Jacobson radical of endomorphisms of direct sum of indecomposables $\newcommand\End{\operatorname{End}}$Let $A$ be a commutative $k$-algebra and $M=N_1\oplus N_2$ be a decomposition of a finite-length $A$-module $M$ into two non-isomorphic indecomposable summands. Consider the endomorphism ring $E=\End(M)$ of $M$. What is $J(E)$?
I would expect that it is of the form
$$\begin{pmatrix}J(\End(N_1)) & *\\ * & J(\End(N_2))\end{pmatrix},$$
but neither am I totally convinced nor can I prove this. At least, $J(\End(N_i))\subseteq J(E)$ as follows: let $a\in J(\End(N_1))$, take $\begin{pmatrix}x_1&*\\{}*&*\end{pmatrix}$ arbitrarily and choose $s$ such that $(1-ax)s=1$. 
$$\left[\begin{pmatrix}1\\&1\end{pmatrix}-\begin{pmatrix}a\\&0\end{pmatrix}\begin{pmatrix}x_1&x_2\\{}*&*\end{pmatrix}\right]\begin{pmatrix}s & sax_2\\&1\end{pmatrix}=\begin{pmatrix}1\\&1\end{pmatrix}.$$
However, this neither shows that $J(\End(N_i))$ exhaust the diagonal entries, nor that any matrix with only off-diagonal entries is contained in $J(E)$. Is this even true?
 A: $\newcommand\End{\operatorname{End}}\newcommand\Aut{\operatorname{Aut}}\newcommand\Hom{\operatorname{Hom}}\newcommand{\Mtx}[1]{\left(\begin{smallmatrix}#1\end{smallmatrix}\right)}$Let $M$ be a module of finite length
    with a decomposition $M=L_1^{n_1}\oplus\dotsb\oplus L_k^{n_k}$ into sums of pairwise non-isomorphic indecomposable modules $L_i$.
    Then $J(\End(M))$ is generated by the Jacobson radicals of the $\End(L_i)$'s
    and all morphisms between distinct $L_i$'s.
Proof.
    It suffices to consider $M=L_1^{n_1}\oplus L_2^{n_2}$.
    It has endomorphism ring
    $$
  \End(M)=\begin{pmatrix}\End(L_1)^{n_1\times n_1} & \Hom(L_2, L_1)^{n_1\times n_2}\\\Hom(L_1, L_2)^{n_2\times n_1}&\End(L_2)^{n_2\times n_2}\end{pmatrix}.
 $$
    For simplicity, assume that $n_1=n_2=1$;
    for the off-diagonal blocks, the general case follows by extending the matrix by zeroes as appropriate,
    and for the diagonal blocks, the general case follows from the fact that $J(S^{n\times n})=J(S)^{n\times n}$ for full matrix rings.  
We claim that matrices with entries only on the diagonal are in $J(\End(M))$ if and only if
    the entries are in the respective Jacobson radicals.
    Consider $\Mtx{1\\&1}-\Mtx{a\\&0}\Mtx{b' & b\\{}*&*}$ and choose $s$ such that $(1-ab')s=1$.
    Then
    $$
  \left[\begin{pmatrix}1\\&1\end{pmatrix}-\begin{pmatrix}a\\&0\end{pmatrix}\begin{pmatrix}b' & b\\{}*&*\end{pmatrix}\right]\begin{pmatrix}s & ab\\&1\end{pmatrix}=\begin{pmatrix}1\\&1\end{pmatrix},
 $$
    so $[\cdots]$ is a unit.
    Conversely, if $$\left[\begin{pmatrix}1\\&1\end{pmatrix}-\begin{pmatrix}a\\&0\end{pmatrix}\begin{pmatrix}b' & b\\{}*&*\end{pmatrix}\right]\begin{pmatrix}s & *\\z&*\end{pmatrix}=\begin{pmatrix}1\\&1\end{pmatrix},$$
    it follows that $z=0$ and $(1-ab')s=1$.
We claim that any matrix with entries only in the off-diagonal blocks belongs to $J(\End(M))$.
    Let $a\in\Hom(L_1,L_2)$, which corresponds to $\Mtx{0\\a&0}\in \End(M)$.
    Since $M$ is of finite length, so is $L_1$.
    Consider an arbitrary matrix $\Mtx{b' & b\\{}*&*}\in\End(M)$ with $b\in\Hom(L_2, L_1)$.
    Fitting's lemma implies that the composition $ba$ is either an isomorphism or nilpotent.
    If it were an isomorphism, $a$ would be an inclusion with retraction $b$,
    implying that $L_2$ contains a summand isomorphic to $L_1$
    and thus contradicting the assumption.
    Thus, $ab$ is nilpotent, say $(ab)^n=0$.
    $$
  \left[\begin{pmatrix}1\\&1\end{pmatrix}-\begin{pmatrix}0\phantom{b'}\\a&0\end{pmatrix}\begin{pmatrix}b' & b\\{}*&*\end{pmatrix}\right]\begin{pmatrix}1 & 0\\(1+\dotsb+(ab)^{n-1})ab' & 1+\dotsb+(ab)^{n-1}\end{pmatrix} = \begin{pmatrix}1\\&1\end{pmatrix},
 $$
    so $[\cdots]$ is a unit.
Recall that no automorphism of $L_i$ factors through a distinct $L_j$,
    and since $L_i$ is of finite length, $\End(L_i)$ is a local ring with maximal ideal consisting of all non-automorphisms.
    Let $J$ be the ideal generated by matrices with entries $\Mtx{J(\End(L_1))&*\\{}*&J(\End(L_2))}$.
    Thus the quotient $M/J=\Aut(L_1)^{n_1\times n_1}\oplus\Aut(L_2)^{n_2\times n_2}$ is a semisimple algebra.
    This shows that $J=J(\End(M))$.
