On prime ideal and irreducible ideal in R[X]. For a noetherian domain $R$, an irreducible ideal $I$ implies $\sqrt{I}$ is a prime ideal. Irreducible implies primary, but not always vice versa. That said, I would like to ask whether the following ideals exist or not for some $R$$\colon$
Question. $P$ (resp. $Q$ and $I$) is a prime (resp. a primary and an irreducible ideal) of $R[X]$ such that

*

*$P$ (resp. $Q$ and $I$) is $not$ monomial. (That is, at least two-generators are necessary.)


*$P \cap R = 0$ (resp. $Q \cap R = 0$, $I \cap R = 0$).


*$R[X]/P$ (resp. $R[X]/Q$ and $R[X]/I$) is finite over $R$.


*By the scalar extension $K[X]$, $P$ (resp. $Q$ and $I$) remains a prime ideal (resp. a primary and an irreducible ideal).
That is, the scheme preserve the same property for its generic fibre.
 A: Let $F$ be a field and let
$$R=F[Y^2,Y^3]=\left\{a_0+\sum_{j=2}^n a_jY^j\mid a_j\in F\right\}.$$
The field of fractions of $R$ is the field $K\cong F(Y)$. Now, let $P=(X-Y)K[X]\cap R[X]$ which is a prime ideal in $R[X]$ since $(X-Y)K[X]$ is a prime ideal in $K[X]$.
Note that $X^2-Y^2=(X-Y)(X+Y)\in P$ but neither $X-Y$ nor $X+Y$ is in $R[X]$ since $Y\notin R$. Assume that $X^2-Y^2=(aX+b)(cX+d)$ for some $a,b,c,d\in R$ with $cX+d\in P$. Then $ac=1$, which yields $a(cX+d)=X+ad\in P$. This implies $X+ad$ is a multiple of $X-Y$ in $K[X]$, and therefore $-ad=Y$. But this is a contradiction since $a,d\in R$ whereas $Y\notin R$. Thus, $P$ cannot be a principal ideal since $P$ contains degree one polynomials (such as $Y^2X-Y^3$), but no degree one polynomial generates $X^2-Y^2$. In fact, with a similar argument, one can show that $X^2-Y^2$ is not any linear sum of degree one polynomials in $P$.
Clearly $P\cap R=0$. Furthermore, because $X^2-Y^2\in P$, $R[X]/P$ is generated by $1$ and $X$ as an $R$-module. Finally, because $Y^2X-Y^3\in P$, we see that the extension of $P$ to $K[X]$ is $(X-Y)K[X]$. Thus, $P$ is a prime ideal which satisfies all the conditions you listed. Since prime ideals are also irreducible and primary, this serves as an irreducible and primary example as well (although, as I showed in a comment, there are simpler examples for primary ideals).
Interestingly, you won't find any examples when $R$ is a UFD. Indeed, suppose $R$ is a UFD and $K$ is its field of fractions, and let $P$ be a prime ideal of $R[X]$ such that $P\cap R=0$. Let $P^e$ denote the extension of $P$ to $K[X]$. That is,
$$P^e=\left\{\sum_{j=1}^nf_jg_j\mid f_j\in K[X],\ g_j\in P\right\}.$$
Since $K[X]$ is a PID, there exists $f\in P^e$ such that $P^e=K[X]f$. By taking out the denominations and any common divisors in the coefficients, we may assume without loss of generality that $f$ is a primitive polynomial in $R[X]$ (i.e. $c(f)=1$ where $c(f)$ denotes the greatest common divisor of the coefficients of $f$, often called the $\textit{content}$ of $f$).
I now claim that $P=R[X]f$. Since $f\in P^e$, we know
$$f=\sum_{j=1}^nf_jg_j$$
for some $f_j\in K[X]$ and $g_j\in P$. Let $r\in R\setminus\{0\}$ such that $rf_j\in R[X]$ for all $j$. Then we see
$$rf=\sum_{j=1}^nrf_jg_j\in P$$
which implies that either $r\in P$ or $f\in P$. But $P\cap R=0$, so we conclude $f\in P$ and therefore $P\supseteq R[X]f$. For the other containment, let $g\in P$. Then $g\in P^e=K[X]f$, so there exists $h\in K[X]$ such that $g=hf$. Let $a,b\in R$ such that $ah\in R[X]$, $c(ah)=b$, and $\gcd(a,b)=1$. Then by Gauss's Lemma,
$$c(ag)=c(ahf)=c(ah)c(f)=b.$$
Since $g\in R[X]$, $b=c(ag)=ac(g)$ which implies that $a$ is a unit, and thus $h=a^{-1}ah\in R[X]$, so $g\in R[X]f$ as desired. From this, we conclude that $P$ is a principal ideal. Thus, if $R$ is a UFD, then there does not exist a prime ideal $P$ of $R[X]$ such that $P$ is not principal and $P\cap R=0$.
