= Primitive idempotents under a surjection of commutative Artinian ideals
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.
Back to article page
