# intersection of non zero prime ideals of polynomial ring R[x] over integral domain R is zero

Let R be an integral domain. Then how to show that intersection of non zero prime ideals of R[x] is zero.

Hint $\$ If $\,f\,$ is in every prime ideal $\ne 0\,$ then $\,1+xf\,$ is in no maximal ideal so is a unit, so $\,f = 0.$
If $p(x) \in R[x]$ is nonzero with leading coefficient $a$, then $p(x)^n\neq 0$ since the leading coefficient of $p(x)^n$ is $a^n$ and $a^n \neq$ as $R$ is an integral domain. Assume for the moment that the constant coefficient of $p(x)$ is nonzero, we have that $p(x) \not \in (x)$ and by a well known application of Zorn's lemma gives a prime ideal extending $(x)$ not containing $p(x)$. We can of course factor $x$ out of $p(x)$ to reduce to the last case, this leaves only the possibility that $p(x)=ax^k$ but then $p(x)\not \in (x-1)$, and the same argument proves that there is a prime ideal not containing $p(x)$
• Easier: any prime ideal containing the nonunit $\,1+x p(x),\,$ does not contain $\,p(x)\ne 0,\,$ see my answer. – Bill Dubuque Jun 12 '14 at 15:47