Why does the quaternion algebra appear both inside M_2(C) and inside M_4(R)?

The 4 × 4 real picture is the regular representation, which exists for any finite-dimensional algebra: A embeds in End_k(A) ≅ M_ A(k). The 2 × 2 complex picture is smaller because H ⊗_R C ≅ M_2(C) — extending scalars splits the division algebra.

Matrix Rings and Endomorphism Rings of Modules

Q: Why is End(R_R) equal to R rather than merely containing it?

Because a right R-linear map is determined by its value at 1: if is right R-linear then (a) = (1)a, so is left multiplication by (1). The module R_R is cyclic and free of rank one, which leaves no room for anything else.

Q: What breaks if I use left modules everywhere?

Nothing mathematically, but the bookkeeping inverts. End(_RR) ≅ R^op, and End(_R R^n) ≅ M_n(R^op) ≅ M_n(R)^op. Wedderburn–Artin can be stated either way; Lam's choice of right modules for this purpose keeps the matrix multiplication order unreversed.

Q: Does M_n(R) ≅ M_n(S) imply R ≅ S?

Yes when R and S are division rings — the number n and the division ring are recovered from the module theory. For general rings the correct statement is weaker: M_n(R) ≅ M_n(S) implies R and S are Morita equivalent, and there exist non-isomorphic Morita equivalent rings.

Q: Is End_R(M) ever commutative for an interesting M?

Yes, and it is a useful signal. Over a commutative ring, End of a module of rank one is commutative; over any ring, End of a simple module is a division ring by Schur's lemma but need not be commutative — End_H(H) ≅ H is the standard example.

Executive Summary

Two identifications underlie the whole structure theory of rings. The first says that a ring loses nothing when regarded as an operator algebra: left multiplications identify $R$ with $End (R_{R})$ , the endomorphism ring of $R$ as a right module over itself. The second says that endomorphisms of a direct sum of $n$ copies of a module are matrices over the endomorphism ring of that module.

Together they explain why matrix rings are unavoidable: $M_{n} (R) ≅ End (R_{R}^{n})$ , and every semisimple ring turns out to be a finite product of such rings over division rings. They also explain the sidedness conventions of the subject. If one insists on left modules and writes maps on the left, the first identification produces $R^{op}$ rather than $R$ , and every subsequent formula acquires an unwanted transpose.

$R ≅ End (R_{R})$ Regular representation

$M_{n} (End M)$

End

M^{n}

$Z (R) \cdot I_{n}$ Centre of

M_{n} (R)

$M_{n} (I)$ Every ideal of

M_{n} (R)

Overview

Let $𝒞$ be an additive category and $A$ an object. The set ${End}_{𝒞} (A)$ of endomorphisms of $A$ is an abelian group under pointwise addition, composition is biadditive and associative, and ${id}_{A}$ is a multiplicative identity. So ${End}_{𝒞} (A)$ is a ring, with no hypotheses at all. Taking $𝒞$ to be right $R$ -modules gives ${End}_{R} (A)$ for any right $R$ -module $A$ .

The interesting case is $A = R_{R}$ , the ring viewed as a right module over itself. Its submodules are exactly the right ideals, and its endomorphisms turn out to be exactly the left multiplications.

L : R ⟶ End (R_{R}), L (r) (a) = r a,

(1.12)

The regular representation. It is a ring isomorphism, not merely an embedding.

Iterating with $A = R^{n}$ produces $M_{n} (R)$ , and the same argument applied to an $n$ -dimensional right vector space $V$ over a division ring $k$ gives ${End}_{k} (V) ≅ M_{n} (k)$ . That last isomorphism is the reason the Wedderburn–Artin theorem can be stated with matrix rings at all.

The internal structure of $M_{n} (R)$ is controlled by the matrix units $e_{i j}$ . They form a $ℤ$ -basis of $M_{n} (ℤ) \subseteq M_{n} (R)$ , multiply by the rule $e_{i j} e_{k l} = δ_{j k} e_{i l}$ , and sum to the identity along the diagonal. Every computation on this page is an exercise in sandwiching an unknown matrix between two matrix units.

