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

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Kernels and images of module homomorphisms are submodules, and injectivity is equivalent to trivial kernel

Statement

For an $R$ -module homomorphism $f : M \to N$ , both $\ker f$ and $im f$ are submodules. Moreover,

$f is injective ⟺ \ker f = {0_{M}} .$

Facts & Assumptions

Given: A homomorphism $f : M \to N$ of left $R$ -modules.

[L1]

A module homomorphism preserves addition and scalar multiplication, and the displayed definitions give its kernel and image (Module homomorphism and isomorphism, kernel, image and cokernel).

[L2]

A nonempty subset of a module is a submodule exactly when it is closed under $r u + v$ (The one-step submodule criterion; intersections and sums of submodules are submodules).

[L3]

The additive underlying function of a module homomorphism is a group homomorphism and therefore sends $0_{M}$ to $0_{N}$ ; moreover $r 0_{N} = 0_{N}$ (Module homomorphism and isomorphism, kernel, image and cokernel, A group homomorphism automatically satisfies $f (e) = e^{'}$ and $f (g^{- 1}) = f (g)^{- 1}$ , and $f (g^{n}) = f (g)^{n}$ for every $n \in Z$ ; for monoid homomorphisms preservation of the identity must be assumed, In a module, $0_{R} m = 0_{M}$ , $r 0_{M} = 0_{M}$ , $(- r) m = - (r m)$ and $r (- m) = - (r m)$ ).

[L4]

A group homomorphism is injective exactly when its kernel is trivial (A group homomorphism is injective if and only if its kernel is trivial).

Proof

technique · direct

1.1L1L3given

The kernel contains $0_{M}$ because $f (0_{M}) = 0_{N}$ . If $u, v \in \ker f$ and $r \in R$ , then $f (r u + v) = r f (u) + f (v) = 0_{N}$ , so $r u + v \in \ker f$ .

1.2L1L3given

The image contains $0_{N} = f (0_{M})$ . If $f (u), f (v) \in im f$ and $r \in R$ , then $r f (u) + f (v) = f (r u + v)$ lies in the image.

1.3L1L4given

The additive underlying function has the same kernel described in [L1], so the group-homomorphism theorem gives $f$ injective exactly when $\ker f = {0_{M}}$ .

2.1step 1.1step 1.2L2

The submodule criterion applied to steps 1.1 and 1.2 proves that $\ker f$ and $im f$ are submodules.

3.1step 1.3step 2.1∎

Together, steps 1.3 and 2.1 prove both assertions.

Depends on

Dependency tree · 2 levels

Sources

McGerty, Algebra II: Rings and Modules, Section 3