TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + claude-sonnet-5)audited 2026-08-17

How statement and proof provenance work

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

${End}_{R} (_{R} R) ≅ R^{o p}$

Statement

For every unital ring $R$ , evaluation at $1$ identifies endomorphisms of the left regular module with right multiplications and gives a ring isomorphism ${End}_{R} (_{R} R) ≅ R^{o p} .$ See The opposite ring $R^{o p}$ .

Facts & Assumptions

Given: The hypotheses and objects in the Statement.

[L1]

For a unital ring $R$ , the opposite ring $R^{o p}$ has the same underlying abelian group, identity, and addition as $R$ , with multiplication $a ⋆ b : = b a$ . Associativity and both distributive laws follow from those of $R$ with the order reversed, and the same element $1$ is a two-sided identity. Thus the displayed operations really form a unital ring, including when $R$ is the zero ring. (The opposite ring $R^{o p}$ ).

[L2]

For every left $R$ -module $M$ , pointwise addition and composition make ${End}_{R} (M)$ a unital ring with identity ${id}_{M}$ . (Module endomorphisms form a ring under pointwise addition and composition).

[L3]

Let $R$ be a ring. A left $R$ -module is an abelian group $(M, +, 0_{M})$ with a scalar action $R \times M \to M$ , $(r, m) \mapsto r m$ , satisfying $r (m + n) = r m + r n, (r + s) m = r m + s m, (r s) m = r (s m), 1_{R} m = m .$ A right $R$ -module has an action $M \times R \to M$ , $(m, r) \mapsto m r$ , with the analogous right-handed axioms. Unless “right” is stated, module means a unital left module. (Unital left and right modules over a ring; unqualified module means left module).

Proof

technique · direct

1.1L1L2L3givenalgebra

Evaluate an endomorphism at $1$ .

2.1step 1.1givenalgebra

Left linearity gives $f (r) = r f (1)$ , so endomorphisms are right multiplications; composition reverses the order of their defining elements.

3.1step 2.1L1L2givenalgebra∎

The map $a \mapsto ρ_{a}$ with $ρ_{a} (x) = x a$ is inverse to $f \mapsto f (1)$ : it is a left $R$ -module homomorphism because $ρ_{a} (r x) = r x a = r ρ_{a} (x)$ , it satisfies $ρ_{f (1)} = f$ by step 2.1, and $ρ_{a} (1) = a$ . It is additive, and $ρ_{a} \circ ρ_{b} (x) = x b a = ρ_{b a} (x) = ρ_{a ⋆ b} (x)$ with $⋆$ the multiplication of [L1], so it is a ring isomorphism onto ${End}_{R} (_{R} R)$ . For the zero ring $1 = 0$ , both ${End}_{R} (_{R} R)$ and $R^{o p}$ are the one-element ring and the correspondence is the unique map between them. This proves the stated claim.

Depends on

Used by

Wedderburn–Artin theorem for semisimple rings Theorem

Dependency tree · two levels

8 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)

End⁡R(RR)≅RopEndR​(R​R)≅Rop