Idempotent matrix over local ring is similar to a diagonal matrix with elements $0$ and $1$

abstract-algebracommutative-algebralocal-ringsmatrices

I want to know why any idempotent matrix $P$ over local ring is similar to a diagonal matrix with elements $0$ and $1$.

I saw a proof in lemma 3.3 of ncatlab, but I don't understand why $P$ is similar to a diagonal matrix if it has an invertible $r$-minor, and why such a matrix can be chosen with entries $0$ and $1$.

Could somebody explain more?

Best Answer

$\newcommand{\Set}[1]{\left\{ #1 \right\}}$Let $V$ be the the underlying free module over the commutative local ring $A$ with unity.

Consider the submodules $$ V_{0} = \Set{ v \in V : P v = 0}, $$ and $$ V_{1} = \Set{ v \in V : P v = v}. $$ Clearly $V_{0} \cap V_{1} = \Set{0}$.

For $v \in V$ one has $$ v = (v - P v) + P v, $$ where

  • $v - P v \in V_{0}$, as $P (v - P v) = P v - P^{2} v = P v - P v = 0$, and
  • $P v \in V_{1}$, as $P (P v) = P^{2} V = P v$.

Therefore $V = V_{0} \oplus V_{1}$, and $P$ is zero on $V_{0}$ and the identity on $V_{1}$.

Addendum The point of $A$ being local is that the direct summands $V_{0}, V_{1}$ are also free,

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