This question claimed (and proved) that if $p$ is a prime ideal of $A=k[x_1,\ldots,x_n]$ with $\operatorname{ht}(p) \in \{0,1,n-1,n\}$, then $A/p$ is Cohen-Macaulay.
Now, let $A$ be a (Noetherian) UFD of Krull dimension $n$ which is Cohen-Macaulay, and let $p$ be a prime ideal of $A$ with $\operatorname{ht}(p) \in \{0,1,n-1,n \}$.
Question 1: Is it true that $A/p$ is Cohen-Macaulay?
My answer: Yes, it is true that $A/p$ is CM, and the same proof holds.
More elaborately:
If $\operatorname{ht}(p)=0$ then $p=(0)$, so trivially $A/p=A/0=A$ is CM.
If $\operatorname{ht}(p)=1$, then $p$ is principal, since a height one prime in a UFD is principal, see wikipedia(11). Then by the nice proposition in this answer, $A/p$ is CM.
If $\operatorname{ht}(p)=n−1$, then $\dim A/p=1$, and one-dimensional integral domains are CM, see wikipedia or MSE.
If $\operatorname{ht}(p)=n$, then $\dim A/p=0$, hence $A/p$ is a field, and fields are CM.
Let us concentrate on $A=k[x_1,\ldots,x_n]$ and $p$ a prime ideal of $A=k[x_1,\ldots,x_n]$ with $\operatorname{ht}(p)=1$.
Now again $p$ is principal, but $A/p$ is not necessarily regular. Indeed, $(x^2-y^3)$ is a height one prime of $k[x,y]$ and $k[x,y]/(x^2-y^3)=k[z^2,z^3]$ is CM but not regular (since regular implies normal, but $k[z^2,z^3]$ is not normal, namely, not integrally closed in its field of fractions $k(z)$).
Question 2: What happens if $\operatorname{ht}(p)=n-1$? In that case we have a one-dimensional integral domain $A/p$; is it necessarily regular?.
Question 3: More generally: Could one guarantee regularity of a quotient of $A=k[x_1,\ldots,x_n]$ by a prime ideal $p$ of height different then $\{0,n\}$? Maybe adding assumptions would help?
Thank you very much!