Primitive idempotent

A nonzero idempotent is primitive when it cannot be written as a sum of two nonzero orthogonal idempotents. For a finite-dimensional algebra $A$, the left module $Ae$ is indecomposable projective exactly when $e$ is primitive.

If idempotents $e_1,\dots ,e_N$ satisfy $e_i{e}_j=0$ for $i>j$, then the left ideals $A{e}_i$ have an internal direct sum. Multiply a relation $\sum z_i=0$ by $e_1$ on the right to obtain $z_1=0$, and continue in order. Pairwise orthogonality in both directions is unnecessary. If the sum has dimension $\dim A$, it is the whole algebra.

Let $A,B$ be finite-dimensional commutative algebras, let $I\subset A$ and $J\subset B$ be ideals, and let $f:I\to J$ be a surjective algebra homomorphism. The ideals need not have identities. Then $f$ gives a bijection between the primitive idempotents of $A$ lying in $I$ with nonzero image and the primitive idempotents of $J$.

Decompose $A$ into its commutative Artinian local rings $A_i$, with identities $e_i$. Each $I\cap A_i$ is either all of $A_i$, in which case $e_i\in I$, or a proper ideal and therefore a nilpotent ideal. The sum of the latter ideals is nilpotent, so its image contributes no idempotents. Each nonzero $f(e_i)$ has local corner algebra $f(e_i)J=f(A_i)$, a quotient of $A_i$, so is primitive. These idempotents are orthogonal, and the remaining part of $J$ is nilpotent; hence they are exactly its primitive idempotents. This statement does not require the entire kernel of $f$ to be nilpotent.