Let $A=\mathbb{Z}[X,Y]/(Y^2-6X^2), B=\mathbb{Z}[X,T]/(T^2-6)$ where $X,Y,T $ are variables and let $x,y$ be the cosets of $X,Y$ in $A$ whilst $x',t$ be the cosets of $X,T$ in $B$.
Consider the ideals $P_1=xA+yA+5A$, $P_2=(x-y)A+5A$, $Q_1=x'B+(t+1)B$.
Show that $P_1, P_2, Q_1$ are prime ideals.
$P_1$ can be done by considering the finitely many elements of $A/P_1$ and checking it is a prime ideal. As for $P_2$ and $Q_1$ it would appear to be "intuitively obvious " that any element in $P_2, Q_1$ wouldn't factorize into ones which aren't, but how do I actually show this?
Compute the quotient rings:
$A / P_1 \cong {\mathbb Z}[X,Y]/(Y^2 - 6X^2, X, Y, 5) = {\mathbb Z}[X,Y]/(X, Y, 5) \cong {\mathbb Z}_5$.
$A / P_2 \cong {\mathbb Z}[X,Y]/(Y^2 - 6X^2, X - Y, 5) \cong {\mathbb Z}[X]/(X^2 - 6X^2, 5) = {\mathbb Z}[X]/(5) \cong {\mathbb Z}_5[X]$.
$B / Q_1 \cong {\mathbb Z}[X, T]/(T^2 - 6, X, T+1) \cong {\mathbb Z}[T]/(-5, T+1) \cong {\mathbb Z}_5$.
In all three cases, the quotient is an integral domain, so $P_1$, $P_2$, and $Q_1$ are prime ideals.