TITLE: Example of a commutative perfect ring that is not artinian
QUESTION [9 upvotes]: I read a result here stating that a commutative perfect ring is artinian if and only if it is $(1,1)$-coherent (see Proposition 5.3).  I'm interested in finding an example of a commutative perfect ring that is not artinian.

REPLY [4 votes]: Any semiprimary ring is perfect so a commutative semiprimary ring must be a ring product of local rings, each of which is semiprimary.  For example, let $F$ be a field and $V$ an infinite dimensional vector space over $F$. Then set $\left<R:+\right>=F\oplus V$ and $\left(a,v\right)\cdot\left(a',v'\right)=\left(aa',av'+a'v\right)$. If you wish a higher index of nilpotency, you can use an example of Camillo, Herzog, and Nielsen in J. Algebra 314 (2007) #1,  pp 471-478 which shows the example  $\left<R':+\right> = F\oplus V\oplus F$ with the radical  $J\left(R'\right)=V\oplus F$ and for $\beta\left(v,v'\right)$ a symmetric nondegenerate bilinear form from $V$ to $F$ set for $v$, $v'\in V$ and $a$, $a'\in F$, $\left(v,a\right) \cdot \left(v',a'\right) = \left(0,\beta\left(v,v'\right)\right)$ gives multiplication on  $J(R')$ and $\left(1,0,0\right)$ acts as the identity of $R'$. This one even is essential over a simple socle but clearly not quasiFrobenius because $V$ is infinite dimensional so $R'$ is not noetherian.