the expression of $x_{s_i}{x}_{s_j+1}$ with $1\le i<j\le n$.

$$\begin{array}{cc}\hfill x_{s_i}{x}_{s_j+1}& {= }_{R_e} (x_{s_{i+1}}+x_{s_{i+1}+1})x_{s_j+1}\hfill \\ \hfill & {= }_{R_e} \cdots {= }_{R_e} (x_{s_j}+x_{s_j+1}+\sum _{k=i+1}^{j-1}{x}_{s_k+1})x_{s_j+1} {= }_{R_e} x_{s_j+1}.\hfill \end{array}$$

Then, $x_{j_1}\cdots x_{j_k} {= }_{R_e} x_{s_{i_1}}\cdots x_{s_{i_l}}{x}_{s_j+1} {= }_{R_e} x_{s_j+1}$.

Now, we see that f can be written as a $\mathbb{Q}$-linear sum of elements in $\left\{x_{s_n}\right\}\cup \{x_{s_j+1}:1\le j\le n\}$, and

$$\begin{array}{cc}\hfill R_e& =R_e{x}_{s_n}\oplus R_e{x}_{s_n+1}\oplus \cdots \oplus R_e{x}_{s_1+1}\hfill \\ \hfill & =\mathbb{Q}{x}_{s_n}\oplus \mathbb{Q}{x}_{s_n+1}\oplus \cdots \oplus \mathbb{Q}{x}_{s_1+1},\hfill \end{array}$$

which is a direct sum of simple $R_e$-modules. Furthermore, $R_e{x}_{s_n}\cong R_e{x}_{s_j+1}\cong \mathbb{Q}$ as rings for $1\le j\le n$, and thus, $R_e=R_e{x}_{s_n}\oplus R_e{x}_{s_n+1}\oplus \cdots \oplus R_e{x}_{s_1+1}\cong {\mathbb{Q}}^{n+1}$, which is the direct product of $n+1$ copies of the field $\mathbb{Q}$ of rational numbers. So, $R_e$ is a semisimple ring. □

Lemma 4.

For all e, ${\mathcal{Q}}_e$ is satisfied.

Proof. 

Assume that $e\notin \mathrm{Fin}$, then $A_e=W_e$ is infinite and ${\mathcal{P}}_e$ requires attention infinitely often. Let $s_1<s_2<\cdots <s_n<\cdots$ be all the consecutive stages at which ${\mathcal{P}}_e$ requires attention. To prove that $R_e$ is not semisimple, we only need to show that $1_R$ cannot be expressed as a sum of nonzero mutually orthogonal idempotents of the form $1_R {= }_{R_e} f_1+f_2\cdots +f_m$ with $m\ge 2$, such that each $R_e{f}_i$ is simple as $R_e$-modules for $1\le i\le m$.

Let $f\in R$ be a nontrivial idempotent in $R_e$, i.e., $f {= }_{R_e} f^2$ and $f {\ne }_{R_e}{0}_R,1_R$. If $R_{e}f$ is as simple as $R_e$-modules, then f is not a sum of two nonzero orthogonal idempotents of $R_e$ because otherwise $R_{e}f$ is decomposed as a sum of two nonzero $R_e$-submodules. Since each $x_i$ with $i<s_1$ equals zero in $R_e$, f is a polynomial in R containing at least one $x_j$ with $j\ge s_1$. Then, each $x_j$ with $j\ge s_1$ is equal to $x_{s_{m_j}}$ or $x_{s_{m_j}+1}$ in $R_e$ for the unique number $m_j$ with $j\in [s_{m_j},s_{m_j+1}-1]$. Assume that each $x_j$ that occurs in $f\in R$ is of the form $x_{s_{m_j}}$ or $x_{s_{m_j}+1}$. Let M be the largest number m, such that $x_{s_m}$ or $x_{s_m+1}$ occurs in f. Then,

$$\begin{array}{cc}\hfill f {= }_{R_e} f{1}_R& {= }_{R_e} f(x_{s_M}+x_{s_M+1}+\cdots +x_{s_1+1})\hfill \\ \hfill & {= }_{R_e} f{x}_{s_M}+f{x}_{s_M+1}+\cdots +f{x}_{s_1+1}.\hfill \end{array}$$

Since f is not a sum of two nonzero orthogonal idempotents of $R_e$, either $f {= }_{R_e} f{x}_{s_M}$ or $f {= }_{R_e} f{x}_{s_j+1}$ for some $1\le j\le M$.

