The Regular Module and the Regular Bimodule

Introduction

A ring acts on its own additive group by multiplication, and the module so obtained, the regular module, is the ring read as a module over itself. It is the smallest module a ring has and at the same time the universal one: its submodules are the ideals, it is cyclic and faithful, it is free of rank one, it is a generator, it is the identity of the tensor product, and its endomorphism ring recovers the ring, its opposite or its centre according to which of the two actions is kept. Every module is a quotient of a direct sum of copies of it, which is why a module theory can be built from the regular module alone.

The article assumes Modules for the definition and the elementary theory, Direct Sums, Free Modules and Rank for free modules, bases, rank and finite generation, The Balanced Product for the tensor product and its universal property, and Projective and Injective Modules for projectivity. The operator reading of the two one-sided multiplications is Left and Right Multiplication in a Ring, in the category Rings; the present article is the module-theoretic counterpart, and the algebra case $A_A$ is used in Algebras and in Centre, Units, Zero Divisors and Division Algebras. The categorification, that the regular bimodule is the identity of the tensor product of bimodules in the language of the module category, is Module Categories, in Linear Algebras, and is named here rather than developed. Flatness of the regular module is Flatness and Exactness; the forms attached to an involution are Hilbert Algebras, in Part II.

Throughout, $R$ is an associative ring with $1 \neq 0$, not assumed commutative. The side of the subscript records the side on which the scalars are written, and a module carrying both actions is a bimodule. No form, no norm and no topology is used.

The Three Regular Modules

The left regular module

Definition. The left regular module ${}_RR$ is the additive group $(R,+)$ with the scalar multiplication

$$ R \times R \to R, \qquad (r,m) \mapsto rm , $$

the multiplication of $R$ itself.

Proposition (the module axioms are the ring axioms). ${}_RR$ is a left $R$-module. The carrier $(R,+)$ is only an abelian group and carries no multiplication of its own; the multiplication of $R$ is not discarded, it is the action. The four module axioms are then the two distributive laws, the associativity of the multiplication, and the unit law, and each is a ring axiom read as a module axiom. In particular $0m = 0$ and $(-r)m = -(rm)$ hold because they already hold in $R$.

Proof. The distributivity of the action in the scalar and in the vector is the two distributive laws of $R$; the compatibility $(rs)m = r(sm)$ is the associativity of $R$; and $1m = m$ is the unit law. The derived identities are the elementary properties of Modules.

The right regular module

Definition. The right regular module $R_R$ is $(R,+)$ with the right scalar multiplication $(m,r) \mapsto mr$.

The regular bimodule

Definition. The left and the right action of $R$ on $(R,+)$ together make it a bimodule ${}_RR_R$; the two actions commute,

$$ a(xb) = (ax)b \qquad (a,b,x \in R), $$

which is the associativity of $R$ once more.

The Submodules Are the Ideals

Theorem. Let $L \subseteq R$ be an additive subgroup. Then $L$ is a submodule of ${}_RR$ if and only if $L$ is a left ideal of $R$; $L$ is a submodule of $R_R$ if and only if $L$ is a right ideal; and $L$ is a submodule of ${}_RR_R$ if and only if $L$ is a two-sided ideal.

Proof. A submodule of ${}_RR$ is an additive subgroup closed under $rl$ for all $r \in R$ and $l \in L$, and that is the definition of a left ideal. The scalar action of ${}_RR$ is by definition the multiplication of $R$, so the two conditions are the same formula and not two equivalent ones. The right-handed statement is the same argument with the multiplication read on the right, and the two-sided statement with both readings at once.

Corollary (the lattice is the lattice of ideals). The identity map on subsets of $R$ is an isomorphism of the lattice of submodules of ${}_RR$ onto the lattice of left ideals of $R$, preserving inclusion, intersection and sum; likewise on the right, and on the two-sided ideals for the bimodule. The one-sided ideal theory of Rings is therefore the submodule theory of the regular module, and the language of ideals is the language of this example.

Four Properties of the Regular Object

Cyclic and faithful

Proposition. ${}_RR$ is cyclic, generated by the unit: ${}_RR = R1$, since $r = r1$. It is faithful: the annihilator $\operatorname{Ann}({}_RR) = \{r : rM = 0\}$ is zero, since $r1 = r$.

