MathOverflow is a question and answer site for professional mathematicians. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I am interested in rings in which every non-unit is a zero divisor. Can you give me an example of such a ring that also has FINITELY many maximal ideals (semilocal), and whose Jacobson radical is not zero? Thanks

P.S. There is a question about this Rings in which every non-unit is a zero divisor but none of the answers satisfy my constrains.

Edit: I am sorry I haven't specified - I am not interested in an example that is a finite ring, or the one that is local (and both the examples in comments are in this category). Also Artinian rings won't do.

share|cite|improve this question
Try $k[x]/\langle x^2 \rangle$ where $k$ is a field. – user91132 May 17 '12 at 13:48
Or any nonreduced finite ring, such as $\mathbb{Z}/n\mathbb{Z}$ if $n$ is not squarefree (this is mentioned in the question referred to). – Laurent Moret-Bailly May 17 '12 at 14:26

Let $A = k[x_1,x_2, \ldots]$ be the polynomial ring over a field $k$ in infinitely many variables and let $\mathfrak{m} = \langle x_1,x_2,\ldots \rangle$. Then $(A / \mathfrak{m}^2)^2$ satisfies all of your conditions.

share|cite|improve this answer
If I am not mistaken this ring has infinitely many maximal ideals... – Niki May 19 '12 at 7:46
If $R = (A / \mathfrak{m}^2)^2$ then $J(R) = \mathfrak{m}R$ has square zero, and $R / J(R) \cong (A / \mathfrak{m}) \times (A / \mathfrak{m}) \cong k \times k$. This ring, and hence $R$ itself, has precisely two maximal ideals. – user91132 May 19 '12 at 12:21

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.