Learning Objectives

Prove that $L : R \to End (R_{R})$ is a ring isomorphism and identify $End (_{R} R)$ with $R^{op}$ .
Derive ${End}_{R} (M^{n}) ≅ M_{n} ({End}_{R} (M))$ from the injections and projections of a direct sum.
Use matrix units to show $Z (M_{n} (R)) = Z (R) \cdot I_{n}$ .
Show that every two-sided ideal of $M_{n} (R)$ is $M_{n} (I)$ for a unique ideal $I$ of $R$ .
Describe the upper triangular subring $T \subseteq M_{n} (k)$ , its nilpotent ideal and its semisimple quotient.
Compute the $4 \times 4$ real matrix representation of the quaternions.

Definitions

Definition(1.12)Endomorphism ring

For a right $R$ -module $A$ , ${End}_{R} (A)$ denotes the set of $R$ -linear maps $A \to A$ , with addition $(f + g) (a) = f (a) + g (a)$ and multiplication $f g = f \circ g$ . Maps are written on the left of their arguments throughout, so $(f g) (a) = f (g (a))$ .

Definition—Matrix ring and matrix units

$M_{n} (R)$ is the ring of $n \times n$ arrays over $R$ with the usual operations; $(A B)_{i j} = \sum_{l} A_{i l} B_{l j}$ . The matrix units $e_{i j}$ have a $1$ in position $(i, j)$ and $0$ elsewhere, and satisfy

e_{i j} e_{k l} = δ_{j k} e_{i l}, \sum_{i = 1}^{n} e_{i i} = 1, A = \sum_{i, j} a_{i j} e_{i j} .

(M)

$R_{R}$ and $_{R} R$: The ring $R$ regarded as a right, respectively left, module over itself. Submodules are right, respectively left, ideals.
$R^{op}$: The opposite ring: same additive group, multiplication reversed. $M_{n} (R)^{op} ≅ M_{n} (R^{op})$ via transposition.
$T_{n} (R)$: The subring of upper triangular matrices in $M_{n} (R)$ .
Free module of rank $n$: $R^{n}$ , with $End (R_{R}^{n}) ≅ M_{n} (R)$ .
Morita invariant: A ring-theoretic property shared by $R$ and $M_{n} (R)$ for every $n$ ; simplicity, primeness and the chain conditions are Morita invariant, commutativity is not.

Vector spaces over a division ring are taken to be right vector spaces, and scalars are written on the right. This is not fussiness: it is what makes the matrix of a composite equal the product of the matrices in the same order.

Core Concepts

Why right modules

Let $V$ be a right $k$ -vector space with basis $e_{1}, \dots, e_{n}$ , so every $v \in V$ is uniquely $\sum_{i} e_{i} c_{i}$ with $c_{i} \in k$ . For $f \in {End}_{k} (V)$ write $f (e_{j}) = \sum_{i} e_{i} a_{i j}$ . Then

f (\sum_{j} e_{j} c_{j}) = \sum_{j} f (e_{j}) c_{j} = \sum_{i} e_{i} (\sum_{j} a_{i j} c_{j}),

(1.13a)

Coordinates transform by $c \mapsto A c$ : matrices act on the left, scalars on the right, and the two never collide.

Composing gives $(f g) (e_{j}) = \sum_{i, l} e_{i} a_{i l} b_{l j}$ , so the matrix of $f g$ is $A B$ — the same order. Had $V$ been a left $k$ -space with maps still written on the left, scalars and matrices would compete for the same side and $End$ would come out as $M_{n} (k)^{op} ≅ M_{n} (k^{op})$ .

Sandwiching with matrix units

For $A = (a_{i j}) \in M_{n} (R)$ the products with matrix units extract and relocate entries:

e_{k i} A e_{j l} = a_{i j} e_{k l} .

(1.13b)

Every entry of $A$ can be moved to any position without leaving the ideal generated by $A$ .

This one identity proves the description of the centre, the description of the two-sided ideals, and the fact that $M_{n} (R)$ is simple whenever $R$ is. It is the workhorse of the subject.

