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.



Finite number of minimal ideals
(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](http://citeseerx.ist.psu.edu/viewdoc/download;jsessionid=0331DDC9E848DF83CBCB901AEAE62A4E?doi=10.1.1.53.7033&rep=rep1&type=pdf). 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](https://mathoverflow.net/questions/22228/what-is-serres-condition-s-n-for-sheaves) $(S\_1)$ 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. 


