Finite number of minimal ideals - MathOverflow

Finite number of minimal ideals

2010-05-27T14:40:58Z

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.

Answer by Greg Marks for Finite number of minimal ideals

2010-06-02T20:49:20Z

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.

Answer by Hailong Dao for Finite number of minimal ideals

2010-06-02T23:51:39Z

(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.

Answer by Greg Marks for Finite number of minimal ideals

2010-06-29T00:01:10Z

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.