(1)

If $f {= }_{R_e} f{x}_{s_M}$, then $f {= }_{R_e} f{x}_{s_M} {= }_{R_e} q{x}_{s_M}$ for some nonzero $q\in \mathbb{Q}$. As f is an idempotent in $R_e$, we have $q=1$ and $f {= }_{R_e} x_{s_M} {= }_{R_e} x_{s_{M+1}}+x_{s_{M+1}+1},$ which is a sum of two nonzero orthogonal idempotents of $R_e$. This case is impossible.

(2)

If $f {= }_{R_e} f{x}_{s_j+1}$, then $f {= }_{R_e} f{x}_{s_j+1} {= }_{R_e} q{x}_{s_j+1}$ for some nonzero $q\in \mathbb{Q}$. Again, we have $q=1$ and $f {= }_{R_e} q{x}_{s_j+1} {= }_{R_e} x_{s_j+1}$.

This shows that a nontrivial idempotent f of $R_e$ is of the form $x_{s_j+1}$ if $R_{e}f$ is a simple $R_e$-module.

The identity $1_R$ of R can be expressed as $1_R {= }_{R_e} x_{s_n}+x_{s_n+1}+\cdots +x_{s_1+1}$ in $R_e$ for any $n\ge 1$. If $1_R {= }_{R_e} f_1+f_2\cdots +f_m$ with $m\ge 2$ is a sum of nonzero mutually orthogonal idempotents of $R_e$, such that each $R_e{f}_i$ is as simple as $R_e$-modules for $1\le i\le m$, then $f_i$ is of the form $x_{s_j+1}$ with $j\ge 1$; however, this is impossible, because any such an expression $1_R {= }_{R_e} f_1+f_2\cdots +f_m$ does not contain one summand of the form $x_{s_n}$. Therefore, $R_e$ is not a direct sum of simple $R_e$-submodules and it is not a semisimple ring. □

We have constructed a uniformly computable sequence of computable rings $\langle R_e:e\in \mathbb{N}\rangle$ such that $e\in \mathrm{Fin}$ if and only if $R_e$ is a computable semisimple ring. Then, $\mathrm{Fin}$ is m-reducible to the index set of computable semisimple rings, and thus, the latter set is ${\mathsf{\Sigma }}_2^0$-hard within the index set of computable rings.

This completes the proof of Theorem 1. □

In the proof of Lemma 3 above, for $e\in \mathrm{Fin}$, $R_e$ is a commutative ring that is isomorphic to the direct product of finitely many copies of the rational field $\mathbb{Q}$. In fact, due to the Wedderburn–Artin Theorem (see page 33, [20]), a commutative semisimple ring is isomorphic to a finite direct product of fields.

Corollary 1.

The index set of finite direct product of computable fields is ${\mathsf{\Sigma }}_2^0$-hard within the index set of computable rings.

The class of semisimple rings forms a subclass of semiperfect rings. The sequence of rings $\langle R_e:e\in \mathbb{N}\rangle$ constructed in the proof of Theorem 1 also implies a corresponding result of semiperfect rings. Indeed, when $e\in \mathrm{Fin}$, by Lemma 3, $R_e$ is semiperfect. Furthermore, when $e\notin \mathrm{Fin}$, suppose that $1_R {= }_{R_e} f_1+\cdots +f_m$ is a sum of mutually orthogonal nontrivial idempotents $f_1,\dots ,f_m$ of $R_e$. For each $1\le i\le m$, we see from the proof of Lemma 4 that there exists a large enough number $M_i$, such that any $x_n$ occurring in $f_i$ satisfying $x_n{\ne }_{R_e}{x}_{M_i}$. Then, $g=x_{M_i}{f}_i$ and $h=(x_{M_i}+1_R)f_i$ are not invertible in the ring $f_i{R}_e{f}_i=R_e{f}_i$ with identity $f_i$, but $g-h {= }_{R_e} f_i$ is the identity of the ring $f_i{R}_e{f}_i$, and thus invertible. By definition, $f_i{R}_e{f}_i$ is not a local ring with identity $f_i$ for any $1\le i\le m$, i.e., $f_i$ is not a local idempotent of $R_e$. So, $R_e$ is not a semiperfect ring for $e\notin \mathrm{Fin}$. This implies the following.

Corollary 2.

The index set of computable semiperfect rings is ${\mathsf{\Sigma }}_2^0$-hard within the index set of computable rings.

4. Local Rings

In this section, we prove the result on the complexity of computable local rings.

Theorem 2.

