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.
[Math] Finite number of minimal ideals
ac.commutative-algebra
Related Solutions
Hochster answered this question in his thesis (by finding such conditions). See: Hochster, M. Prime ideal structure in commutative rings. Trans. Amer. Math. Soc. 142 1969 43–60.
Let $A$ be your ring and $X=\mathrm{Spec} A$. The minimal primes of $A$ correspond to the irreducible components of $X$. An element of $f\in A$ induces a function $\widehat f:X\to \coprod_{\mathfrak p\in \mathrm{Spec} A}\kappa(\mathfrak p)$ and this function vanishes everywhere if and only if $f\in\mathfrak p$ for all $\mathfrak p\in \mathrm{Spec} A$, that is, when it is nilpotent. For an $f\in A$ contained in a minimal prime the induced function $\widehat f$ vanishes on the irreducible component corresponding to the minimal prime containing $f$.
If $X$ has a single irreducible component, then the functions induced by the elements of $A$ in the corresponding single prime ideal are vanishing everywhere hence they are nilpotent (in particular zero-divisors).
If $X$ has more than one irreducible component, then do the following: Choose $f_1,\dots,f_m$ such that each $f_i$ is contained in exactly one minimal prime ideal and for each minimal prime there is (exactly) one $f_i$ contained in it. This choice ensures that the product of any proper subset of these functions will not vanish on at least one irreducible component and hence it is not nilpotent. (The point is, that if their product vanished on an irreducible component, then one of them would have to, but then it would be the one that vanishes on that component). Now take an arbitrary element $g$ in any of the minimal primes and for simplicity assume it is from the one corresponding to $f_1$. (We allow $g$ to be contained in other primes as well, but that has no consequence). Then the following claim implies that $g$ is a zero-divisor.
Claim Let $g,f_2,\dots,f_t$ be such that $g\cdot f_2\cdots f_t$ is nilpotent, but $f_2\cdots f_t$ is not nilpotent, then $g$ is a zero-divisor.
Proof Let $n$ be the smallest non-negative integer for which there exists a $t-1$-uple $(n_2,\dots,n_t)\in\mathbb N^{t-1}$ such that $g^n\cdot f_2^{n_2}\cdots f_t^{n_t}=0$. Observe that $n$ exists and $n\geq 1$ by the assumptions. Then $g\cdot (g^{n-1}\cdot f_2^{n_2}\cdots f_t^{n_t})=0$, but $g^{n-1}\cdot f_2^{n_2}\cdots f_t^{n_t}\neq 0$. $\square$
Comment Apparently this is essentially the same proof as the one Graham included in the comments, but I can't let go any chance of giving a geometric proof of an algebra question. Also, it clarifies the unclear step pointed out by Georges. In fact, it seems one needs to do a little yoga to get the result. In any case algebra=geometry. :)
Best Answer
(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.