Idempotents and corners

A decomposition $M = M_{1} \oplus M_{2}$ of a right $R$ -module corresponds to an idempotent $e \in {End}_{R} (M)$ , namely the projection onto $M_{1}$ along $M_{2}$ . In particular $R = e R \oplus (1 - e) R$ for any idempotent $e \in R$ , and

{End}_{R} (e R) ≅ e R e, ϕ ⟼ ϕ (e),

(1.13c)

The corner ring. Taking $e = e_{11}$ in $M_{n} (R)$ recovers $e_{11} M_{n} (R) e_{11} ≅ R$ .

The corner construction is developed on the Corner Rings page; the point here is that matrix rings and corner rings are inverse operations, $R ⇝ M_{n} (R) ⇝ e_{11} M_{n} (R) e_{11} ≅ R$ .

The upper triangular subring

Let $k$ be a division ring and $T \subseteq M_{n} (k)$ the upper triangular matrices, $I \subseteq T$ those with zero diagonal. Then $I$ is a two-sided ideal of $T$ , the diagonal map gives $T / I ≅ k \times \dots \times k$ ( $n$ factors), and $I$ is nilpotent of exact index $n$ : the product $e_{12} e_{23} \dots e_{n - 1, n} = e_{1 n}$ is nonzero, so $I^{n - 1} \neq 0$ , while $I^{n} = 0$ because each factor raises the distance from the diagonal by at least one.

Key Results

Theorem(1.12)The regular representation

Let $R$ be a ring. The map $L : R \to End (R_{R})$ sending $r$ to left multiplication $L (r) : a \mapsto r a$ is a ring isomorphism. Dually, right multiplications give a ring isomorphism $R^{op} ≅ End (_{R} R)$ .

Proof

$L (r)$ is right $R$ -linear: $L (r) (a b) = r (a b) = (r a) b = (L (r) (a)) b$ . Additivity of $L$ is clear, and $L (r s) (a) = (r s) a = r (s a) = L (r) L (s) (a)$ , so $L$ is a ring homomorphism with $L (1) = id$ .

Injective. If $L (r) = 0$ then $r = r \cdot 1 = L (r) (1) = 0$ .

Surjective. Let $ϕ \in End (R_{R})$ and set $r = ϕ (1)$ . For every $a \in R$ , right $R$ -linearity gives $ϕ (a) = ϕ (1 \cdot a) = ϕ (1) a = r a = L (r) (a)$ , so $ϕ = L (r)$ .

For the dual statement, right multiplication $ρ_{r} : a \mapsto a r$ is left $R$ -linear and $ρ_{r} ρ_{s} (a) = ρ_{r} (a s) = a s r = ρ_{s r} (a)$ . So $r \mapsto ρ_{r}$ reverses products: it is an isomorphism from $R^{op}$ , not from $R$ .

Theorem—Endomorphisms of a direct sum of copies

Let $M$ be a right $R$ -module, $n \geq 1$ , and $M^{n}$ the direct sum of $n$ copies of $M$ . Write $E = {End}_{R} (M)$ . Then

{End}_{R} (M^{n}) ≅ M_{n} (E), f ⟼ (π_{i} f ι_{j})_{i, j},

(Ex. 20)

$ι_{j}$ and $π_{i}$ are the canonical injection into and projection from the $j$ -th and $i$ -th summands.

In particular, taking $M = R_{R}$ and using $(1.12)$ , $End (R_{R}^{n}) ≅ M_{n} (R)$ ; and for an $n$ -dimensional right vector space $V$ over a division ring $k$ , ${End}_{k} (V) ≅ M_{n} (k)$ .

Proof

The structural maps satisfy $π_{i} ι_{j} = δ_{i j} {id}_{M}$ and $\sum_{l = 1}^{n} ι_{l} π_{l} = {id}_{M^{n}}$ . Define $Φ (f)_{i j} = π_{i} f ι_{j} \in E$ . Additivity is immediate, and

Φ (f g)_{i j} = π_{i} f g ι_{j} = π_{i} f (\sum_{l} ι_{l} π_{l}) g ι_{j} = \sum_{l} (π_{i} f ι_{l}) (π_{l} g ι_{j}) = (Φ (f) Φ (g))_{i j},

—

so $Φ$ is multiplicative; $Φ (id) = I_{n}$ because $π_{i} ι_{j} = δ_{i j} id$ . The map $Ψ ((f_{i j})) = \sum_{i, j} ι_{i} f_{i j} π_{j}$ is a two-sided inverse: $Φ Ψ = id$ by the orthogonality relations, and $Ψ Φ (f) = \sum_{i, j} ι_{i} π_{i} f ι_{j} π_{j} = f$ by the completeness relation applied on both sides.

