Is there a non-artinian noetherian ring whose non-units are zero-divisors?

Equivalent formulation:

Is there a noetherian ring of positive dimension whose non-units are zero-divisors?

[In this post, "ring" means "commutative ring with one", and "dimension" means "Krull dimension".]

Here is the motivation:

Let $A$ be a ring whose non-units are zero-divisors.

If $A$ is not noetherian, then $A$ can have positive dimension: see this answer of user18119.

If $A$ is noetherian and reduced, then $\dim A\le0$: see this answer of user26857.

[Recall that a noetherian ring is artinian if and only if its dimension is $\le0$. Recall also that a ring has the property that its non-units are zero-divisors if and only if it is isomorphic to its total ring of fractions.]

  • $\begingroup$ See also mathoverflow.net/questions/42647/… $\endgroup$ – lhf Oct 25 '18 at 13:07
  • $\begingroup$ @lhf - Thanks! I looked carefully at this great thread, but didn't find an answer to my question. What am I missing? $\endgroup$ – Pierre-Yves Gaillard Oct 25 '18 at 13:11
  • $\begingroup$ If $R$ is Artinian then every non-zero divisor in $R$ is a unit. $\endgroup$ – 1ENİGMA1 Oct 25 '18 at 13:12
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    $\begingroup$ @Pierre-YvesGaillard, I just wrote a comment to help :). $\endgroup$ – 1ENİGMA1 Oct 25 '18 at 13:19
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    $\begingroup$ @1ENİGMA1 - Thanks a lot! I apologize if my comment was rude. This post of Qing Liu shows that, more generally, rings of dimension $\le0$ have this property. $\endgroup$ – Pierre-Yves Gaillard Oct 25 '18 at 13:35

Yes. Example: Let B = K[[X]], with K a field. Let M = B/(X), the residue module. Let A = B \oplus M, with product natural on B, action of B on M and a^2 = 0, all a\in M. Then A is clearly noetherian, Dim A = 1 and hence not Artinian. The non units of A are N = (X) \oplus M. Any element in N times any element in M is 0

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This is a community wiki answer whose purpose is to rephrase R. C. Cowsik's beautiful answer with a slightly different notation.

Here is an example of a non-artinian noetherian ring $A$ whose non-units are zero-divisors.

Let $K$ be a field, $X$ an indeterminate, and $A$ the additive group $K[[X]]\times K$. One checks that the formula $$ (f,\lambda)(g,\mu)=(fg,\lambda g(0)+\mu f(0)) $$ defines on $A$ a structure of commutative ring with one. [See this answer of user26857 for a generalization of this construction.]

One verifies that $A$ is noetherian, that $A$ has exactly two prime ideals, namely $\mathfrak p:=(0)\times K$ and $\mathfrak m:=(X)\times K$, that they satisfy $\mathfrak p\subset\mathfrak m$, and that we have $(f,\lambda)(0,1)=(0,0)$ for all $(f,\lambda)\in\mathfrak m$.

This shows that $A$ is indeed a non-artinian noetherian ring whose non-units are zero-divisors.


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