## Finite number of minimal ideals

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What is the necessary condition on a ring that guarantees the number of minimal non-zero ideals to be finite? Neither Noetherian or Artinian condition seems sufficient, and the ring being semisimple seems too strong.

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Noetherian is sufficient (although obviously not necessary). Just take a primary decomposition of the ideal (0). – JT May 27 2010 at 15:08

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I believe kwan is asking for finitely many minimal ideals, not minimal prime ideals. If so, then every ring has a unique minimal ideal: $(0)$. If instead you're asking for minimal non-zero ideals, it's much harder. Can you give us an idea of why you're asking? – Graham Leuschke May 27 2010 at 15:22
Thanks for point that out. Yes, I meant non-zero ideals, not necessarily prime. I was reading about the annihilator of a maximal ideal in a ring being a direct sum of non-zero minimal ideals (socle), and wondered if it's common for a ring to have infinitely many non-zero minimal ideals. – ashpool May 27 2010 at 20:31

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Here are a couple things to notice. (1) if $R$ is noetherian local and reduced, then it has no minimal non-zero ideals, since the square of any ideal is a strictly smaller one. So that's an example of what you're asking about, but not a very satisfying one. (I needed $R$ to be local to rule out things like $I=I^2$.) (2) Let $R = k[x,y]/(x^2,xy,y^2)$. Then the non-maximal non-zero ideals of $R$ are all of the form $(\alpha x - \beta y)$ for field elements $\alpha, \beta$. Thus $R$ has your property if and only if $k$ is a finite field, which again is not a very satisfying answer. – Graham Leuschke May 27 2010 at 21:33

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@kwan, for your recent question: If $\text{dim}_kI$ is at least $2$, then the set of lines will not be finite unless $k$ is finite. By the way, I just notice you have not accepted any of my answers and many others, for example the one by BCrnd on flatness. If you think the answers are correct, please accept them. – Hailong Dao Aug 16 2010 at 2:29
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## 3 Answers

(Inspired by Graham's comment) For simplicity I will consider the case $(R,m)$ is a Noetherian, local ring.

A non-zero minimal ideal better be principal. Also, if $(x)$ is such ideal, then for any $y\in m$, the ideal $(xy)$ has to be $0$, so $mx=0$.

Let $I= \{x\in R|mx=0\}$, the socle of $R$. Since $Im=0$, $I$ is a vector space over $k=R/m$. You want to know when the set of $1$-dimensional subspaces of $I$ is finite. This happens if and only if $\dim_kI\leq 1$ or $k$ is a finite field.

If $R$ is also Artinian (i.e. $\dim R=0$), then $\dim_kI\leq 1$ means precisely that $R$ is Gorenstein. So in this case (Noetherian, local of dimension $0$) one has a particularly nice answer: the set of nonzero minimal ideals is finite iff $R$ is Gorenstein or $k$ is finite.

Note that if $\dim R\geq 1$, then $I=0$ iff $\text{depth}\ R\geq 1$.

In general one can localize to get at least necessary conditions. For example, if the height of all maximal ideals is at least $1$, then $R$ satisfying Serre's condition $(S_1)$ is certainly sufficient, since the socle when you localize at any maximal ideal will then be $0$, so there is no non-zero minimal ideals.

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I'd like to be credited with an assist on this answer :) . – Graham Leuschke Jun 3 2010 at 12:38
@Graham: Done, thanks!  – Hailong Dao Jun 4 2010 at 19:55
Thanks for your answer. How can I see that $2\leq \mbox{dim }_{k}I<\infty$ is impossible? – ashpool Aug 15 2010 at 21:12

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Please clarify: are you asking about commutative rings or noncommutative rings? (The tag "ac.commutative-algebra" suggests the former; your reference to semisimplicity suggests the latter.) If the question is about noncommutative rings, then presumably by "ideals" you mean two-sided ideals.

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Sorry, I should have mentioned it was commutative ring. – ashpool Jun 3 2010 at 11:37

I'm not sure how to answer your exact question ("the necessary condition that guarantees..."?), but here are a few minor observations.  A direct product of commutative rings has only finitely many minimal (by definition nonzero) ideals if and only if each component ring has only finitely many and all but finitely many of the component rings have none.  So it suffices to consider indecomposable commutative rings with only finitely many ideals.  There are all sorts of indecomposable commutative rings with no minimal ideals.  Now suppose the indecomposable commutative ring has a positive finite number of minimal ideals.  The socle of such a ring has square zero; thus, the socle is a nonunital subring with the structure of an additive abelian group with zero multiplication on it (of course, this additive abelian group need not have only finitely many minimal subgroups).  One can construct all sorts of examples of this sort.  For instance, let $A$ be an indecomposable commutative ring with no minimal ideals, and let $M$ be an $A$-module with only finitely many simple submodules.  (For example, one might take $M$ to be a uniserial $A$-module.)  Let $R = A \oplus M$ as an additive group, with multiplication given by $(a_1, m_1) (a_2, m_2) = (a_1 a_2, a_1 m_2 + a_2 m_1)$.  Then $R$ is an indecomposable commutative ring with only finitely many minimal ideals.

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