ExampleConstruction: AI-adaptedVerification: AI-adaptedprecheck passaudited 2026-08-17

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$M_{n} (F)$ as a direct sum of minimal left ideals

Example

For every field $F$ and $n \geq 1$ , the left regular module of $M_{n} (F)$ is the internal direct sum $M_{n} (F) = ⨁_{j = 1}^{n} M_{n} (F) e_{j j}$ of simple left ideals. See Matrix rings over division rings are semisimple.

Facts & Assumptions

Given: The hypotheses and objects in the Example.

[L1]

For a division ring $D$ and $n \geq 1$ , the left regular module of $M_{n} (D)$ is the direct sum of its simple column ideals $M_{n} (D) e_{j j} ≅ D^{n}$ . (Matrix rings over division rings are semisimple).

[L2]

For $r \geq 1$ , $n_{i} \geq 1$ , and division rings $D_{i}$ , every simple left module over $\prod_{i} M_{n_{i}} (D_{i})$ is supported on exactly one factor and is isomorphic to that factor's column module $D_{i}^{n_{i}}$ . (Simple modules over a product of matrix rings over division rings).

[L3]

The matrix unit $E_{i j} \in M_{m \times n} (F)$ has entries $(E_{i j})_{r s} = δ_{r i} δ_{s j},$ so it has entry $1$ in position $(i, j)$ and $0$ everywhere else. (Matrix units $E_{i j}$ and the Kronecker delta).

Verification

technique · direct

1.1L1L2L3givenalgebra

The left ideal $M_{n} (F) e_{j j}$ consists of matrices supported in column $j$ . Reading that column identifies it with the natural column module $F^{n}$ . If a nonzero vector lies in a submodule of $F^{n}$ , matrix units send it to every standard basis vector, so the submodule is all of $F^{n}$ ; each column ideal is therefore simple.

2.1step 1.1givenalgebra

Every matrix is the sum of its column matrices, and matrices supported in distinct columns have zero intersection. Hence $M_{n} (F) = ⨁_{j = 1}^{n} M_{n} (F) e_{j j}$ as a left module.

3.1step 2.1givenalgebra∎

For $n = 1$ this is the single simple left ideal $F$ . The decomposition selects $n$ minimal left ideals but makes no assertion that these are the only minimal left ideals. This proves the stated claim.

Depends on

Used by

False statement: every semisimple ring is commutative False statement

Dependency tree · two levels

11 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

Arvind Nair, Algebra I, Lecture 5 (standard reference, not scraped)

Mn(F)Mn​(F) as a direct sum of minimal left ideals