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.

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