## The kernels and images of module homomorphisms are submodules

Let $R$ be a ring with 1. Let $M$ and $N$ be left $R$-modules, and let $\varphi : M \rightarrow N$ be an $R$-module homomorphism. Prove that $\mathsf{ker}\ \varphi$ and $\mathsf{im}\ \varphi$ are $R$-submodules of $M$ and $N$, respectively.

We use the Submodule Criterion for both sets.

Note that $\varphi(0) = 0$, so that $\mathsf{ker}\ \varphi$ and $\mathsf{im}\ \varphi$ are both nonempty.

Let $x,y \in \mathsf{ker}\ \varphi$ and let $r \in R$. Note that $\varphi(x + r \cdot y) = \varphi(x) + r \cdot \varphi(y) = 0$, so that $x + r \cdot y \in \mathsf{ker}\ \varphi$. Thus $\mathsf{ker}\ \varphi$ is an $R$-submodule of $M$.

Let $x,y \in M$ and let $r \in R$. Note that $\varphi(x) + r \cdot \varphi(y) = \varphi(x + r \cdot y)$; in particular, since $\varphi(x)$ and $\varphi(y)$ are arbitrary in $\mathsf{im} \varphi$, by the Submodule criterion we have that $\mathsf{im}\ \varphi$ is an $R$-submodule of $N$.