The Endomorphism Algebra of a Module

Introduction

The endomorphisms of a module form a ring, and because the base ring acts centrally the ring is an algebra. This article studies that algebra in its own right: its structure when the module is free or semisimple, and its centre, the commutative part of the ring of operators.

The article assumes the identifications of Left and Right Multiplication of a Module and Module Endomorphisms — the endomorphism ring as the commutant of the action, its units, and the density theorem — and it assumes the matrix and division-ring vocabulary of Modules over an Algebra and Automorphisms of Modules over an Algebra. It adds the algebra structure and the centre. It stays inside Part I: no distance, norm, form or limit occurs. Throughout, $R$ is a commutative ring with $1 \neq 0$, $A$ is a unital associative $R$-algebra, $M$ is a left $A$-module, $E=\operatorname{End}_A(M)$, and $Z(E)$ is its centre.

The Algebra Structure

The endomorphism algebra

The endomorphism ring is an $R$-algebra, and this is the object of the article.

Proposition. With $E=\operatorname{End}_A(M)$, the multiplication by $r \in R$ given by $(rf)(m)=r f(m)$ turns $E$ into a unital associative $R$-algebra, and the inclusion $E \subseteq \operatorname{End}_R(M)$ is a homomorphism of $R$-algebras.

Proof. The map $rf$ is $A$-linear because $r$ acts centrally, so $E$ is closed under the $R$-action; the algebra axioms are inherited from $\operatorname{End}_R(M)$, and the inclusion preserves the product, the unit and the scalars. $\square$

The word algebra rather than ring records that the base ring acts; when $R=F$ is a field and $M$ is finite-dimensional over $F$, the endomorphism algebra is finite-dimensional over $F$, and its dimension is at most $(\dim_F M)^2$.

The algebra of a free module

For a free module the endomorphism algebra is a matrix algebra, and the matrix is read in the standard basis.

Theorem. For $n \geq 1$ there is an isomorphism of $R$-algebras

$$ \operatorname{End}_A({}_A A^n) \xrightarrow{\ \sim\ } M_n(A^{\mathrm{op}}), $$

given on the row $e_i$ of the standard basis by $f(e_j)=\sum_i e_i\,a_{ij}$, so that $f(v)=vX$ for the matrix $X=(a_{ij})$, and the product on the right is the matrix product in $A^{\mathrm{op}}$.

Proof. An $A$-linear map is determined by its values on the basis, and $A$-linearity forces $f(a e_j)=a f(e_j)$, which is the row-vector computation $f(av)=(av)X=a(vX)$ by associativity. Thus every such $f$ is right multiplication by a unique matrix $X$ with entries in $A$. Composition is $f_X \circ f_Y(v)=(vY)X=v(YX)$, so the ring of endomorphisms is the opposite of the matrix ring $M_n(A)$ with the usual product, that is $M_n(A^{\mathrm{op}})$. The $R$-linearity is entrywise. $\square$

For $n=1$ this is the identification $\operatorname{End}_A({}_A A)\cong A^{\mathrm{op}}$ of Automorphisms of Modules over an Algebra, and it is the base case of the theorem.

The algebra of an isotypic module

For a sum of copies of one simple module the endomorphism algebra is a matrix algebra over a division ring, by the density theorem.

Proposition. Let $S$ be a simple left $A$-module, $D=\operatorname{End}_A(S)$, and $M=S^n$ a finite direct sum. Then

$$ E \cong M_n(D) $$

as $R$-algebras, with the entries acting on the components.

Proof. This is the isotypic computation of Module Endomorphisms: an endomorphism of $S^n$ is a matrix of maps $S \to S$, each in $D$, and composition is matrix composition. $\square$

The general semisimple structure

The two descriptions combine into the structure theorem for the endomorphism algebra of a semisimple module of finite length.

Theorem. Let $M=S_1^{n_1}\oplus\cdots\oplus S_k^{n_k}$ with the $S_i$ pairwise non-isomorphic simple modules and $D_i=\operatorname{End}_A(S_i)$. Then

$$ E \cong M_{n_1}(D_1) \times \cdots \times M_{n_k}(D_k) $$

as $R$-algebras, and $M$ is the direct sum of the natural modules $D_i^{\,n_i}$ over the factors.

Proof. By Schur's lemma there are no nonzero maps between non-isomorphic simple modules, so an endomorphism preserves each isotypic component $S_i^{n_i}$, and on that component it is a matrix over $D_i$ by the proposition; the product decomposition follows. $\square$

The theorem is the operator form of the semisimple structure theorem, and it exhibits $E$ as a product of matrix algebras over division rings, hence as a semisimple ring.