Proposition—Centre of a matrix ring

For every ring $R$ and every $n \geq 1$ , $Z (M_{n} (R)) = {z I_{n} : z \in Z (R)}$ ; that is, the centre of $M_{n} (R)$ consists of the scalar matrices with entry in the centre of $R$ .

Proof

Let $A = (a_{i j})$ be central and fix $k \neq l$ . Comparing $A e_{k l}$ , whose $(i, l)$ entry is $a_{i k}$ and whose other columns vanish, with $e_{k l} A$ , whose $(k, j)$ entry is $a_{l j}$ and whose other rows vanish, forces $a_{i k} = 0$ for $i \neq k$ and $a_{k k} = a_{l l}$ . Hence $A = z I_{n}$ for a single $z \in R$ . Commuting with $r I_{n}$ for all $r \in R$ then gives $z r = r z$ , i.e. $z \in Z (R)$ . The converse inclusion is immediate.

Proposition(3.1)Ideals of a matrix ring

Let $R$ be a ring and $n \geq 1$ . The map $I \mapsto M_{n} (I)$ is an inclusion-preserving bijection from the two-sided ideals of $R$ onto the two-sided ideals of $M_{n} (R)$ . In particular $M_{n} (R)$ is simple if and only if $R$ is simple.

Proof

That $M_{n} (I)$ is an ideal is a direct check. Conversely let $𝔄$ be an ideal of $M_{n} (R)$ and put $I = {a \in R : a e_{11} \in 𝔄}$ . This is an ideal of $R$ : it is additively closed, and for $r \in R$ we have $(r a) e_{11} = (r e_{11}) (a e_{11}) \in 𝔄$ and $(a r) e_{11} = (a e_{11}) (r e_{11}) \in 𝔄$ .

If $A \in 𝔄$ then $(1.13 b)$ gives $e_{1 i} A e_{j 1} = a_{i j} e_{11} \in 𝔄$ , so every entry of $A$ lies in $I$ and $𝔄 \subseteq M_{n} (I)$ . Conversely if $a \in I$ then $a e_{i j} = e_{i 1} (a e_{11}) e_{1 j} \in 𝔄$ for all $i, j$ , and summing shows $M_{n} (I) \subseteq 𝔄$ . Uniqueness of $I$ follows since $I$ is recovered as the set of $(1, 1)$ entries.

Corollary—What matrix rings inherit

For any ring $R$ and any $n \geq 1$ : $M_{n} (R)$ is simple iff $R$ is; $M_{n} (R)$ is left noetherian (respectively left artinian) iff $R$ is; and $rad M_{n} (R) = M_{n} (rad R)$ . By contrast $M_{n} (R)$ is never commutative and never a domain for $n \geq 2$ and $R \neq 0$ , since $e_{12} e_{21} \neq e_{21} e_{12}$ and $e_{12}^{2} = 0$ .

Worked Example

The quaternions inside $M_{2} (ℂ)$

Let $ℍ = ℝ \oplus ℝ i \oplus ℝ j \oplus ℝ ℓ$ with $i^{2} = j^{2} = - 1$ , $i j = - j i = ℓ$ . The assignment

a + b i + c j + d ℓ ⟼ (\begin{matrix} a + b i & c + d i \\ - c + d i & a - b i \end{matrix})

(E.1)

An $ℝ$ -algebra isomorphism of $ℍ$ onto ${(\begin{matrix} α & β \\ - \bar{β} & \bar{α} \end{matrix}) : α, β \in ℂ} \subseteq M_{2} (ℂ)$ .

is an isomorphism onto that subring. Checking the generators is enough: $i \mapsto (\begin{matrix} i & 0 \\ 0 & - i \end{matrix})$ , $j \mapsto (\begin{matrix} 0 & 1 \\ - 1 & 0 \end{matrix})$ , and their product is $(\begin{matrix} 0 & i \\ i & 0 \end{matrix})$ , the image of $ℓ$ . The determinant of the image of $q$ is $a^{2} + b^{2} + c^{2} + d^{2}$ , the reduced norm — which is why nonzero quaternions are invertible.

