Extension of a maximal ideal

Question

Let $R$ be a commutative ring, with the natural ring homomorphism $f:R \to R\left[ x \right]$, let $M$ be a maximal ideal of $R$.

For each ideal $I$ of $R$, the ideal $f(I)R[x]$ of $R[x]$ generated by $f(I)$ is called the extension of $I$ to $R[x]$, denoted by ${I^e}$.

Is ${M^e}$ a maximal ideal of $R\left[ x \right]$?

Answer 1

No , because $$R[X]/M^{\text e}=R[X]/MR[X]\simeq(R/M)[X],$$ which is a P.I.D., not a field. All you can say is that $M^{\text e}$ is a prime ideal.

A maximal ideal containing $M^{\text e}$ would be, for instance, $(M^{\text e}, X-r)$, where $r\in R$.