The index set of computable local rings is ${\mathsf{\Pi }}_2^0$-complete within the index set of computable rings.

Proof. 

Since the index set $\mathrm{Inf}=\{e\in \mathbb{N}:|W_e|=\infty \}$ of infinite computably enumerable sets is ${\mathsf{\Pi }}_2^0$-complete, we only need to construct a uniformly computable sequence of computable rings $\langle R_e:e\in \mathbb{N}\rangle$ such that for all e, $e\in \mathrm{Inf}$ if and only if $R_e$ is local. Consider the polynomial ring $R=\mathbb{Z}[x_n:n\in \mathbb{N}]$ with infinitely many indeterminates $\{x_n:n\in \mathbb{N}\}$. Let

$$F=\{{\displaystyle \frac{f}{g}}:f,g\in R,g\ne 0,(f,g)=1\}$$

be the fraction field of R with the following usual operations:

$${\displaystyle \frac{f_1}{g_1}}+F{\displaystyle \frac{f_2}{g_2}}={\displaystyle \frac{f_1{g}_2+g_1{f}_2}{g_1{g}_2}} \mathrm{and} {\displaystyle \frac{f_1}{g_1}}·F{\displaystyle \frac{f_2}{g_2}}={\displaystyle \frac{f_1{f}_2}{g_1{g}_2}}.$$

Here, for $f,g\in R$ with $g\ne 0$, we identify ${\displaystyle \frac{f}{g}}={\displaystyle \frac{f^{\prime }}{g^{\prime }}}\in F$ for the unique $f^{\prime },g^{\prime }\in R$ with $(f^{\prime },g^{\prime })=1$, i.e., $f^{\prime },g^{\prime }$ coprime. The rings R and F are computable with fixed codings into natural numbers.

Fix a standard enumeration $\langle W_{e,s}:e,s\in \mathbb{N}\rangle$ of all c.e. sets $\langle W_e:e\in \mathbb{N}\rangle$ with $W_e={\bigcup }_{s\in \mathbb{N}}{W}_{e,s}$, $W_{e,0}=\varnothing$ and $|W_{e,s}-W_{e,s-1}|\le 1$ for all $s\ge 1$. That is, at each stage s, there is at most one number enumerating into $W_e$.

We first enumerate a uniformly computably enumerable sequence of c.e. subrings $\langle T_e={\bigcup }_{s\in \mathbb{N}}{T}_{e,s}:e\in \mathbb{N}\rangle$ of the field F by stages, such that for all e,

$$e\in \mathrm{Inf}\iff T_e \mathrm{is} \mathrm{a} \mathrm{local} \mathrm{ring}.$$

To make sure that $T_e$ is local for $e\in \mathrm{Inf}$, rather than using the ${\mathsf{\Pi }}_2^0$ definition of local rings in terms of noninvertible elements, we use the fact that fields are natural examples of local rings. So, we enumerate the sequence of c.e. subrings $\langle T_e:e\in \mathbb{N}\rangle$ of F such that $T_e$ is exactly the whole field F when $W_e$ is infinite.

Fix an effective listing of the polynomial ring $R=\mathbb{Z}[x_n:n\in \mathbb{N}]$ without repetitions as $r_0=0,r_1=1,r_2,\dots ,r_n,\dots .$ That is, $R\setminus \{0,1\}=\{r_i:i\ge 2\}$. We view R as a subring of the fraction field F and identify an element $r\in R$ with ${\displaystyle \frac{r}{1}}$ in F. For any $r\in R$, let $N\left(r\right)$ denote the largest index i such that $x_i$ occurs in r if $r\notin \mathbb{Z}$; let $N\left(r\right):=0$ if $r\in \mathbb{Z}$. For example, if $r=2+3x_1{x}_3^6+x_2^4{x}_5^2$, then $N\left(r\right)=5$.

Construction of $\langle T_e:e\in \mathbb{N}\rangle$.

Stage 0. For all e, let $T_{e,0}=\{0,1\}$.

For $s\ge 1$, assume that we have obtained $T_{e,s-1}$ for all e. Stage $s\ge 1$ proceeds as follows. There are three cases.

Stage $s=3n$. For all $e>n$, let $T_{e,s}=T_{e,s-1}$.

For all $e\le n$, check whether some number goes into $W_e$ at stage n.

