Kernel of Linear Transformation is Submodule

Theorem
Let $\struct {R, +_R, \times_R}$ be a ring.

Let $\struct {G, +_G, \circ_G}_R$ and $\struct {H, +_H, \circ_H}_R$ be $R$-modules.

Let $\phi: G \to H$ be a linear transformation.

Let $\map \ker \phi$ denote the kernel of $\phi$.

Then $\map \ker \phi$ is a submodule of $G$.

Proof
By definition, the kernel of $\phi$ is defined as:
 * $\map \ker \phi := \phi^{-1} \sqbrk {\set {e_H} }$

where $e_H$ is the identity of $\struct {H, +_H}$.

where $\phi^{-1} \sqbrk S$ denotes the preimage of $S$ under $\phi$.

From Null Module is Module:
 * $\struct {\set {e_H}, +_H, \circ_H}_R$ is a module

where $\struct {\set {e_H}, +_H, \circ_H}_R$ is the null module.

We have that $\struct {\set {e_H}, +_H, \circ_H}_R$ is a module which is a subset of $H$.

It follows that $\struct {\set {e_H}, +_H, \circ_H}_R$ is a submodule of $H$.

The result follows from Preimage of Submodule under Linear Transformation is Submodule.