On Rings of Weak Global Dimension at Most One

Abstract

A ring R is of weak global dimension at most one if all submodules of flat R-modules are flat. A ring R is said to be arithmetical (resp., right distributive or left distributive) if the lattice of two-sided ideals (resp., right ideals or left ideals) of R is distributive. Jensen has proved earlier that a commutative ring R is a ring of weak global dimension at most one if and only if R is an arithmetical semiprime ring. A ring R is said to be centrally essential if either R is commutative or, for every noncentral element $x\in R$, there exist two nonzero central elements $y,z\in R$ with $xy=z$. In Theorem 2 of our paper, we prove that a centrally essential ring R is of weak global dimension at most one if and only is R is a right or left distributive semiprime ring. We give examples that Theorem 2 is not true for arbitrary rings.

1. Introduction

We consider only nonzero associative unital rings. For a ring R, we write w.gl.dim. $R\le 1$ if R is a ring of weak global dimension at most one, i.e., R satisfies the following equivalent (The equivalence of the conditions is well known; e.g., see the conditions in [1] (Theorem 6.12)).

• For every finitely generated right ideal X of R and each finitely generated left ideal Y of R, the natural group homomorphism $X{\otimes }_{R}Y\to XY$ is an isomorphism.
• Every finitely generated right (resp., left) ideal of R is a flat right (resp., left) R-module.
• Every right (resp., left) ideal of R is a flat right (resp., left) R-module.
• Every submodule of any flat right (resp., left) R-module is flat.
• ${\mathrm{Tor}}_2^R(A,B)=0$ for all right (resp., left) R-modules A and B.

Since every projective module is flat, any right or left (semi)hereditary ring is of weak global dimension at most one. (a module M is said to be hereditary (resp., semihereditary) if all submodules (resp., finitely generated submodules) of M are projective.) We also recall that a ring R is of weak global dimension zero if and only if R is a Von Neumann regular ring, i.e., $r\in rRr$ for every element r of R. Von Neumann regular rings are widely used in mathematics; see [2,3].

A ring R is said to be arithmetical if the lattice of two-sided ideals of R is distributive, i.e., $X\cap (Y+Z)=X\cap Y+X\cap Z$ for any three ideals $X,Y,Z$ of R. A ring R is said to be semiprime (resp., prime) if R does not have nilpotent nonzero ideals (resp., the product of any two nonzero ideals of R are nonzero).

Theorem 1.

(C.U.Jensen ([4], Theorem)). A commutative ring R is a ring of weak global dimension at most one if and only if R is an arithmetical semiprime ring.

A ring R with center C is said to be centrally essential if $R_C$ is an essential extension of the module $C_C$, i.e., for every nonzero element $r\in R$, there exist two nonzero central elements $x,y\in R$ with $rx=y$. Centrally essential rings are studied in many papers; e.g., see [5].

There are many noncommutative centrally essential rings. For example, if F is the field $\mathbb{Z}/2\mathbb{Z}$ and $Q_8$ is the quaternion group of order 8, then the group algebra $F{Q}_8$ is a finite noncommutative centrally essential ring; see [5].

Let F be the field $\mathbb{Z}/3\mathbb{Z}$, and let V be a vector F-space with basis $e_1,e_2,e_3$. It is known that the exterior algebra of the space V is a finite centrally essential noncommutative ring. It is known that there exists a centrally essential ring R such that the factor ring $R/J\left(R\right)$ with respect to the Jacobson radical is not a PI ring (in particular, the ring $R/J\left(R\right)$ is not commutative).

A module M is said to be distributive (resp., uniserial) if the submodule lattice of M is distributive (resp., is a chain). It is clear that a commutative ring is right (resp., left) distributive if and only if the ring is arithmetical.

The main result of this work is Theorem 2.

Theorem 2.

For a centrally essential ring R, the following conditions are equivalent.

1. 

R is a ring of weak global dimension at most one.

2. 

R is a right (resp., left) distributive semiprime ring.

3. 

R is an arithmetical semiprime ring

2. Remarks and Proof of Theorem 2

Example 1.

The implication (1) ⇒ (2) of Theorem 2 is not true for arbitrary rings. There exists a right hereditary ring R of weak global dimension at most one that is neither right distributive nor semiprime; in particular, the right hereditary ring R is of weak global dimension at most one. Let F be a field, and let R be the 5-dimensional F-algebra consisting of all $3\times 3$ matrices of the following form: $\left(\begin{array}{ccc}{f}_{11}& f_{12}& f_{13}\\ 0& f_{22}& 0\\ 0& 0& f_{33}\end{array}\right)$, where $f_{ij}\in F$. The ring R is not semiprime, since the following set is a nonzero nilpotent ideal of R: $\left\{\left(\begin{array}{ccc}0& f_{12}& f_{13}\\ 0& 0& 0\\ 0& 0& 0\end{array}\right)\right\}$. Let $e_{11}$, $e_{22}$, and $e_{33}$ be ordinary matrix units. The ring R is not right or left distributive, since every idempotent of a right or left distributive ring is central (see [6]), but the matrix unit $e_{11}$ of R is not central. To prove that the ring R is right hereditary, it is sufficient to prove that $R_R$ is a direct sum of hereditary right ideals. We have that $R_R=e_{11}R\oplus e_{22}R\oplus e_{33}R$, where $e_{22}R$ and $e_{33}R$ are projective simple R-modules; in particular, $e_{22}R$ and $e_{33}R$ are hereditary R-modules. Any direct sum of hereditary modules is hereditary; see ([7], 39.7, p. 332). Therefore, it remains to show that the R-module $e_{11}R=e_{11}F+e_{12}F+e_{13}F$ is hereditary, which is directly verified.

