# Example of a commutative perfect ring that is not artinian

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.

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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 $\mathbb{R}$.

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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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
The answer to this follow-up question is no, due to them being zero-dimensional. (A commutative ring R is perfect iff R is zero-dimensional and for every sequence $\{t_k\}_{k=1}^\infty \subseteq J(R)$ there is some some $n$ with $t_1\cdots t_n = 0$.) – Jason Juett Aug 25 '15 at 23:12
Dear @JasonJuett : that characterization is not correct. an infinite product of fields is both zero dimensional and has radical zero, but is not perfect. But of course you're spot on that commutative perfect rings are zero dimensional. Regards – rschwieb Aug 26 '15 at 2:52
Oops, I was thinking of Bass rings (rings over which every nonzero module has a maximal submodule) when I said that. Perfect rings can be characterized as completely decomposable Bass rings. Thanks for the correction. – Jason Juett Aug 26 '15 at 5:09

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.

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Just 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! – rschwieb Jan 25 '13 at 16:21