TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-17

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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Matrix rings over division rings are semisimple

Statement

Let $D$ be a division ring and $n \geq 1$ . On the set of $n \times n$ arrays over $D$ , use entrywise addition and the product $(A B)_{i j} : = \sum_{k = 1}^{n} a_{i k} b_{k j} .$ These operations make a ring $M_{n} (D)$ , and this ring is semisimple. More precisely, its left regular module is the direct sum of the simple column ideals $M_{n} (D) e_{j j} ≅ D^{n}$ for $1 \leq j \leq n$ . See A semisimple ring as a ring whose left regular module is semisimple.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

A unital ring $R$ is semisimple when its left regular module $_{R} R$ is semisimple. This is a left-module definition and uses no Jacobson radical. For the zero ring, the regular module is zero and hence semisimple; the Wedderburn-Artin theorem below is stated for nonzero rings. (A semisimple ring as a ring whose left regular module is semisimple).

[L2]

A division ring is a ring $D$ with $1 \neq 0$ in which every nonzero element has a two-sided multiplicative inverse. (Division ring: a ring with $1 \neq 0$ in which every nonzero element is a unit).

[L3]

A ring has an abelian-group addition, an associative multiplication with identity, and both distributive laws; multiplication need not be commutative. (Ring: an abelian group under addition and a monoid under multiplication, with multiplication distributing over addition on both sides).

[L4]

Finite sums in a commutative monoid are independent of the chosen enumeration, and the empty sum is $0$ . (A finite sum in a commutative monoid indexed by an arbitrary finite set).

[L5]

A left $R$ -module $M$ is simple if $M \neq 0$ and its only submodules are $0$ and $M$ . Equivalently, $M$ has no proper nonzero submodule. (Simple module: a nonzero module with no proper nonzero submodule).

Proof

technique · direct

1.1L2L3L4givenalgebra

Entrywise addition makes the arrays an abelian group. Associativity of multiplication follows by expanding both $(A B) C$ and $A (B C)$ and reassociating the finite double sum; the two distributive laws follow entrywise from those of $D$ . The matrix $I = (δ_{i j} 1_{D})$ is a two-sided identity. Thus the displayed operations make the unital ring $M_{n} (D)$ without any commutativity assumption on $D$ .

2.1step 1.1L4givenalgebra

Let $e_{j} = e_{j j}$ . Every matrix has the unique decomposition $A = \sum_{j = 1}^{n} A e_{j}$ , and $M_{n} (D) e_{i} \cap \sum_{j \neq i} M_{n} (D) e_{j} = 0$ because the two sides have disjoint possible nonzero columns. Hence $_{M_{n} (D)} M_{n} (D) = ⨁_{j = 1}^{n} M_{n} (D) e_{j} .$

3.1L2L5step 1.1step 2.1givenalgebra

Sending a matrix in $M_{n} (D) e_{j}$ to its $j$ -th column identifies that left ideal with $D^{n}$ under left matrix multiplication. If $0 \neq v \in D^{n}$ , choose $k$ with $v_{k} \neq 0$ . For any $w \in D^{n}$ , the matrix whose only possibly nonzero column is column $k$ , with entry $a_{i k} = w_{i} v_{k}^{- 1}$ , sends $v$ to $w$ . Thus every nonzero submodule of $D^{n}$ is all of $D^{n}$ , so each column ideal is simple.

4.1L1step 2.1step 3.1given∎

The decomposition in step 2.1 is therefore a finite direct sum of simple left modules, so [L1] makes $M_{n} (D)$ semisimple. For $n = 1$ it is the single simple column ideal, and the hypothesis $n \geq 1$ excludes an empty decomposition.

Depends on

Used by

Dependency tree · two levels

22 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

William Crawley-Boevey, Noncommutative Algebra, Chapter 1 Sections 1.1-1.9 (standard reference, not scraped)