Ring

Peirce decomposition

In a ring 𝑅, let 𝑒 be an idempotent and 𝑓 =(1 −𝑒) its complement. Then #m/thm/ring

  1. 𝑅 =𝑅𝑒 ⊕𝑅𝑓 as left ideals
  2. 𝑅 =𝑒𝑅 ⊕𝑓𝑅 as right ideals;
  3. 𝑅 =𝑒𝑅𝑒 ⊕𝑒𝑅𝑓 ⊕𝑓𝑅𝑒 ⊕𝑓𝑅𝑓 as abelian groups.

In ^D3, we have that 𝑒𝑅𝑒 and 𝑓𝑅𝑓 are in fact subrings, called the corner rings, where

𝑒𝑅𝑒={𝑟∈𝑅:∣:𝑒𝑟=𝑟=𝑟𝑒}𝑓𝑅𝑓={𝑟∈𝑅:∣:𝑓𝑟=𝑟=𝑟𝑓}.

These are named after the case where 𝑒 =diag⁡(1,…,1,0,…,0) in a matrix ring, since then

𝑒𝑅𝑒={[∗000]},𝑒𝑅𝑓={[0∗00]}𝑓𝑅𝑒={[00∗0]}𝑓𝑅𝑓={[000∗]}

Properties

  1. 𝑒 is central iff 𝑒𝑅𝑓 =𝑓𝑅𝑒 =0.
Proof

For a given 𝑟 ∈𝑅,

𝑒𝑟𝑓=𝑒𝑟(1−𝑒)=𝑒𝑟−𝑒𝑟𝑒,𝑓𝑟𝑒=(1−𝑒)𝑟𝑒=𝑟𝑒−𝑒𝑟𝑒,

so both are zero iff 𝑒𝑟 =𝑒𝑟𝑒 =𝑟𝑒, proving ^P1.


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