Example of indecomposable self injective ring

Question

Is there any example of an indecomposable self-injective commutative ring with 4 or more maximal ideals?$$$$$$$$

Answer 1

If $R$ is noetherian, then $R$ is gorenstein of dimension zero, hence artinian. In particular, $R$ is a finite direct product of local artinian rings. If you mean $R$ is indecomposable as a direct product of rings, then this means $R$ has to be local, hence has a unique maximal ideal.

I don't know what we can say in the non-noetherian case.

Answer 2

You may be interested in the following example. Let $R$ be the graded algebra over $\mathbb{Z}/2$ with generators $y_0,y_1,y_2,\dotsc$ subject to relations $y_i^3=y_iy_{i+1}$, where $y_i$ has degree $2^i$. In http://arxiv.org/abs/1206.0137, Leigh Shepperson and I prove that $R$ is self-injective in the appropriate sense for graded rings, and not Noetherian. It is easily seen to be indecompsable. It only has one graded maximal ideal (consisting of the elements of strictly positive degree). However, it has infinitely many inhomogeneous maximal ideals. The same paper has a number of other interesting examples of non-Noetherian rings that are self-injective or have related properties.

Answer 3

The answer is no. In fact, even if $R$ is a (possibly noncommutative) right self-injective ring with no nontrivial idempotents, then $R$ is a local ring, and the unique maximal (left or right) ideal is the set of left zero-divisors! The proof is found as Exercise 3.2 in "Exercises in Modules and Rings" by T.Y. Lam.