(1)

If the answer is yes, i.e., $W_{e,n}\ne W_{e,n-1}$, then for all $r_i$ with $2\le i\le n$ and $N\left(r_i\right)\le n$, add ${\displaystyle \frac{1}{r_i}}$ into $T_{e,s}$ if it was not enumerated into $T_e$ before. In this case,

$$T_{e,s}=T_{e,s-1}\cup \{{\displaystyle \frac{1}{r_i}}:2\le i\le n,N\left(r_i\right)\le n\}.$$

(2)

If the answer is no, i.e., $W_{e,n}=W_{e,n-1}$, then continue to check whether there was a largest stage $m<n$ such that $W_{e,m}\ne W_{e,m-1}$.

(a)

If such a stage m existed, then we have

$$\{{\displaystyle \frac{1}{r_i}}:2\le i\le m,N\left(r_i\right)\le m\}\subseteq T_{e,3m}.$$

Now for all $r_i\ne 0,1$ with $i\le n$ and $N\left(r_i\right)\le m$, add ${\displaystyle \frac{1}{r_i}}$ into $T_{e,s}$ if it was not enumerated into $T_e$ before. That is,

$$T_{e,s}=T_{e,s-1}\cup \{{\displaystyle \frac{1}{r_i}}:2\le i\le n,N\left(r_i\right)\le m\}.$$

(b)

Otherwise, do nothing. Set $T_{e,s}=T_{e,s-1}$.

Stage $s=3n+1$. For $e\le s$, enumerate $r_n$ into $T_{e,s}$ if $r_n\notin T_{e,s-1}$; in this case, $T_{e,s}=T_{e,s-1}\cup \left\{r_n\right\}.$ For $e>s$, let $T_{e,s}=T_{e,s-1}$.

Stage $s=3n+2$. For $e\le s$, for all $y_1,y_2\in T_{e,s-1}$, add $y_1{+}_F{y}_2 \mathrm{and} y_1{·}_F{y}_2$ into $T_{e,s}$ if they are not in $T_{e,s-1}$. In this case,

$$T_{e,s}=T_{e,s-1}\cup \{y_1{+}_F{y}_2:y_1,y_2\in T_{e,s-1}\}\cup \{y_1{·}_F{y}_2:y_1,y_2\in T_{e,s-1}\}.$$

For $e>s$, let $T_{e,s}=T_{e,s-1}$.

This ends the construction.

For all e, let $T_e={\bigcup }_{s\in \mathbb{N}}{T}_{e,s}$. Clearly, $T_e$ is a subring of F that contains the polynomial ring R in the sense that an element $r\in R$ is identified with ${\displaystyle \frac{r}{1}}\in F$. A sequence $\langle T_e:e\in \mathbb{N}\rangle$ of c.e. subrings of a computable ring R is uniformly computably enumerable if and only if the coding set $X=\{\langle e,r\rangle \in \mathbb{N}\times R:r\in T_e\}$ is a c.e. subset of $\mathbb{N}\times R$. To see that the sequence $\langle T_e:e\in \mathbb{N}\rangle$ is a uniformly computably enumerable sequence of c.e. subrings of the field F, we only need to show that the coding set

$$T=\{\langle e,{\displaystyle \frac{f}{g}}\rangle :e\in \mathbb{N},{\displaystyle \frac{f}{g}}\in T_e\}$$

is a c.e. subset of the computable set $\mathbb{N}\times F$. As in the construction, list R as $r_0=0$, $r_1=1,r_2,\cdots$. Let $e\in \mathbb{N},{\displaystyle \frac{f}{g}}\in F$ with $f=r_i,g=r_j$. As $r_i\in T_{e,3i+1}\subseteq T_e$ and $r_i,r_j$ are coprime, we have

$$\begin{array}{cc}\hfill {\displaystyle \frac{f}{g}}\in T_e\iff {\displaystyle \frac{r_i}{r_j}}\in T_e& \iff {\displaystyle \frac{1}{r_j}}\in T_e\hfill \\ \hfill & \iff \exists n,m[n\ge j\wedge N\left(r_j\right)\le m\le n\wedge W_{e,m}\ne W_{e,m-1}]\hfill \end{array}$$

in which case ${\displaystyle \frac{1}{r_j}}\in T_{e,3n}$. Again, $N\left(r_j\right)$ is the largest index k, such that $x_k$ appears in $r_j$ if $r_j\notin \mathbb{Z}$ and $N\left(r_j\right)=0$ otherwise. Hence, $\langle e,{\displaystyle \frac{r_i}{r_j}}\rangle \in T$ is a ${\mathsf{\Sigma }}_1^0$ property on $e\in \mathbb{N},{\displaystyle \frac{r_i}{r_j}}\in F$ and T is a c.e. subset of $\mathbb{N}\times F$. Note that a set is c.e. if and only if it is ${\mathsf{\Sigma }}_1^0$. For more equivalent characterizations of c.e. sets, refer to Chapter II, [2].

