Let $R=\mathbb{Z}[X,Y]$ and $S=(YX^2-Y)$.
Question:
a) Show that the ideal $I$ generated by $Y-4$ in $R/S$ is not prime.
b) Find complete list of prime ideals of $R/S$ containing $I$.
c) Which one of the ideals found in (b) are maximal?
Answer:
For part (a), I showed that as $(Y-4)(X^2-1)=YX^2-Y-4X^2+4=4-4X^2$ in $R/S$, $Y-4$ divides $4(1-X)(1+X)$ but it cannot divide any of those factors. Thus, $Y-4$ is not prime, whence $I$ is not a prime ideal.
For part (b) and (c), prime ideals of $R/S$ containing $I$ are in one to one correspondence with the prime ideals of $R$ containing $YX^2-Y$ and $Y-4$. Let $P$ be such a prime ideal of $R$. Then by $P$ being prime, it has to contain at least one of the factors of $YX^2-Y$. Thus, we have the following options. $P$ contains either the ideal $$(Y-4,X-1),\quad (Y-4,X+1),\quad (Y,Y-4), \quad (Y-4,Y,X+1), \quad (Y-4,Y,X-1), \quad (Y-4,X+1,X-1), \quad \text{or} \quad (Y-4,Y,X-1,X+1).$$
Now I know that if $P$ contains $(X,Y)$, then $P=R$ but for the other options I don't know what to do. Also, if you spot any mistake on my answer up to this point, please let me know. Thanks in advance...