To simplify my question posted at 1 and later on at 2, I repost my question here by changing arbitrary polynomial by a quadratic one.
Let $\mathbb{Z}[X]$ be the ring of polynomials in one variable, $X^2+d\in \mathbb{Z}[X]$ be an irreducible polynomial, $A = \mathbb{Z}[X]/(X^2+d)$ and $x$ = $X$ (mod $X^2+d$). Can we classify all the primary ideals of $A$?
$(a)$. If a prime ideal $P$ of $A$ is principal then $P$-primary ideals are a prime power.
$(b)$. If $P=(p, x-a)$ where $p$ is a prime integer and $X-a$ is a factor of $X^2+d$ mod $p$ (e.g. see 3), then can we describe $P$-primary ideals in this case?
Through the back to back discussion in the comments section of 2, David Lampert helped me to catch up on primary ideals of $A$ when $d=4$. Indeed, in this case, the only non-principal prime ideal is $M=(2, x)$ and we located the all possible indispensable $M$-primary ideals, I mentioning in the following list.
$(1)$. $2^{n+1}A$
$(2)$. $(2^n\cdot x)A$
$(3)$. $2^nM$
$(4)$. $2^n(2+x)A$
$(5)$. $2^n(2-x)A$
$(6)$. $2^n(4, 2+x)$
where $n\geq 0$.
My try: Prime ideals of $A$ are either $pA$ or $(p, x-a)$ where $p$ is a prime integer and $x-a$ is a factor of $X^2+d$ mod $p$, i.e. $a^2\equiv -d$ mod $p$. In the first case primary ideals are prime power. Let $P=(p, x-a)$. For $p=2$ the prime ideal is either $P=(2, x)$ ($d$ is even) or $P=(2, x+1)$ ($d$ is odd).
In the first case $P^2=2P$. If $d$ is the power of $2$ then primary ideals are the same as listed above. If $d$ is not a power of $2$ then $P$-primary ideals are the following ones.
$(1)$. $2^{n+1}A$
$(2)$. $2^nP$
where $n\geq 0$.