An infinite c.e. set is always the range of a one-to-one computable function. For the c.e. subset T of $\mathbb{N}\times F$, let $\alpha :\mathbb{N}\to \mathbb{N}\times F$ be a one-to-one computable function such that the range of $\alpha$ is T, i.e., $T=\left\{\alpha \right(m):m\in \mathbb{N}\}$. Then, $\alpha$ effectively induces a computable sequence $\langle R_e:e\in \mathbb{N}\rangle$ of computable rings such that $R_e$ is isomorphic to $T_e$.

Defining $\langle R_e:e\in \mathbb{N}\rangle$.

Step 1. For all e, let $X_e=\mathrm{\{}n\in \mathbb{N}:\alpha \left(n\right)=\langle e,{\displaystyle \frac{f}{g}}\rangle \in T\mathrm{\}}$ be the domain of $R_e$.

Step 2. Define ring operations on $R_e$ according to corresponding operations in F. For any $m,n,k\in X_e$, let $\alpha \left(m\right)=\langle e,{\displaystyle \frac{f_1}{g_1}}\rangle$, $\alpha \left(n\right)=\langle e,{\displaystyle \frac{f_2}{g_2}}\rangle$ and $\alpha \left(k\right)=\langle e,{\displaystyle \frac{f_3}{g_3}}\rangle$ with ${\displaystyle \frac{f_i}{g_i}}\in F$ for $i=1,2,3$. Now, define the following operations:

$$\begin{array}{cc}\hfill & m=0_{R_e}\iff {\displaystyle \frac{f_1}{g_1}}=0_F\iff f_1=0_R,g_1=1_R;\hfill \\ \hfill & n=1_{R_e}\iff {\displaystyle \frac{f_2}{g_2}}=1_F\iff f_2=g_2=1_R;\hfill \\ \hfill & n{+}_{R_e}m=k\iff {\displaystyle \frac{f_1}{g_1}}+F{\displaystyle \frac{f_2}{g_2}}={\displaystyle \frac{f_3}{g_3}}\iff (f_1{g}_2+g_1{f}_2)g_3=g_1{g}_2{f}_3;\hfill \\ \hfill & n{·}_{R_e}m=k\iff {\displaystyle \frac{f_1}{g_1}}·F{\displaystyle \frac{f_2}{g_2}}={\displaystyle \frac{f_3}{g_3}}\iff f_1{f}_2{g}_3=g_1{g}_2{f}_3.\hfill \end{array}$$

Since F is a computable field and $\alpha :\mathbb{N}\to \mathbb{N}\times F$ is a one-to-one computable function with range $T=\{\langle e,{\displaystyle \frac{f}{g}}\rangle :e\in \mathbb{N},{\displaystyle \frac{f}{g}}\in T_e\}$, $R_e=(X_e,{+}_{R_e},{·}_{R_e},0_{R_e},1_{R_e})$ is a computable ring isomorphic to $T_e$, and the sequence $\langle R_e:e\in \mathbb{N}\rangle$ of computable rings is uniformly computable.

Lemma 5.

For all e, $e\in \mathrm{Inf}$ if and only if $R_e$ is a local ring.

Proof. 

For all e, $R_e$ is isomorphic to $T_e={\bigcup }_{s\in \mathbb{N}}{T}_{e,s}$, the constructed c.e. subring of the fraction field F of the polynomial ring $R=\mathbb{Z}[x_n:n\in \mathbb{N}]$. It suffices to prove that $e\in \mathrm{Inf}$ if and only if $T_e$ is local. Let R be listed as $r_0=0,r_1=1,r_2,\dots$. Then,

$$F=\{{\displaystyle \frac{r_i}{r_j}}:i,j\in \mathbb{N},r_j\ne 0,(r_i,r_j)=1\}$$

with the usual operations on fractions of polynomials of R. Each $r_i\in R$ is identified with ${\displaystyle \frac{r_i}{1}}\in F$ and R is a subring of F.

