not every matrix is sum of two idempotent matrices i'm reading the proof of theory show that the matrix ring $M_{n}(R)$ over $R$ contains an element which cannot be written as sum of two idempotents. ($R$ is a ring with $1$, and $n$ is a positive integer greater than $1$)
Lemma: Let $a$ be an element of $R$ with $a^{2} = 0$. if $a = e + f$ for idempotents $e, f$ then $4a = 0$.
theory's proof: We write $M_{n}(R) = \sum_{i,j=1}^{n}Re_{ij} $ where $e_{ij}$ are matrix units. Suppose,
to the contrary, that every element of $M_{n}(R)$ is the sum of two idempotents. Then, by Lemma, $4e_{12} = 0 $ and so $4R = 0$. Consider the element $a = e_{11} + e_{12} + e_{21}$, and choose idempotents $e = \sum r_{ij}e_{ij}$ and $f$ such that $a = e + f$. Since $a - e = f = f^{2} = a^{2} — ae — ea + e$ , we get $a^{2}= a + ae + ea — 2e$ . Comparing the coefficients of $e_{11}$, $e_{12}$ and $e_{21}$ on both sides, we get $1 = r_{12} + r_{21}$, $0 = r_{11} + r_{22} - r_{12}$ and $0 = r_{11} + r_{22} - r_{21}$ therefore $1 = 2r_{12}$. Then $4R = 0$ implies that $1 = 4r_{12}^{2} = 0$, which is a contradiction.
Please explain me why we got $4R=0$? And compare the coefficients of $e_{11}$, $e_{12}$ and $e_{21}$?
(Sorry, English is not my native language)


Why we got $4R=0$

Because $\begin{bmatrix}0&4r\\0&0\end{bmatrix}=r4\begin{bmatrix}0&1\\0&0\end{bmatrix}=\begin{bmatrix}0&0\\0&0\end{bmatrix}$ for any $r$, so $4R=\{0\}$.

And compare the coefficients of $e_{11}$, $e_{12}$ and $e_{21}$?

Just write out $a^{2}= a + ae + ea — 2e$:
$\begin{bmatrix}2&1\\ 1&1\end{bmatrix}= \begin{bmatrix}1&1\\1&0\end{bmatrix}^2=\begin{bmatrix}1&1\\1&0\end{bmatrix}+\begin{bmatrix}1&1\\1&0\end{bmatrix}\begin{bmatrix}r_{11}&r_{12}\\r_{21}&r_{22}\end{bmatrix}+\begin{bmatrix}r_{11}&r_{12}\\r_{21}&r_{22}\end{bmatrix}\begin{bmatrix}1&1\\1&0\end{bmatrix}-2\begin{bmatrix}r_{11}&r_{12}\\r_{21}&r_{22}\end{bmatrix}$
$\begin{bmatrix}2&1\\ 1&1\end{bmatrix}= \begin{bmatrix}1&1\\1&0\end{bmatrix}+\begin{bmatrix}r_{11}+r_{21}&r_{12}+r_{22}\\r_{11}&r_{12}\end{bmatrix}+\begin{bmatrix}r_{11}+r_{12}&r_{11}\\r_{21}+r_{22}&r_{21}\end{bmatrix}-\begin{bmatrix}2r_{11}&2r_{12}\\2r_{21}&2r_{22}\end{bmatrix}$
$\begin{bmatrix}2&1\\ 1&1\end{bmatrix}= \begin{bmatrix}1+r_{21}+r_{12}&1+r_{22}+r_{11}-r_{12}\\1+r_{11}+r_{22}-r_{21}&r_{12}+r_{21}-2r_{22}\end{bmatrix}$