The regular representation over $ℝ$

Now apply $(1.12)$ with $R = ℍ$ : left multiplication embeds $ℍ ≅ {End}_{ℍ} (ℍ_{ℍ})$ inside ${End}_{ℝ} (ℍ) ≅ M_{4} (ℝ)$ . Using the ordered basis $1, i, j, ℓ$ and recording the images of the basis vectors as columns, $q = a + b i + c j + d ℓ$ maps to

(\begin{matrix} a & - b & - c & - d \\ b & a & - d & c \\ c & d & a & - b \\ d & - c & b & a \end{matrix}) .

(E.2)

The image is a $4$ -dimensional $ℝ$ -subalgebra of $M_{4} (ℝ)$ isomorphic to $ℍ$ .

Verify the second column: $q i = a i + b i^{2} + c j i + d ℓ i = - b + a i + d j - c ℓ$ , using $j i = - ℓ$ and $ℓ i = j$ — exactly the entries $(- b, a, d, - c)$ . The remaining columns are the same computation with $j$ and $ℓ$ .

A calculation with matrix units

Take $R = ℤ$ and $n = 2$ . Every ideal of $M_{2} (ℤ)$ is $M_{2} (m ℤ)$ for some $m \geq 0$ , by the proposition above. Concretely, the ideal generated by $(\begin{matrix} 0 & 6 \\ 0 & 0 \end{matrix})$ contains $e_{11} (6 e_{12}) e_{21} = 6 e_{11}$ , hence contains $6 e_{i j}$ for all $i, j$ , and equals $M_{2} (6 ℤ)$ . By contrast the left ideal $M_{2} (ℤ) e_{11}$ , consisting of the matrices whose second column is zero, is not of the form $M_{2} (I)$ : the correspondence is for two-sided ideals only.

Frameworks and Models

Matrix rings over a field are a laboratory: almost every phenomenon in finite-dimensional ring theory can be produced by choosing a subring of $M_{n} (k)$ carefully. The standard families are worth knowing by name.

Subrings of $M_{n} (k)$
- Triangular families
  - $T_{n} (k)$ , all upper triangular matrices — nilpotent radical, semisimple quotient $k^{n}$
  - upper triangular matrices whose last column vanishes above the diagonal
  - diagonal matrices whose first and last entries agree, plus an arbitrary $(1, n)$ corner entry
- Congruence subrings of $M_{2} (ℤ)$
  - $(\begin{matrix} ℤ & n ℤ \\ 0 & ℤ \end{matrix})$
  - matrices with $a \equiv d$ and $b \equiv c (\mod n)$
  - matrices with $a \equiv d (\mod n)$ and $b, c \in n ℤ$
- Field and division subalgebras
  - $ℚ [α] ≅ ℚ (\sqrt{- 3})$ inside $M_{2} (ℚ)$ , for $α = (\begin{matrix} 0 & - 1 \\ 1 & - 1 \end{matrix})$ with characteristic polynomial $t^{2} + t + 1$
  - $ℍ \subseteq M_{2} (ℂ)$ and $ℍ \subseteq M_{4} (ℝ)$
  - $ℂ \subseteq M_{2} (ℝ)$ via $i \mapsto (\begin{matrix} 0 & - 1 \\ 1 & 0 \end{matrix})$
- Mixed-coefficient triangular rings
  - $(\begin{matrix} ℚ & ℝ \\ 0 & ℚ \end{matrix})$ and $(\begin{matrix} ℚ & ℝ \\ 0 & ℝ \end{matrix})$ — see the Triangular Rings page

The last family leaves matrix rings behind: its entries come from different rings, and its systematic treatment is the triangular ring construction $(\begin{matrix} R & M \\ 0 & S \end{matrix})$ .

