PropositionStatement: Literature-sourcedProof: AI-adaptedprecheck passaudited 2026-08-03

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.

The canonical map $M \to M / N$ is a surjective module homomorphism with kernel $N$ ; thus every submodule is a kernel

For a submodule $N \leq M$ , the canonical map

$π : M ⟶ M / N, π (m) : = m + N,$

is a surjective $R$ -module homomorphism and has kernel $N$ . Hence every submodule is the kernel of a module homomorphism.

Given: A left $R$ -module $M$ and a submodule $N \leq M$ .

[L1]

[L2]

[L3]

A module homomorphism preserves addition and scalar multiplication, and its kernel is the inverse image of zero (Module homomorphism and isomorphism, kernel, image and cokernel).

[L4]

The additive group of a submodule is closed under inverses, and the coset criterion consequently gives $m + N = N$ if and only if $m \in N$ (Submodule of a module, $x \in a H$ iff $a^{- 1} x \in H$ , and $a H = b H$ iff $a^{- 1} b \in H$ ).

technique · direct

1.1L2given

The underlying additive map $π$ is a homomorphism and is surjective.

1.2L1given

The quotient action gives $π (r m) = r m + N = r (m + N) = r π (m)$ .

1.3L3L4given

Since $π (m) = 0_{M / N} = N$ exactly when $m \in N$ , its kernel is $N$ .

2.1step 1.1step 1.2L3

Steps 1.1--1.2 show that $π$ is a surjective module homomorphism.

3.1step 1.3step 2.1∎

Together with step 1.3, this proves the claim and shows that every submodule is a kernel.

Depends on