[proofplan]
We prove both assertions by directly verifying the submodule axioms. For the kernel, the defining condition $f(x)=0_N$ is preserved under addition, additive inverses, and scalar multiplication because $f$ is $R$-linear. For the image, each element is represented as $f(x)$ for some $x\in M$, and the module operations in $N$ can be pulled back to the corresponding operations in $M$.
[/proofplan]
[step:Verify that the kernel is closed under the module operations]
Define the subset $K\subset M$ by
\begin{align*}
K=\ker f=\{x\in M:f(x)=0_N\}.
\end{align*}
Since $f$ is additive,
\begin{align*}
f(0_M)=f(0_M+0_M)=f(0_M)+f(0_M).
\end{align*}
Adding the additive inverse of $f(0_M)$ in $N$ gives $f(0_M)=0_N$, so $0_M\in K$.
Let $x,y\in K$. Then $f(x)=0_N$ and $f(y)=0_N$. By additivity of $f$,
\begin{align*}
f(x+y)=f(x)+f(y)=0_N+0_N=0_N,
\end{align*}
so $x+y\in K$.
Let $x\in K$. Since $x+(-x)=0_M$, additivity gives
\begin{align*}
f(x)+f(-x)=f(x+(-x))=f(0_M)=0_N.
\end{align*}
Thus $f(-x)$ is the additive inverse of $f(x)$. Because $f(x)=0_N$, it follows that $f(-x)=0_N$, so $-x\in K$.
Finally, let $r\in R$ and $x\in K$. Since $f$ is $R$-linear,
\begin{align*}
f(rx)=r f(x)=r0_N=0_N.
\end{align*}
Hence $rx\in K$. Therefore $K=\ker f$ is a submodule of $M$.
[guided]
We need to prove that $\ker f$ is not merely a subset of $M$, but a subset stable under all operations inherited from the left $R$-module $M$. Define
\begin{align*}
K=\ker f=\{x\in M:f(x)=0_N\}.
\end{align*}
The submodule verification requires four checks: $K$ contains the zero element of $M$, is closed under addition, is closed under additive inverses, and is closed under scalar multiplication by elements of $R$.
First, we show that $0_M\in K$. Since $f$ is a homomorphism of the additive groups underlying the modules, it is additive. Therefore
\begin{align*}
f(0_M)=f(0_M+0_M)=f(0_M)+f(0_M).
\end{align*}
Adding the additive inverse of $f(0_M)$ in the abelian group underlying $N$ to both sides gives $f(0_M)=0_N$. Thus $0_M$ satisfies the defining condition for membership in $K$, and hence $0_M\in K$.
Next, take arbitrary elements $x,y\in K$. The definition of $K$ gives $f(x)=0_N$ and $f(y)=0_N$. Additivity of the [module homomorphism](/page/Module%20Homomorphism) gives
\begin{align*}
f(x+y)=f(x)+f(y)=0_N+0_N=0_N.
\end{align*}
Thus $x+y$ again satisfies the defining equation for $K$, so $x+y\in K$.
Now take an arbitrary $x\in K$. We need to prove that $-x\in K$. Since $x+(-x)=0_M$, applying additivity of $f$ gives
\begin{align*}
f(x)+f(-x)=f(x+(-x))=f(0_M)=0_N.
\end{align*}
This equation says exactly that $f(-x)$ is the additive inverse of $f(x)$ in $N$. But $x\in K$, so $f(x)=0_N$, and the additive inverse of $0_N$ is $0_N$. Therefore $f(-x)=0_N$, which proves $-x\in K$.
Finally, let $r\in R$ and $x\in K$. The $R$-linearity of $f$ gives
\begin{align*}
f(rx)=r f(x).
\end{align*}
Since $x\in K$, we have $f(x)=0_N$, and scalar multiplication by $r$ sends $0_N$ to $0_N$ in the left $R$-module $N$. Hence
\begin{align*}
f(rx)=r0_N=0_N.
\end{align*}
Thus $rx\in K$. The subset $K=\ker f$ contains $0_M$ and is closed under addition, additive inverses, and scalar multiplication, so it is a submodule of $M$.
[/guided]
[/step]
[step:Verify that the image is closed under the module operations]
Define the subset $I\subset N$ by
\begin{align*}
I=\operatorname{im} f=\{f(m):m\in M\}.
\end{align*}
From the previous step, $f(0_M)=0_N$, so $0_N=f(0_M)\in I$.
Let $a,b\in I$. By definition of $I$, there exist $x,y\in M$ such that $a=f(x)$ and $b=f(y)$. By additivity of $f$,
\begin{align*}
a+b=f(x)+f(y)=f(x+y).
\end{align*}
Since $x+y\in M$, this shows $a+b\in I$.
Let $a\in I$. Choose $x\in M$ such that $a=f(x)$. As in the kernel step, additivity gives $f(-x)=-f(x)$, and therefore
\begin{align*}
-a=-f(x)=f(-x).
\end{align*}
Since $-x\in M$, we have $-a\in I$.
Finally, let $r\in R$ and $a\in I$. Choose $x\in M$ such that $a=f(x)$. By $R$-linearity of $f$,
\begin{align*}
ra=r f(x)=f(rx).
\end{align*}
Since $rx\in M$, this shows $ra\in I$. Therefore $I=\operatorname{im} f$ is a submodule of $N$.
[/step]
[step:Conclude both submodule assertions]
The first step proves that $\ker f$ is a submodule of $M$. The second step proves that $\operatorname{im} f$ is a submodule of $N$. These are exactly the two conclusions asserted in the theorem.
[/step]
