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.
Example of a commutative perfect ring that is not artinian
2 Answers
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 $\mathbb{R}$.
(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.
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1@rschwieb : I haven't registered yet, but I will if necessary ;-). Anyway thank you for this helpful website! – 2016-11-10
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
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0Just to draw a connection: the $F\oplus V$ example is isomorphic to the one in my solution, but it is nice to have a bigger family of examples with higher nilpotency! – 2013-01-25