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In the Wikipedia irreducible ring page (revision at the time when I posted this question), a (meet-)irreducible ring is defined as a ring in which the zero ideal is irreducible. However, there is no reference provided for this definition, and I have been unable to find the same definition or any references to (meet-)irreducible rings in mathematical papers or other sources. I could only find a definition for meet-irreducible modules, which is similar to the above definition (Irreducible Module - Wolfram Math World).

And also it's said that the following three conditions for $A$ are equivalent:

  1. $A$ is meet-irreducible.
  2. $A$ has only one minimal prime ideal.
  3. $\mathrm{Spec} A$ is irreducible.

I'm sure that (2) and (3) are equivalent. But I guess that (1) and (2) are not equivalent. The implication (1) ⇒ (2) seems to be correct when $A$ is noetherian but not general, and I have no idea about the the implication (3) ⇒ (2).

I suspect that the person who edited the Wikipedia page may have confused between meet-irreducibility and irreducibility of spectra. Can someone confirm if my suspicion is correct? Or it's possible to prove the equivalence between (1) and (2)? And Additionally, I am curious if there is any existing definition of meet-irreducible rings or any references to such rings outside of Wikipedia.

postscript

I had a counter-example of the implication (2) ⇒ (1) from a reply to my tweet which I'd posted while I was trying to prove (2) ⇔ (1). $R = k[x,y]/(x^2, xy)$ has only one minimal prime ideal $\sqrt {(0)} = (x)$, while $(x) \cap (y) = (0).$ And they say a ideal which has only one isolated prime and any embedded prime generates a counter-example.

I also had a counter-example and of the implication (2) ⇒ (1) and a reference to "meet-irreducible ring" in a comment below. So I now want counter-example or proof of the implication (1) ⇒ (2).

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    $\begingroup$ These guys, remark "... meet irreducible (i.e. any finite number of non zero ideals have a non zero intersection) ring R ..." $\endgroup$
    – NDB
    Commented Aug 2, 2023 at 0:29
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    $\begingroup$ Your suspicion is correct: any Artinian local ring has irreducible spectrum (consisting of a single point), but in such a ring the zero ideal is irreducible if and only if the ring is Gorenstein. So take e.g. $k[x,y]/(x,y)^2$, which has only one prime $(x,y)$, and $(x) \cap (y) = 0$ $\endgroup$
    – math54321
    Commented Aug 2, 2023 at 17:27
  • $\begingroup$ @math54321 thank you for your counter-example. Is there also any counter-example to the implication (1) ⇒ (2)? $\endgroup$
    – linuxmetel
    Commented Aug 2, 2023 at 17:38
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    $\begingroup$ Someone from this discussion should edit the wikipedia page. $\endgroup$ Commented Aug 2, 2023 at 17:42
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    $\begingroup$ (1) $\Rightarrow$ (2) is trickier: it certainly holds if the ring is Noetherian. But there are counterexamples in the non-Noetherian case: the ring in this MO answer is one $\endgroup$
    – math54321
    Commented Aug 2, 2023 at 18:41

1 Answer 1

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In general, the zero ideal being irreducible neither implies nor is implied by the spectrum being irreducible. Every Artinian local ring has irreducible spectrum (consisting of a single point), but in such a ring the zero ideal is irreducible if and only if the ring is Gorenstein. Thus e.g. $k[x,y]/(x,y)^2$ has a single prime $(x, y)$, although $(x) \cap (y) = 0$. (This is the simplest non-Gorenstein ring, having a 2-dimensional socle with basis $x, y$.)

For Noetherian rings, the zero ideal being irreducible does imply the spectrum is irreducible: if $R$ is Noetherian with distinct minimal primes $\mathfrak{p} \ne \mathfrak{q}$, then there exist $x, y \in R$ with $\mathfrak{p} = \operatorname{ann} x$, $\mathfrak{q} = \operatorname{ann} y$ (since minimal primes are associated primes in a Noetherian ring). Then $Rx \cong R/\mathfrak{p}, Ry \cong R/\mathfrak{q}$, and $\operatorname{Ass}(Rx \cap Ry) \subseteq \operatorname{Ass}(R/\mathfrak{p}) \cap \operatorname{Ass}(R/\mathfrak{q}) = \{\mathfrak{p}\} \cap \{\mathfrak{q}\} = \emptyset \implies (x) \cap (y) = 0 \implies 0$ is not irreducible.

This does not extend to the non-Noetherian case, see here for an explicit counterexample. (The ring there has 2 minimal primes $(x) \ltimes E(k)$ and $(y) \ltimes E(k)$, although the zero ideal is irreducible.)

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