Comparison and Classification

Which properties pass from $R$ to $M_{n} (R)$ , $n \geq 2$
	Passes to $M_{n} (R)$	Reflects back to $R$
Simple	yes	yes
Prime	yes	yes
Left noetherian	yes	yes
Left artinian	yes	yes
Semisimple	yes	yes
Commutative	no	no
Domain	no	no
Local	no	no
Division ring	no	no

Which properties pass from $R$ to $M_{n} (R)$ , $n \geq 2$

Endomorphism rings of familiar modules
Module	Endomorphism ring	Why
$R_{R}$	$R$	left multiplications, $(1.12)$
$_{R} R$	$R^{op}$	right multiplications reverse products
$R_{R}^{n}$	$M_{n} (R)$	matrix of components
$e R$ , $e^{2} = e$	$e R e$	$ϕ \mapsto ϕ (e)$
$V$ , an $n$ -dimensional right $k$ -space	$M_{n} (k)$	choose a basis
$V$ , infinite-dimensional over $k$	not a matrix ring	contains a non-Dedekind-finite pair, $a b = 1 \neq b a$
A simple module $S$	a division ring	Schur's lemma

The penultimate row is the standard warning: the shift operator on a countably infinite-dimensional space is left-invertible without being invertible, so ${End}_{k} (V)$ is not Dedekind-finite and cannot be $M_{n}$ of anything.

Relationship Map

module

M

{End}_{R} (M)

M_{n} ({End}_{R} M)

{End}_{R} (M^{n})

The chain is a closed loop: passing to endomorphisms of powers of a module, and passing to corners of a matrix ring, undo one another. That loop is the elementary shadow of Morita equivalence.

All rings

R ≅ End (R_{R})

Matrix rings $M_{n} (R)$

≅ End (R_{R}^{n})

; ideals are

M_{n} (I)

$M_{n} (D)$ , $D$ a division ringsimple artinian — the model of Wedderburn–Artin

$M_{n} (k)$ , $k$ a fieldcentral simple over

k

of dimension

n^{2}

Downstream: Ideals of Matrix Rings and the Correspondence Theorem proves the ideal correspondence in the generality needed for Wedderburn–Artin; Matrix Rings over Division Rings identifies $M_{n} (D)$ as the model simple artinian ring; Corner Rings runs the construction backwards.

Design Considerations

Design considerations here means the choices made when modelling a problem with these algebraic structures.

Which side for modules? If you want $End$ of the free module of rank $n$ to be $M_{n} (R)$ with the usual multiplication, use right modules and write maps on the left. Any other combination introduces an opposite ring or a transpose, and the two errors do not cancel.
Rows or columns? With right modules, coordinates are columns and matrices act on the left. Switching to row vectors is legitimate but silently replaces $M_{n} (R)$ by its opposite.
Scalars on which side? For a division ring $k$ , right $k$ -vector spaces make ${End}_{k} (V) ≅ M_{n} (k)$ ; left ones give $M_{n} (k^{op})$ . For $k$ commutative the distinction evaporates, which is why it is invisible in linear algebra courses.
**When to pass to $M_{n}$ .** Morita-invariant questions — simplicity, primeness, chain conditions, the radical — can be moved to whichever of $R$ and $M_{n} (R)$ is easier. Questions about units, commutativity, or zero-divisors cannot.
Idempotents as design tools. A decomposition of the identity into orthogonal idempotents $1 = e_{1} + \dots + e_{n}$ turns $R$ into a ring of $n \times n$ 'generalised matrices' with $(i, j)$ block $e_{i} R e_{j}$ . Choosing the idempotents well is the practical form of choosing a matrix presentation.

Standards and Notation

Standards here covers notation, symbol and markup standards, and reference implementations, rather than material or design codes.