If $e\in \mathrm{Inf}$, $W_e$ is infinite. Let ${\displaystyle \frac{f}{g}}\in F$ with $f=r_i,g=r_j\ne 0$. As in the construction, if $g\notin \mathbb{Z}$, let $N\left(g\right)$ be the largest index k, such that $x_k$ occurs in g; if $g\in \mathbb{Z}$, let $N\left(g\right)=0$. Since $W_e$ is infinite, there is a least stage $n\ge j+N\left(g\right)$ such that some new number enumerates into $W_{e,n}$. Then, ${\displaystyle \frac{1}{r_j}}$ has been added into $T_e$ by stage $3n$, i.e., ${\displaystyle \frac{1}{r_j}}\in T_{e,3n}$. Note that $r_i\in T_{e,3i+1}$. Therefore, we have

$${\displaystyle \frac{f}{g}}=r_i·F{\displaystyle \frac{1}{r_j}}\in T_{e,3(n+i+1)+2}\subseteq T_e.$$

In this case, $T_e=F$ is a field, which is a local ring.

If $e\notin \mathrm{Inf}$, $W_e$ is finite. There are two cases.

(1)

If $W_e=\varnothing$, the empty set, no number enumerates into $W_e$ during the construction. Then, no fractions of the form ${\displaystyle \frac{r_i}{r_j}}\in F$ with $r_j\notin \{0,1\}$ and $r_i,r_j$ coprime were enumerated into $T_e$. In this case, $T_e$ equals the polynomial ring $\mathbb{Z}[x_n:n\in \mathbb{N}]$.

(2)

Otherwise, $W_e\ne \varnothing$, let $s_e$ be the largest stage at which some number enumerates into $W_e$. Then, no number goes into $W_e$ at any stage $s>s_e$. During the construction, we only added fractions of the form ${\displaystyle \frac{r_i}{r_j}}\in F$ with $N\left(r_j\right)\le s_e$ into $T_e$. That is, $T_e$ equals the fraction ring

$$\{{\displaystyle \frac{f}{g}}\in F:f\in \mathbb{Z}[x_n:n\in \mathbb{N}],g\in \mathbb{Z}[x_0,\dots ,x_{s_e}],g\ne 0\}.$$

In both cases, $T_e$ is not a local ring.

This completes the proof of the lemma. □

We have obtained a uniformly computable sequence $\langle R_e:e\in \mathbb{N}\rangle$ of computable rings such that $e\in \mathrm{Inf}$ if and only if $R_e$ is local. So, the index set of computable local rings is ${\mathsf{\Pi }}_2^0$-complete. This finishes the proof of Theorem 2.

Finally, we remark that a domain (i.e., a commutative ring with no zero divisors) is semisimple if and only if it is a field, which is a ${\mathsf{\Pi }}_2^0$ property. Since the sequence of computable rings $\langle R_e:e\in \mathbb{N}\rangle$ constructed in Theorem 2 also satisfies the condition that $e\in \mathrm{Inf}$ if and only if $R_e$ is a field, we have already obtained the exact complexity for semisimple domains.

Corollary 3.

The index set of computable semisimple domains (= fields) is ${\mathsf{\Pi }}_2^0$-complete within the index set of computable rings.

Theorem 2 also implies corresponding results for more general classes of rings, including fields such as semisimple rings and semiperfect rings.

Corollary 4.

The index set of computable semisimple rings (resp., semiperfect rings) is ${\mathsf{\Pi }}_2^0$-hard within the index set of computable rings.

5. Conclusions

We have conducted the study of the computational complexity of the typical subclasses of computable semiperfect rings such as semisimple rings and local rings by the techniques of computability theory. First, we obtained an upper bound that the index set of computable semisimple rings is ${\mathsf{\Sigma }}_3^0$ and a lower bound that this index set can be ${\mathsf{\Sigma }}_2^0$-hard within the index set of computable rings. Second, we obtained the exact complexity that the index set of computable local rings is ${\mathsf{\Pi }}_2^0$-complete. Third, as corollaries, we see that the index set of computable semiperfect rings can be both ${\mathsf{\Sigma }}_2^0$-hard and ${\mathsf{\Pi }}_2^0$-hard within the index set of computable rings. Finally, we conjecture that the index set of computable semisimple rings (resp., computable semiperfect rings) is ${\mathsf{\Sigma }}_3^0$-complete within the index of computable rings.

A ring R is semilocal if the quotient ring $R/Jac\left(R\right)$ of R modulo $Jac\left(R\right)$, the Jacobson radical of R, is semisimple. A commutative ring is semilocal if and only if it has finitely many maximal ideals. Semilocal rings form a big class of rings that includes semiperfect rings. Similar to the study of the complexity of computable semiperfect rings developed here, we can further consider the complexity of computable semilocal rings.