The Centre

Definition and first properties

The centre of the endomorphism algebra is the set of operators that commute with every endomorphism.

Definition. The centre of $E$ is

$$ Z(E)=\{z \in E : zf=fz \text{ for all } f \in E\}. $$

It is a commutative subalgebra of $E$ containing the image of $R$.

Proposition. $Z(E)=E\cap E'$, where $E'$ is the centralizer of $E$ in $\operatorname{End}_R(M)$; consequently $Z(E)$ is a commutative $R$-algebra, and $M$ is a left $Z(E)$-module on which every element of $E$ acts $Z(E)$-linearly.

Proof. An element lies in the centre exactly when it belongs to $E$ and commutes with all of $E$, which is $E\cap E'$. For the module statement, $z \in Z(E)$ acts on $M$ through its action as an endomorphism, and $f(zm)=z f(m)$ because $z$ commutes with $f$. $\square$

Since $E=L_A'$ and $L_A \subseteq E'$, the centre sits between the two commutants: it is the part of the endomorphism algebra that is also in the bicommutant of the action.

The centre of a matrix algebra

The centre of a matrix algebra is the scalars, which is the computation the two structure theorems need.

Proposition. For any unital ring $B$ and $n \geq 1$,

$$ Z(M_n(B)) = Z(B)\cdot I_n = \{ \lambda I_n : \lambda \in Z(B) \}. $$

Proof. A matrix $C$ commutes with every matrix unit $E_{ij}$; commuting with $E_{ij}$ forces the entries of $C$ to be constant across each row and column, so $C=\lambda I_n$ for some $\lambda \in B$. Such a scalar matrix commutes with an arbitrary $X=(x_{ij})$ exactly when $\lambda x_{ij}=x_{ij}\lambda$ for all $i,j$; as the entries range over all of $B$, this says $\lambda \in Z(B)$. $\square$

The centre of a free module's endomorphism algebra

Corollary. For a free module ${}_A A^n$,

$$ Z(\operatorname{End}_A(A^n)) \cong Z(A^{\mathrm{op}}) = Z(A), $$

the scalar matrices with entries in the centre of the algebra.

Proof. Combine the matrix theorem with the identification $\operatorname{End}_A(A^n)\cong M_n(A^{\mathrm{op}})$ and $Z(A^{\mathrm{op}})=Z(A)$. $\square$

Thus for the regular module $E=A^{\mathrm{op}}$ and $Z(E)=Z(A)$, and enlarging the rank does not enlarge the centre: the centre of the endomorphism algebra of a free module is the centre of the algebra, whatever the rank.

The centre of an isotypic module's endomorphism algebra

Corollary. With $S$ simple, $D=\operatorname{End}_A(S)$ and $M=S^n$,

$$ Z(E) \cong Z(D), $$

the set of central endomorphisms of the simple module.

Proof. $E\cong M_n(D)$ and the matrix theorem gives $Z(M_n(D))=Z(D)\cdot I_n$. $\square$

In the semisimple case the centre is therefore $\prod_i Z(D_i)$, a product of fields when the $D_i$ are division rings finite-dimensional over a field, and the centre is a field exactly when there is a single isotypic component with $Z(D)$ a field.

The centre as the algebra of $A$-linear scalars

The centre has an intrinsic description that does not mention matrices.

Proposition. $Z(E)$ is exactly the subalgebra of $E$ consisting of the maps $z$ such that $z$ is $A$-linear and commutes with every $A$-linear map; equivalently $z \in E'$, the centralizer of $E$ inside $\operatorname{End}_R(M)$ that happens to lie in $E$.

Proof. Restatement of $Z(E)=E\cap E'$. $\square$

The description shows that the centre is the largest subalgebra of $\operatorname{End}_R(M)$ that is centralised by the whole endomorphism algebra; it is the algebra of scalars that the module's own symmetry forces.

Commutative Endomorphism Algebras

When the endomorphism algebra is commutative

The endomorphism algebra is commutative exactly when all its operators commute, and the structure theorem makes this a condition on the decomposition.

Proposition. Let $M$ be semisimple of finite length, $M=S_1^{n_1}\oplus\cdots\oplus S_k^{n_k}$. Then $E$ is commutative if and only if $n_i=1$ for all $i$ and each $D_i$ is a field.