Preferred notation

M_{n} (R)

for the matrix ring (Lam);

e_{i j}

for matrix units

Common variants

{Mat}_{n} (R)

R^{n \times n}

, and

𝕄_{n} (R)

all occur; ISO 80000-2 prefers upright symbols for operators

Endomorphisms

{End}_{R} (M)

; some authors write

M_{R}

-linear maps on the right as

(m) f

, which changes composition order

Triangular

T_{n} (R)

{UT}_{n} (R)

for upper triangular matrices

GAPMatrixAlgebra, FullMatrixAlgebra; RadicalOfAlgebra for the radical

Sage / MagmaMatrixSpace(R, n), MatrixAlgebra(R, n); both act on column vectors by default

Failure Modes and Common Mistakes

Do not assume $M_{n} (R) ≅ M_{m} (S)$ forces $n = m$ and $R ≅ S$ . It does when $R$ and $S$ are division rings, but not in general — Morita equivalence is coarser than isomorphism.
Do not compute the centre by 'diagonal matrices commute with everything'. They do not: $diag (1, 0)$ fails to commute with $e_{12}$ . Only scalar matrices with central entry are central.
Do not confuse $M_{n} (R)^{op}$ with $M_{n} (R^{op})$ as sets of matrices — they are isomorphic, but the isomorphism is transposition, not the identity map.
Do not expect the trace to be a ring homomorphism or even multiplicative; over a noncommutative $R$ it is only well defined modulo the additive commutator subgroup.

Quick Reference

Regular representation

R ≅ End (R_{R})

r \mapsto (a \mapsto r a)

Left version

R^{op} ≅ End (_{R} R)

r \mapsto (a \mapsto a r)

Powers

{End}_{R} (M^{n}) ≅ M_{n} ({End}_{R} M)

Free module

End (R_{R}^{n}) ≅ M_{n} (R)

Matrix units

e_{i j} e_{k l} = δ_{j k} e_{i l}

\sum_{i} e_{i i} = 1

e_{k i} A e_{j l} = a_{i j} e_{k l}

Centre

Z (M_{n} (R)) = Z (R) I_{n}

Ideals

𝔄 ◁ M_{n} (R) iff 𝔄 = M_{n} (I)

I ◁ R

Corner

e_{11} M_{n} (R) e_{11} ≅ R

; generally

{End}_{R} (e R) ≅ e R e

Standard subrings of $M_{n} (k)$ , $k$ a division ring
Subring	Radical	Semisimple quotient
$M_{n} (k)$	$0$	itself
$T_{n} (k)$ , upper triangular	strictly upper triangular, nilpotent of index $n$	$k \times \dots \times k$ , $n$ factors
Scalars $k I_{n}$	$0$	$k$
$(\begin{matrix} k & k \\ 0 & k \end{matrix})$	$(\begin{matrix} 0 & k \\ 0 & 0 \end{matrix})$ , square zero	$k \times k$
$e_{11} M_{n} (k) e_{11}$	$0$	$k$

Frequently Asked Questions

Why is

End (R_{R})

equal to

R

rather than merely containing it?

Because a right $R$ -linear map is determined by its value at $1$ : if $ϕ$ is right $R$ -linear then $ϕ (a) = ϕ (1) a$ , so $ϕ$ is left multiplication by $ϕ (1)$ . The module $R_{R}$ is cyclic and free of rank one, which leaves no room for anything else.

What breaks if I use left modules everywhere?

Nothing mathematically, but the bookkeeping inverts. $End (_{R} R) ≅ R^{op}$ , and $End (_{R} R^{n}) ≅ M_{n} (R^{op}) ≅ M_{n} (R)^{op}$ . Wedderburn–Artin can be stated either way; Lam's choice of right modules for this purpose keeps the matrix multiplication order unreversed.

Are all the ideals of

M_{n} (R)

really of the form

M_{n} (I)

All two-sided ones, yes, and the correspondence is a lattice isomorphism. The proof is a two-line calculation with matrix units. One-sided ideals are a different story: $M_{n} (k)$ over a field is simple but has a full lattice of left ideals, one for each subspace of $k^{n}$ .

Does

M_{n} (R) ≅ M_{n} (S)

imply

R ≅ S

Yes when $R$ and $S$ are division rings — the number $n$ and the division ring are recovered from the module theory. For general rings the correct statement is weaker: $M_{n} (R) ≅ M_{n} (S)$ implies $R$ and $S$ are Morita equivalent, and there exist non-isomorphic Morita equivalent rings.

