Let $R$ be a commutative Noetherian, local, reduced (no non-zero nilpotent) ring; is it true that $R$ has only one minimal prime ideal?
If $R$ is Noetherian and local, then I can show that $(0)$ is the only minimal ideal of $R$ in the sense that $(0)$ is the only ideal minimal among the set of all proper ideals of $R$ , but I am unable to say anything about minimal prime ideals.
Consider $(k[x,y]/(xy))_{(x,y)}$ for a field $k$, this has two minimal prime ideals: $(x)$ and $(y)$.
This example can be motivated by algebraic geometry: Suppose for simplicity that $k$ is algebraically closed. Then $\operatorname{Spec}(k[x,y]/(xy))$ corresponds to the union of the two lines $\{(x,0) \mid x \in k\}$ and $\{(0,y) \mid y \in k\}$ in $\Bbb A^2(k)$. Note that these lines intersect at the point $(0,0)$. By the Nullstellensatz, this point corresponds to the ideal $(x,y)$. This closed point is contained in two irreducible components of $\operatorname{Spec}(k[x,y]/(xy))$, that's why we get two minimal prime ideals in the localization.
This construction, along with the geometric motivation can be generalized to $n$ variables in the obvious way, giving an example of a commutative noetherian local reduced ring with exactly $n$ minimal prime ideals for any $n$. (Of course, there can be only finitely many if the ring is Noetherian.)
Edit: it's probably worth pointing out that a commutative reduced ring $R$ has only one minimal prime ideal iff it is a domain. If $R$ is a domain, then $(0)$ is the unique minimal prime ideal. If $R$ has a unique minimal prime ideal, then this minimal prime ideal is equal to the intersection of all prime ideals, i.e. the nilradical, which is $(0)$ by assumption, thus $(0)$ is a prime ideal and it follows that $R$ is a domain.
(In geometric language, a scheme is integral iff it is reduced and irreducible. As irreducible components correspond to minimal primes for affine schemes, $\operatorname{Spec}(R)$ is irreducible iff $R$ has a unique minimal prime ideal.)