A prime ideal in a polynomial ring whose intersection with the base ring is zero

Question

It seems very straightforward but I can't prove or disprove this statement :

Let $R $ be an integral domain and $P $ be a prime ideal of $R [x] $ such that $P \cap R $ is zero. Then $P$ must be principal? Or is there any counterexamples?

What about $R $ is UFD?

Approach : we can localize $R [x] $ using $R^{\times} $, then $P $ coincides with the prime ideal in this localized ring, that is, $K[x] $ where $K $ is a field of fraction of $R $. So in this ring, corresponding prime ideal is principal, but it seems not give a nice information.

Answer 1

An answer to your question is given in Sharma, P.K., A note on Ideals in Polynomial Rings, Arch. Math. 37 (1981), 325-329.

The main theorem of the article is

Theorem 1: Let $R[x]$ be a polynomial ring over an integral domain $R$ and let $P\neq 0$ be a prime ideal in $R[x]$ such that $P\cap R=0$. Let $$\varphi(X)=a_{0}X^{d}+a_{1}X^{d-1}+\ldots+a_{d-1}X+a_{d}$$ be a polynomial of least positive degree in $P$. Then $P=(\varphi(X))$ if and only if there does not exist $t\notin (a_{0})$ such that $ta_{i}\in (a_{0})$ for $1\leq i\leq d$.

One of the corollaries of the article is

Corollary 1: Let $R$ be a unique factorization domain or a valuation ring and $P\neq 0$ a prime ideal in $R[x]$ with $P\cap R=0$. Then there exists an irreducible polynomial $\varphi(X)$ with least positive degree among non-zero elements of $P$ such that $P=(\varphi(X))$.

To fully answer your question, define $\operatorname{ct}(f(x))$ to be the ideal generated by the coefficients of $f(x)$ in $R$. Then we have,

Corollary: Let $R$ be an integral domain and $P\neq 0$ a prime ideal in $R[x]$ with $P\cap R=0$. If there exists a polynomial $\varphi(X)\in P$ such that $\varphi(X)$ is of least positive degree in $P$ and $\operatorname{ct}(\varphi(X))=R$ then $P=(\varphi(X))$.

Proof: Write $\varphi(X)=a_{0}X^{d}+\ldots+a_{d-1}X+a_{d}$. Let $t\in R$ be given such that $ta_{i}\in (a_{0})$ for $1\leq i\leq d$. Since $\operatorname{ct}(\varphi(X))=R$ there exists $b_{i}\in R$ such that $$\sum\limits_{0\leq i\leq d}a_{i}b_{i}=1.$$ Write $ta_{i}=a_{0}r_{i}$ for each $1\leq i\leq d$. Then $$t=t\left(\sum\limits_{0\leq i\leq d}a_{i}b_{i}\right)=a_{0}\left(\sum\limits_{1\leq i\leq d}r_{i}b_{i}+tb_{0}\right)\in(a_{0}).$$ Thus $P=(\varphi(X))$ by the theorem.$$\tag*{$\blacksquare$}$$