Let $(R, \mathfrak m)$ be a Noetherian local ring such that $\mathfrak m \notin \text{Ass}(R)$ and $\text{Spec}(R)=\text{Ass}(R)\cup \{\mathfrak m \}$. Then, must $R$ be Cohen-Macaulay?
Of course the hypothesis implies $R$ has positive depth, but I think the ring may not necessarily be Cohen-Macaulay. Unfortunately, I cannot come up with a counterexample. Please help.
Yes, $R$ must be Cohen-Macaulay. In fact, $R$ is a local ring of dimension 1 and $\operatorname{Ass} R = \operatorname{Min} R$.
If the dimension of $R$ is greater than 1, the hypotheses cannot be satisfied because there will be infinitely many height 1 primes by prime avoidance. In particular, $\mathfrak{m}$ is not the union of primes of smaller height, so $\mathfrak{m}\setminus \cup_j \mathfrak{p}_j$ is nonempty for any finite set of nonmaximal primes $\{\mathfrak{p}_1,\cdots,\mathfrak{p}_t\}$, so any nonzero element in $\mathfrak{m}\setminus \cup_j \mathfrak{p}_j$ will have a height one prime containing it that cannot be one of the $\mathfrak{p}_j$. Since the set of associated primes of a Noetherian ring is finite, you cannot have $\operatorname{Spec} R$ is the disjoint union of a finite set of primes and $\mathfrak{m}$.
So, your ring $R$ must be of dimension $\le 1$. If $\dim R = 0$, $\mathfrak{m}$ is associated, so this cannot be the case either. Thus, $\dim R = 1$ or there are no such $R$. However, any reduced local ring of dimension 1 will satisfy your hypotheses; reduced implies $\cup_n (0:\mathfrak{m}^n) \subset \sqrt{0} = 0$, so $1 \le \operatorname{depth} R \le \dim R = 1$, so $R$ is Cohen-Macaulay, and thus $\mathfrak{m}$ contains a nonzerodivisor, which is equivalent to $\mathfrak{m}$ not being an associated prime. So $$\operatorname{Spec} R = \{\mathfrak{m}\} \cup \operatorname{Ass} R = \{\mathfrak{m}\} \cup \operatorname{Min} R.$$ Further, if the conditions are satisfied, as you note $\operatorname{depth} R>0$ but we always have $\operatorname{depth} R \le \dim R = 1$, so $\operatorname{depth} R = 1$ and $R$ is Cohen-Macaulay.