Why does the quaternion algebra appear both inside

M_{2} (ℂ)

and inside

M_{4} (ℝ)

The $4 \times 4$ real picture is the regular representation, which exists for any finite-dimensional algebra: $A$ embeds in ${End}_{k} (A) ≅ M_{\dim A} (k)$ . The $2 \times 2$ complex picture is smaller because $ℍ \otimes_{ℝ} ℂ ≅ M_{2} (ℂ)$ — extending scalars splits the division algebra.

{End}_{R} (M)

ever commutative for an interesting

M

Yes, and it is a useful signal. Over a commutative ring, $End$ of a module of rank one is commutative; over any ring, $End$ of a simple module is a division ring by Schur's lemma but need not be commutative — ${End}_{ℍ} (ℍ) ≅ ℍ$ is the standard example.

References

T. Y. Lam, A First Course in Noncommutative Rings, Graduate Texts in Mathematics 131, Springer-Verlag, 1991, §1, Examples (1.10), (1.12)–(1.13) (pp. 11–14), with Exercises 9 and 20; the ideal correspondence is (3.1) in §3.
F. W. Anderson and K. R. Fuller, Rings and Categories of Modules, 2nd edition, Graduate Texts in Mathematics 13, Springer-Verlag, 1992, §§4 and 22.
N. Jacobson, Basic Algebra II, 2nd edition, W. H. Freeman, 1989, Chapter 3.
L. H. Rowen, Ring Theory, Volume I, Academic Press, 1988, Chapter 1.
T. Y. Lam, Lectures on Modules and Rings, Graduate Texts in Mathematics 189, Springer-Verlag, 1999, §17 (Morita theory).

AI Suggested Questions

Prove that $M_{n} (R)$ and $R$ have isomorphic lattices of two-sided ideals directly from Morita theory.
Show that ${End}_{k} (V)$ for infinite-dimensional $V$ has exactly three two-sided ideals when $\dim V$ is countable.
Given orthogonal idempotents summing to $1$ , write $R$ as a generalised matrix ring and identify the blocks.
For which pairs $(n, R)$ is $M_{n} (R)$ isomorphic to a triangular ring $(\begin{matrix} A & M \\ 0 & B \end{matrix})$ ?
Work out the automorphism group of $M_{n} (k)$ for $k$ a field and relate it to the Skolem–Noether theorem.
Compute the centre and the ideals of the congruence subring of $M_{2} (ℤ)$ with $b, c \in n ℤ$ .
Explain why the trace form on $M_{n} (k)$ is nondegenerate and what that says about the radical.

Matrix and Endomorphism Rings

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Why right modules

Sandwiching with matrix units

Idempotents and corners

The upper triangular subring

Key Results

Worked Example

The quaternions inside $M_{2} (ℂ)$

The regular representation over $ℝ$

A calculation with matrix units

Frameworks and Models

Comparison and Classification

Relationship Map

Design Considerations

Standards and Notation

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

References

AI Suggested Questions

Continue learning

01Executive Summary

02Overview

03Learning Objectives

04Definitions

05Core Concepts

Why right modules

Sandwiching with matrix units

Idempotents and corners

The upper triangular subring

06Key Results

07Worked Example

The quaternions inside M2(ℂ)

The regular representation over ℝ

A calculation with matrix units

08Frameworks and Models

09Comparison and Classification

10Relationship Map

11Design Considerations

12Standards and Notation

13Failure Modes and Common Mistakes

14Quick Reference

15Frequently Asked Questions

16Related KEVOS Topics

17References

18AI Suggested Questions

Continue learning

Executive Summary

Overview

Learning Objectives

Definitions

Core Concepts

Key Results

Worked Example

The quaternions inside $M_{2} (ℂ)$

The regular representation over $ℝ$

Frameworks and Models

Comparison and Classification

Relationship Map

Design Considerations

Standards and Notation

Failure Modes and Common Mistakes

Quick Reference

Frequently Asked Questions

Related KEVOS Topics

References

AI Suggested Questions