The following lemma is well known; e.g., see ([1], Assertion 6.13).

Lemma 1.

Let R be a ring in which the principal right ideals are flat. If r and s are two elements of R with $rs=0$, then there exist two elements $a,b\in R$ such that $a+b=1$, $ra=0$, and $bs=0$.

Lemma 2.

Let R be a centrally essential ring in which the principal right ideals are flat. Then, the ring R does not have nonzero nilpotent elements.

Proof. 

Indeed, let us assume that there exists a nonzero element $r\in R$ with $r^2=0$. Since the ring R is centrally essential, there exist two nonzero central elements $x,y\in R$ with $rx=y$. Since $r^2=0$, we have that $y^2={\left(rx\right)}^2=r^2{x}^2=0$. Since $y^2=0$, it follows from Lemma 1 that there exist two elements $a,b\in R$ such that $a+b=1$, $ry=0$, and $by=yb=0$. Then, $y=y(a+b)=ya+yb=0$. This is a contradiction. □

Lemma 3.

There exists right and left uniserial prime rings R that habe a non-flat principal right ideal.

Proof. 

There exists right and left uniserial prime rings R with two nonzero elements $r,s\in R$ such that $rs=0$; see ([8], p. 234, Corollary). The uniserial ring R is local; therefore, the invertible elements of R form the Jacobson radical $J\left(R\right)$ of R. The ring R is not a ring in which the principal right ideals are flat. Indeed, let us assume the contrary. By Lemma 1, there exist two elements $a,b\in R$ such that $a+b=1$, $ra=0$, and $bs=0$. We have that either $aR\subseteq bR$ or $bR\subseteq aR$; in addition, $aR+bR=R=Ra+Rb$. Therefore, at least one of the elements $a,b$ of the local ring R is invertible; in particular, this invertible element is not a right or left zero-divisor. This contradicts to the relations $ra=0$ and $bs=0$. □

Remark 1.

It follows from Lemma 3 that the implication (2) ⇒ (1) of Theorem 2 is not true for arbitrary rings.

Lemma 4.

Every centrally essential semiprime ring R is commutative.

Proof. 

Assume the contrary. Then, the ring R does not coincide with its center C and $xy-yx\ne 0$ for some $x,y\in R$. We note that $A=\{c\in C:xc\in C\}$ is an ideal of the ring C. The set $d\in C|dA=0$ is not empty, since we can take $d=0$. We take any element $d\in C$ with $dA=0$. If $xd\ne 0$, then $xdz\in C\backslash \left\{0\right\}$ for some $z\in C$. Hence $dz\in A$, and therefore, $d\left(dz\right)=0$ and ${\left(dz\right)}^2=0$. Thus, $dz=0$ and $xdz=0$; this is a contradiction. Therefore, $xd=0$, and thus, $d\in A$. Therefore, $d^2=0$ and $d=0$. This implies that ${Ann}_C\left(A\right)=0$. For any $a\in A$, we have that $xa=ax\in C$. Thus,

$$(xy-yx)a=x\left(ya\right)-y\left(xa\right)=xay-xay=0$$

and $(xy-yx)A=0$. However, $c_1(xy-yx)=c_2$ for some nonzero elements $c_1,c_2\in C$, so $c_{2}A=0$ and, hence, ${Ann}_C\left(A\right)\ne 0$; this is a contradiction. Thus, R is commutative. □

The Completion of the Proof of Theorem 2

Proof. 

(1) ⇒ (2). Since R is a centrally essential ring of weak global dimension at most one, it follows from Lemma 2 that the ring R does not have nonzero nilpotent elements. By Lemma 4, the centrally essential semiprime ring R is commutative. By Theorem 1, R is an arithmetical semiprime ring. Any commutative arithmetical ring is right and left distributive.

The implication (2) ⇒ (3) follows from the property that every right or left distributive ring is arithmetical.

(3) ⇒ (1). Since R is a centrally essential semiprime ring, it follows from Lemma 4 that the ring R is commutative; in particular, R is centrally essential. In addition, R is arithmetical. By Theorem 1, the ring R is of weak global dimension at most one. □