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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 http://mathoverflow.net/questions/42647/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.

edited May 17 2012 at 20:19

asked May 17 2012 at 13:15

Niki
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Try $k[x]/\langle x^2 \rangle$ where $k$ is a field. – Konstantin Ardakov May 17 2012 at 13:48

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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 2012 at 14:26

Konstantin Ardakov · Answer 1 · 2012-05-17 23:55:24Z UTC

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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.

answered May 17 2012 at 23:55

Konstantin Ardakov
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	If I am not mistaken this ring has infinitely many maximal ideals... – Niki May 19 2012 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. – Konstantin Ardakov May 19 2012 at 12:21

semilocal total quotient ring whose J(R) is not zero

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