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LemmaStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (deepseek-v4-pro + gpt-5.6-terra)audited 2026-08-26
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Over a commutative ring the homomorphism group Hom⁡R(M,N) is an R-module

Statement

Let R be a commutative ring and let M,N be R-modules. Then the abelian group Hom⁡R(M,N) of The abelian group Hom⁡R(M,N) and maps induced by pre- and postcomposition becomes an R-module under the pointwise scalar action

(rf)(m):=r f(m)(r∈R, f∈Hom⁡R(M,N), m∈M).

The underlying additive group is the published one, unchanged: this extends the abelian-group structure rather than replacing it. Commutativity of R is used, and is used only to see that rf is again R-linear.

Facts & Assumptions

Given: A commutative ring R and R-modules M,N.

[L1]

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) (The abelian group Hom⁡R(M,N) and maps induced by pre- and postcomposition).

[L2]

A ring R is commutative when its multiplication is commutative, xy=yx for all x,y∈R (Commutative ring).

[L3]

A left R-module is an abelian group (M,+,0M) with an action satisfying r(m+n)=rm+rn, (r+s)m=rm+sm, (rs)m=r(sm) and 1Rm=m (Unital left and right modules over a ring; unqualified module means left module).

[L4]

A function f:M→N between left R-modules is an R-module homomorphism if f(m+m′)=f(m)+f(m′) and f(rm)=rf(m) for all m,m′∈M and r∈R (Module homomorphism and isomorphism, kernel, image and cokernel).

Proof

technique · direct
1.1L1given

The set Hom⁡R(M,N) already carries the pointwise addition making it an abelian group, with the zero homomorphism as neutral element and (−f)(m)=−f(m) as inverse. Nothing below alters that addition.

2.1L2L3L4step 1.1algebra

For r∈R and f∈Hom⁡R(M,N) the function rf defined by (rf)(m)=r f(m) is again an R-module homomorphism. Additivity is pointwise: (rf)(m+m′)=r(f(m)+f(m′))=rf(m)+rf(m′). Homogeneity is where commutativity enters: for s∈R, (rf)(sm)=r f(sm)=r(s f(m))=(rs)f(m)=(sr)f(m)=s(r f(m))=s (rf)(m).

3.1L1L3step 2.1algebra

The module axioms hold pointwise, each being an identity in N evaluated at an arbitrary m∈M: ((r+r′)f)(m)=(r+r′)f(m)=rf(m)+r′f(m), (r(f+g))(m)=r(f(m)+g(m))=rf(m)+rg(m), ((rr′)f)(m)=(rr′)f(m)=r(r′f(m)) and (1Rf)(m)=1Rf(m)=f(m).

4.1step 1.1step 3.1∎

So Hom⁡R(M,N) with the addition of step 1.1 and the action of step 2.1 satisfies the definition of an R-module, and its additive group is the published abelian group unchanged.

Remarks

Depends on

Used by

Dependency tree · two levels

9 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