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

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.]

• – lhf
Oct 25, 2018 at 13:07
• @lhf - Thanks! I looked carefully at this great thread, but didn't find an answer to my question. What am I missing? Oct 25, 2018 at 13:11
• If $R$ is Artinian then every non-zero divisor in $R$ is a unit. Oct 25, 2018 at 13:12
• @Pierre-YvesGaillard, I just wrote a comment to help :). Oct 25, 2018 at 13:19
• @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. Oct 25, 2018 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 $$0$$.
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.