[proofplan] The proof has two directions. First, an idempotent endomorphism $e$ decomposes each element $m \in M$ as its $e$-fixed part $e(m)$ plus the error term $m-e(m)$, and idempotence shows that the error term is killed by $e$. The intersection of these two pieces is zero because an element in the image of $e$ is fixed by $e$, while an element in the kernel is killed by $e$. Conversely, a direct sum decomposition gives a well-defined projection onto the first summand, and direct verification shows that this projection is $R$-linear, idempotent, and has the required image and kernel. [/proofplan] [step:Decompose every element into an image part and a kernel part] Let $m \in M$. Define \begin{align*} u &:= e(m), & v &:= m-e(m). \end{align*} Then $u \in \operatorname{im}(e)$ by definition of image. Since $e$ is $R$-linear and $e \circ e = e$, we have \begin{align*} e(v) &= e(m-e(m)) \\ &= e(m)-e(e(m)) \\ &= e(m)-e(m) \\ &= 0. \end{align*} Thus $v \in \ker(e)$. The equality \begin{align*} m = e(m) + (m-e(m)) = u+v \end{align*} therefore shows that every element of $M$ lies in $\operatorname{im}(e)+\ker(e)$. [/step] [step:Show the image and kernel have zero intersection] Let $x \in \operatorname{im}(e) \cap \ker(e)$. Since $x \in \operatorname{im}(e)$, there exists $m \in M$ such that $x=e(m)$. Since $x \in \ker(e)$, we also have $e(x)=0$. Using $e \circ e=e$, we compute \begin{align*} x &= e(m) \\ &= e(e(m)) \\ &= e(x) \\ &= 0. \end{align*} Hence $\operatorname{im}(e) \cap \ker(e)=\{0\}$. Together with the previous step, this proves \begin{align*} M = \operatorname{im}(e) \oplus \ker(e). \end{align*} [/step] [step:Construct the projection associated to a direct sum decomposition] Assume conversely that $A,B \le M$ and $M=A \oplus B$. By the definition of internal direct sum, each $m \in M$ has a unique expression \begin{align*} m = a+b \end{align*} with $a \in A$ and $b \in B$. Define the map \begin{align*} p: M &\to M \\ a+b &\mapsto a, \end{align*} where $a \in A$ and $b \in B$ are the unique summands of the given element. We verify that $p$ is $R$-linear. Let $m_1,m_2 \in M$, and write \begin{align*} m_1 &= a_1+b_1, & m_2 &= a_2+b_2 \end{align*} with $a_1,a_2 \in A$ and $b_1,b_2 \in B$. Since $A$ and $B$ are submodules, $a_1+a_2 \in A$ and $b_1+b_2 \in B$, so the uniqueness of the direct sum decomposition gives \begin{align*} p(m_1+m_2) &= p((a_1+a_2)+(b_1+b_2)) \\ &= a_1+a_2 \\ &= p(m_1)+p(m_2). \end{align*} For $r \in R$, since $A$ and $B$ are submodules, $ra_1 \in A$ and $rb_1 \in B$, and hence \begin{align*} p(rm_1) &= p(ra_1+rb_1) \\ &= ra_1 \\ &= r p(m_1). \end{align*} Therefore $p \in \operatorname{End}_R(M)$. [/step] [step:Verify the projection is idempotent with image $A$ and kernel $B$] Let $m \in M$, and write $m=a+b$ with $a \in A$ and $b \in B$. Then $p(m)=a$. Since $a=a+0$ with $a \in A$ and $0 \in B$, we have $p(a)=a$. Therefore \begin{align*} (p \circ p)(m) &= p(p(m)) \\ &= p(a) \\ &= a \\ &= p(m). \end{align*} Since this holds for every $m \in M$, we have $p \circ p=p$. The image of $p$ is $A$: for every $m=a+b \in M$, $p(m)=a \in A$, so $\operatorname{im}(p) \subset A$; conversely, if $a \in A$, then $a=a+0$ and $p(a)=a$, so $A \subset \operatorname{im}(p)$. Hence $\operatorname{im}(p)=A$. The kernel of $p$ is $B$: if $m=a+b \in \ker(p)$, then \begin{align*} 0 = p(m)=a, \end{align*} so $m=b \in B$, and therefore $\ker(p) \subset B$. Conversely, if $b \in B$, then $b=0+b$ with $0 \in A$, so \begin{align*} p(b)=0, \end{align*} and thus $b \in \ker(p)$. Hence $\ker(p)=B$. This proves the converse and completes the proof. [/step]