Ideal generated by maximal ideal and some other element of integral domain. Let $R$ be an integral domain, $M$ a maximal ideal of $R[x]$, $P$ be a prime ideal of $R$, and $M \cap R=(0)$. Let $0\neq p\in P $. Then prove that $(M,p)=R[x]$.
If $R$ is just a field it is done as each of its nonzero elements is a unit so is $p$. But how to show this as a whole for supposing $R$ to be an integral domain as all integral domains are not fields.
Thank you.

Since $0 \neq p \in R$ and $M \cap R = (0)$, it follows that $p$  is not in $M$. Thus, the ideal $(p, M)$ is strictly larger than $M$. By maximality of $M$, it follows that $(p, M) = R[x]$.
As pointed out in the comments, $P$ was superfluous. Its only role was to tell that us $p$ is an element of $R$.
Moreover, we didn't even require $R$ to be an integral domain.
