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

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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Endomorphisms of a finite direct sum are matrices of Hom-groups

Statement

For $n \in N$ and left $R$ -modules $M_{1}, \dots, M_{n}$ , endomorphisms of $⨁_{j} M_{j}$ correspond to $n \times n$ matrices $(f_{i j})$ with $f_{i j} \in {Hom}_{R} (M_{j}, M_{i})$ , and composition is matrix multiplication using composition in the entries. For $n = 0$ , both sides are the one-element zero ring. See The endomorphism ring ${End}_{R} (M)$ under addition and composition.

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For a left $R$ -module $M$ , define ${End}_{R} (M) : = {Hom}_{R} (M, M) .$ Addition is pointwise and multiplication is composition, $(f g) (m) : = f (g (m))$ . The ring laws and the identity endomorphism are established in prop-endomorphisms-form-a-ring. (The endomorphism ring ${End}_{R} (M)$ under addition and composition).

[L2]

For left $R$ -modules $M, N$ , the set ${Hom}_{R} (M, N)$ of module homomorphisms is an abelian group under pointwise addition, with zero the zero homomorphism and inverse $(- f) (m) = - f (m)$ (def-module-homomorphism-kernel-image-and-cokernel, def-group). (The abelian group ${Hom}_{R} (M, N)$ and maps induced by pre- and postcomposition).

[L3]

Let $(M_{i})_{i \in I}$ be left $R$ -modules and $N$ a left $R$ -module. For every family of homomorphisms $f_{i} : M_{i} \to N$ , there is a unique homomorphism $f : ⨁_{i \in I} M_{i} ⟶ N$ such that $f \circ ȷ_{i} = f_{i}$ for every $i$ . It is given by $f ((m_{i})) = \sum_{i \in supp (m)} f_{i} (m_{i}) .$ For $I = \emptyset$ , this is the unique map $0 \to N$ . (Universal property of a direct sum of modules).

Proof

technique · direct

1.1L1L2L3givenalgebra

We use inclusions and projections to send $f$ to entries $f_{i j} = π_{i} f ι_{j}$ , and reconstruct $f$ by finite sums.

2.1step 1.1L3algebra

Composition becomes matrix multiplication because $\sum_{k} ι_{k} π_{k} = id$ on a finite direct sum, so the $(i, j)$ entry of $f \circ g$ is $π_{i} (f \circ g) ι_{j} = π_{i} f (\sum_{k} ι_{k} π_{k}) g ι_{j} = \sum_{k} (π_{i} f ι_{k}) (π_{k} g ι_{j}) = \sum_{k} f_{i k} \circ g_{k j},$ which is the matrix product with composition in the entries; the sum is finite because $n$ is.

3.1step 1.1step 2.1L1algebra∎

For $n = 0$ the direct sum is the zero module, ${End}_{R} (0)$ has one element by [L1], and the set of $0 \times 0$ matrices also has exactly one element, so both sides are the one-element zero ring as the Statement records. For $n = 1$ the matrix is the single entry $f_{11} = π_{1} f ι_{1} = f$ , and the correspondence is the identity on ${End}_{R} (M_{1})$ . This proves the stated claim.

Depends on

Used by

Wedderburn–Artin theorem for semisimple rings Theorem

Dependency tree · two levels

7 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)