Example of a commutative perfect ring that is not artinian

Question

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.

Answer 1

I believe such rings are easily constructed using subrings of triangular rings. Here is my attempt:

$R=\left\{ \begin{pmatrix} a &b\\ 0 & a \end{pmatrix} \ : \ a \in \mathbb{Q}, \ b \in \mathbb{R}\right\}$

It's easy to confirm this is a commutative ring. By observing that the subset of $R$ with zeros on the diagonal (call it $J$) is a nilpotent ideal, and that $R/J\cong \mathbb{Q}$, you can see that $J=rad(R)$.

Since $R/rad(R)$ is Artinian with nilpotent radical, it is a semiprimary ring, and so a perfect ring. Pick any infinite strictly ascending or descending chain of $\mathbb{Q}$ submodules of $\mathbb{R}$, and index them as $M_i$. Check that the sets $K_i=\left\{ \begin{pmatrix} 0 &b\\ 0 & 0 \end{pmatrix} \ : \ b \in M_i\right\}$ form an infinite chain of right ideals.

If there is no problem with my construction, you can pick any field to replace $\mathbb Q$ and any infinite dimensional vector space $B$ over your field to replace $\mathbb{R}$.

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(Additional answer to follow up in comments below)

Since the ring in the above construction is not Noetherian, I am curious about the following question: Is there any Noetherian perfect commutative ring with identity which fails to be Artinian? – Joy-Joy Jun 7 '13 at 2:12

There is no Noetherian example. The radical of a right perfect ring is nil, and with Noetherian that implies the radical is a nilpotent ideal. Then the Hopkins-Levitzki theorem kicks in and says the ring is Artinian too. In brief, if $R/J(R)$ is Artinian and $J(R)$ is nil, then $R$ is Artinian iff Noetherian.

Answer 2

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.