Intersection of a domain $R$ with an ideal $\mathfrak a$ of height $2$ in $R[X]$

Question

Let $R$ be an integral domain, and consider the polynomial ring $R[X]$ over $R$. Suppose that ${\mathfrak a}$ is an ideal of $R[X]$ such that $\operatorname{ht}(\mathfrak a) = 2$.

Q. Is the intersection $R \cap \mathfrak a \neq 0$? That is, does the intersection $R \cap \mathfrak a$ contain an element other than $0$?

Either case of $R$ being noetherian or nonnoetherian should be answered in the affirmative. Still, I am not sure whether it is true or not when ${\mathfrak a}$ is not a prime.

Answer 1

Let R be Noetherian ring. if $R \cap \mathfrak{a}=0$,then $R \subset R[X]/\mathfrak{a}$ and hence $R^{\times} = R[X]^{\times} \subset (R[X]/\mathfrak{a})^{\times} $. Therefore $\dim R \leq \dim R[X]/\mathfrak{a} $ . But this is contradiction since $\ht \mathfrak{a}=2$ and $\dim R[X]=\dim R +1$.

Answer 2

Let $K$ be the fraction field of $R$. The technique of working over $K$ gives the affirmative answer regardless of whether $\mathfrak a$ is prime. My understanding is that the height of a prime ideal $P$ in a not necessarily noetherian ring is defined to be the length of a maximal chain of primes between $(0)$ and $P$. The height of a non-prime ideal is the minimal height of its minimal primes.

Assume that $\mathrm{ht}(\mathfrak a)=2$ and $\mathfrak a\cap R=(0)$, aiming for contradiction. Take a non-zero element $f\in\mathfrak a$. Let $f=f_1^{\alpha_1}\cdots f_r^{\alpha_r}$ be the factorization of $f$ into irreducible factors in $K[X]$. Then $P_i:=f_iK[X]\cap R[X]$ is a prime ideal of height 1. Hence $\mathfrak a\not\subset P_i$ for any $i\in\{1,\dots,r\}$. Take an element $g\in\mathfrak a\setminus\bigcup\limits_{i=1}^rP_i$. Then $g$ is relatively prime to $f$ in $K[X]$. Take $a,b\in K[X]$ such that $af+bg=1$. Multiplying $af+bg$ by all the denominators of all the coefficients of $a$ and $b$ yields a non-zero element of $\mathfrak a\cap R$. Contradiction.