Proof. The two statements are the computations $r = r \cdot 1$ and $r \cdot 1 = 0 \Rightarrow r = 0$.

Free of rank one

Proposition. ${}_RR$ is free of rank one, with basis $\{1\}$: the expansion $r = r \cdot 1$ is unique, because a linear combination of the single basis element $1$ with coefficient $r$ is $r1$ and determines $r$. Hence ${}_RR$ is finitely generated, projective and free of rank one, and in the same way $R_R$ and ${}_RR_R$ are free of rank one, so the rank of the regular module in the sense of Direct Sums, Free Modules and Rank is defined and equals $1$.

Proof. Linear independence of $\{1\}$ is $r1 = 0 \Rightarrow r = 0$; generation is $r = r1$. A module with a basis is free and is therefore projective, by Projective and Injective Modules.

A generator

Definition. A left $R$-module $G$ is a generator if every left $R$-module is a quotient of a direct sum $G^{(I)}$ of copies of $G$.

Proposition. ${}_RR$ is a generator.

Proof. By Modules §9 every module $M$ is a quotient of a free module, $M \cong F/\ker \varphi$ with $F$ free; by the universal property of a basis $F$ is a direct sum of copies of ${}_RR$, and a direct sum of copies of ${}_RR$ is a direct sum of copies of ${}_RR$. The defining quotient is $\varphi$.

Corollary (the regular module sees everything). Every left $R$-module is a quotient of a direct sum of copies of ${}_RR$, so a property preserved by quotients and direct sums and true of ${}_RR$ is true of every module; the regular module is the single test object of the category. The dual statement, with submodules and products, holds for the injective envelope of ${}_RR$, which is not the regular module in general.

The identity of the tensor product

Proposition. For a right $R$-module $M$ there is a natural isomorphism

$$ M \otimes_R {}_RR \cong M, \qquad m \otimes r \mapsto mr , $$

and for a left $R$-module $N$ a natural isomorphism ${}_RR \otimes_R N \cong N$, $r \otimes n \mapsto rn$. For bimodules the same element is the two-sided identity, ${}_SM_R \otimes_R {}_RR_R \cong {}_SM_R$ and ${}_RR_R \otimes_R {}_RN_T \cong {}_RN_T$, naturally in the outer module.

Proof. The map $m \otimes r \mapsto mr$ is well defined because $(m,r) \mapsto mr$ is $R$-balanced: it is additive in each variable and $(ms) \otimes r$ and $m \otimes (sr)$ both go to $m(sr) = (ms)r$. It is $R$-linear on the right of $M$ when $M$ is a bimodule, and the assignment $m \mapsto m \otimes 1$ is its inverse, since $m \otimes r = (m \cdot r) \otimes 1 = (m \otimes 1) r$; uniqueness is the universal property of The Balanced Product. The left-handed statement is the mirror, and the bimodule statements follow by carrying the outer actions through both maps.

Remark (the categorical reading). The identities above say that ${}_RR_R$ is the unit of the tensor product of bimodules: tensoring a bimodule with ${}_RR_R$ on either side returns it. This is the module-level form of the fact that the tensor product is a monoidal structure whose unit is the base object, and its categorical statement is Module Categories.

The Endomorphisms of the Regular Object

The one-sided endomorphisms

Proposition. There are ring isomorphisms

$$ \operatorname{End}_R(R_R) \cong R, \qquad \operatorname{End}_R({}_RR) \cong R^{\mathrm{op}} , $$

the first given by the left multiplications $b \mapsto L_b$ and the second by the right multiplications $b \mapsto R_b$, where $L_b(x) = bx$ and $R_b(x) = xb$.