Proof. $E\cong\prod_i M_{n_i}(D_i)$ is commutative exactly when each factor is: $M_{n_i}(D_i)$ is commutative only for $n_i=1$ with $D_i$ commutative, and for $n_i=1$ it is $D_i$, commutative when it is a field. $\square$

In the commutative case $E$ is a product of fields, so every endomorphism is a scalar on each isotypic component, and the module is multiplicity-free.

The division-ring case

Proposition. If $M$ is simple then $E=\operatorname{End}_A(M)$ is a division ring and $Z(E)$ is a field.

Proof. Schur's lemma gives the division ring, and the centre of a division ring is a field. $\square$

Examples

(a) The regular module over a commutative algebra. For $A$ commutative and $M=A$, $E=A^{\mathrm{op}}=A$ and $Z(E)=Z(A)=A$: every endomorphism is a scalar and the endomorphism algebra is the algebra itself.

(b) A matrix algebra on its defining module. For $A=M_n(F)$ and $M=F^n$ simple, $D=\operatorname{End}_A(F^n)=F$, so $E=F$ and $Z(E)=F$; the endomorphism algebra is one-dimensional and equal to its centre.

(c) A matrix algebra on itself. For $A=M_n(F)$ and $M=A$, $E=M_n(F)^{\mathrm{op}}$, and $Z(E)=F\cdot I$: the centre is the scalars, one-dimensional, and the endomorphism algebra is not commutative for $n \geq 2$.

(d) A free module of higher rank. For $M=A^n$ with $A$ a field $F$, $E=M_n(F)$ and $Z(E)=F$, whatever $n$ is.

(e) The quaternions. For $A=\mathbb{H}$ and $M=\mathbb{H}$, $E\cong\mathbb{H}^{\mathrm{op}}\cong\mathbb{H}$ and $Z(E)=\mathbb{R}$, the real scalars; the endomorphism algebra is a division ring and its centre is the base field.

(f) A semisimple module with two components. For $A=F$ a field and $M=F\oplus F$, $E\cong M_2(F)$, with centre $F$; for two non-isomorphic simple modules, $A=M_n(F)\times M_m(F)$ and $M=F^n\oplus F^m$, $E\cong F\times F$ is commutative and equal to its centre.

Summary

The endomorphism ring $E=\operatorname{End}_A(M)$ of a left $A$-module is a unital $R$-algebra, and it is the subject here together with its centre. For a free module, $E\cong M_n(A^{\mathrm{op}})$, the matrices over the opposite algebra, so for the regular module $E\cong A^{\mathrm{op}}$; for a finite isotypic module $S^n$ over a simple $S$, $E\cong M_n(D)$ with $D=\operatorname{End}_A(S)$ a division ring; and for a semisimple module of finite length $M=S_1^{n_1}\oplus\cdots\oplus S_k^{n_k}$, $E\cong\prod_i M_{n_i}(D_i)$. The centre of a matrix algebra is the scalar matrices with entries in the centre of the coefficient ring, so the centre of the endomorphism algebra of a free module of any rank is $Z(A)$, that of an isotypic module is $Z(D)$, and in the semisimple case it is $\prod_i Z(D_i)$. The centre is $E\cap E'$, the operators that are $A$-linear and commute with every $A$-linear operator, and $M$ is a module over it on which $E$ acts by $Z(E)$-linear maps. The endomorphism algebra is commutative exactly when the module is multiplicity-free with commutative division-ring endomorphisms, and it is a division ring exactly when the module is simple.

Summary of Notation

Symbol	Meaning
$R$	commutative ring with identity, the base ring
$A$	unital associative $R$-algebra, generally noncommutative
$M$	left $A$-module
$S$, $S_i$	simple left $A$-modules
$E=\operatorname{End}_A(M)$	the endomorphism algebra of $M$
$Z(E)$	the centre of $E$
$E'$	the centralizer of $E$ in $\operatorname{End}_R(M)$
$A^{\mathrm{op}}$	opposite algebra, product $a \cdot_{\mathrm{op}} b = ba$
$M_n(A^{\mathrm{op}})$	matrices over the opposite algebra, $=\operatorname{End}_A(A^n)$
$D=\operatorname{End}_A(S)$	the division ring of a simple module
$M_n(D)$	matrix ring over $D$, $=\operatorname{End}_A(S^n)$
$Z(A)$, $Z(D)$	the centres of $A$ and of $D$
$n_i$	the multiplicity of $S_i$ in $M$