Proof. Let $T : R_R \to R_R$ be right $R$-linear, $T(xr) = T(x)r$. With $x = 1$ and the right action $1 \cdot r = r$, one has $T(r) = T(1)r = L_{T(1)}(r)$, so $T = L_b$ with $b = T(1)$; conversely $L_b(xr) = b(xr) = (bx)r = L_b(x)r$, so $L_b$ is right $R$-linear. The assignment $b \mapsto L_b$ is additive and satisfies $L_{bc} = L_b L_c$, hence is a ring isomorphism onto $\operatorname{End}_R(R_R)$, injective because $L_b(1) = b$. The left-module case is the mirror: a left $R$-linear $T$ has $T(r) = T(r \cdot 1) = rT(1) = R_{T(1)}(r)$, and $b \mapsto R_b$ is additive with $R_{bc} = R_cR_b$, hence a ring isomorphism $R^{\mathrm{op}} \to \operatorname{End}_R({}_RR)$.

Corollary. The endomorphism ring of the regular module is the ring itself, read on the same side for the representation and on the opposite side for the anti-representation: the left multiplications are the endomorphisms of the right regular module and reproduce $R$, the right multiplications are the endomorphisms of the left regular module and reproduce $R^{\mathrm{op}}$. This is the module-level form of the composition law $L_aL_b = L_{ab}$, $R_aR_b = R_{ba}$ of Left and Right Multiplication in a Ring.

The bimodule endomorphisms are the centre

Proposition. $\operatorname{End}_{R\text{-}R}(R) \cong Z(R)$, the centre of $R$; the isomorphism sends $z \in Z(R)$ to $L_z = R_z$.

Proof. An $R$-$R$-linear $T$ is in particular right $R$-linear, so $T = L_b$, and in particular left $R$-linear, so $T = R_c$; then $L_b = R_c$ gives $b x = x c$ for all $x$, and with $x = 1$ one gets $b = c$ and then $bx = xb$ for all $x$, so $b \in Z(R)$. Conversely a central $b$ has $L_b = R_b$ and this map is a bimodule endomorphism. The assignment is additive and multiplicative and injective because $L_z(1) = z$.

Corollary. The bimodule ${}_RR_R$ has no nontrivial endomorphisms exactly when $Z(R)$ is a field and the only central scalars are the multiples of $1$; in particular, for a central simple algebra over a field $F$ one has $\operatorname{End}_{R\text{-}R}(R) \cong F$. This is the ring-level statement verified below on the matrix ring.

Examples

A field

For $R = F$ a field, ${}_FF$ is the one-dimensional vector space, its submodules are $0$ and $F$, and every $F$-module is free. The regular module is the trivial case from which the linear algebra of Vector Spaces is assembled: an $F$-module is free of rank $n$, and the free module of rank $n$ is $(F_F)^n$.

The ring of integers

For $R = \mathbb{Z}$, the submodules of $\mathbb{Z}_{\mathbb{Z}}$ are the ideals $n\mathbb{Z}$, $n \geq 0$, and they are pairwise distinct; in particular the regular module is not semisimple, since $n\mathbb{Z}$ has no complement in $\mathbb{Z}$ for $n \geq 2$: a decomposition $\mathbb{Z} = n\mathbb{Z} \oplus C$ would give a torsion-free complement of rank $1 - 1 = 0$, that is $C = 0$. The regular module therefore need not be semisimple, and the failure is the very thing that makes the structure theory of Modules over a PID non-trivial.

A product of rings and the idempotents

For $R = \mathbb{Z}/6\mathbb{Z}$ the element $e = 3$ satisfies $e^2 = 9 = 3$, so it is idempotent, and $1 - e = 4$ is idempotent as well. The two principal ideals are

$$ Re = \{0,3\} \cong \mathbb{Z}/2\mathbb{Z}, \qquad R(1-e) = \{0,2,4\} \cong \mathbb{Z}/3\mathbb{Z} , $$

and together they give the direct sum decomposition ${}_RR = Re \oplus R(1-e) \cong \mathbb{Z}/2\mathbb{Z} \oplus \mathbb{Z}/3\mathbb{Z}$, which is the Chinese remainder theorem read as a decomposition of the regular module. In general an idempotent $e$ produces the direct summand $Re$ of ${}_RR$, with complement $R(1-e)$; the regular module is the universal source of the idempotents and of the projectivity they create.

The matrix ring

Let $R = M_n(F)$ be a full matrix ring over a field. Then $R$ has no two-sided ideal other than $0$ and $R$, so the regular bimodule ${}_RR_R$ is simple: its only sub-bimodules are $0$ and $R$. The left regular module is not simple but semisimple: it is the direct sum of $n$ copies of the simple module $S = F^n$, the column space of $R$,

$$ {}_RR \cong S \oplus \cdots \oplus S \quad (n \text{ summands}), \qquad \dim_F S = n , $$

because the matrices vanishing outside the $j$-th column form a minimal left ideal $L_j$, the $n$ of them are the simple left ideals, and ${}_RR = L_1 \oplus \cdots \oplus L_n$. For $n = 2$ this is ${}_RR \cong S \oplus S$ with $\dim_F S = 2$, the decomposition whose existence makes the double centraliser computation of Algebras of Endomorphisms and the sharpening of Involutions of the Endomorphism Ring possible. The centre is $Z(R) = F1$, so $\operatorname{End}_{R\text{-}R}(R) \cong F$ by the corollary above.

The group algebra

Let $G$ be a finite group and $F$ a field in which $|G|$ is invertible, so that $F[G]$ is semisimple by Maschke's theorem. Then every left $F[G]$-module is semisimple, and the regular module decomposes as

$$ {}_FF[G] \; \cong \; \bigoplus_i S_i^{\oplus \dim S_i} , $$

the direct sum running over the simple modules $S_i$ up to isomorphism, each occurring with multiplicity its own dimension; comparing dimensions gives the identity $\sum_i (\dim S_i)^2 = |G|$. The decomposition is produced by the idempotents $e_i$ that the simple modules define, one for each irreducible, and the centre $Z(F[G])$ has dimension the number of the $e_i$: for $G$ abelian of order $n$, and $F$ containing the $n$-th roots of unity, all $\dim S_i = 1$ and the decomposition is into $n$ copies of $F$. The case $G$ cyclic of order four over $\mathbb{Q}(i)$ is computed below.

Summary

A ring $R$ acts on its own additive group by multiplication, giving the left regular module ${}_RR$, the right regular module $R_R$, and the regular bimodule ${}_RR_R$ whose two actions commute. The submodules of these three objects are exactly the left, the right and the two-sided ideals of $R$, so the lattice of ideals is the lattice of submodules of the regular object. The regular module is cyclic on the unit, faithful, free of rank one, a generator, and the identity of the tensor product, $M \otimes_R {}_RR \cong M$ and ${}_RR \otimes_R N \cong N$; every module is a quotient of a direct sum of copies of it. Its endomorphism ring is the ring itself in the two one-sided readings, $\operatorname{End}_R(R_R) \cong R$ and $\operatorname{End}_R({}_RR) \cong R^{\mathrm{op}}$, and its bimodule endomorphism ring is the centre, $\operatorname{End}_{R\text{-}R}(R) \cong Z(R)$. The examples range from the field, where the regular module is the trivial case of linear algebra, and the integers, where it is not semisimple, through the product of rings, where the idempotents split it, to the matrix ring, where it is simple as a bimodule and a direct sum of copies of the simple module as a left module, and to the group algebra, where it is the direct sum of the irreducibles with multiplicity the dimension.

Summary of Notation

Symbol	Meaning
${}_RR$	the left regular module, $(R,+)$ with $r m = rm$
$R_R$	the right regular module, $(R,+)$ with $m r = mr$
${}_RR_R$	the regular bimodule, the two commuting actions
$L \subseteq R$ submodule of ${}_RR$	left ideal of $R$
$L \subseteq R$ submodule of $R_R$	right ideal of $R$
$L \subseteq R$ submodule of ${}_RR_R$	two-sided ideal of $R$
${}_RR = R1$, $\operatorname{Ann}({}_RR) = 0$	cyclic on the unit, faithful
$\{1\}$	the basis, of rank one
$M \otimes_R {}_RR \cong M$, ${}_RR \otimes_R N \cong N$	the regular bimodule is the tensor unit
$\operatorname{End}_R(R_R) \cong R$, $b \mapsto L_b$	left multiplications
$\operatorname{End}_R({}_RR) \cong R^{\mathrm{op}}$, $b \mapsto R_b$	right multiplications
$\operatorname{End}_{R\text{-}R}(R) \cong Z(R)$	the bimodule endomorphisms are the centre