Original Question: Let $R$ be a commutative ring with identity and $I$ maximal ideal in $R$. Show that $I[x]$ is a prime ideal in $R[x]$ and is not maximal ideal in $R[x]$, find two distinct maximal ideals that contain $I[x]$.
The fact that $R$ is a commutative ring with identity gives us: $R[x]$ is commutative and that $I$ is also prime. I know that $I[x]$ is generated by $I$ generator and by $x$ but have trouble making anything useful with this information
The question is can we conclude something about $I[x]$ from $I$?
-In general and in this specific case.
And it would be nice to get a hint or general idea on how can we prove the existence of two different
maximal ideals over $R[x]$ just from the fact that $I[x]$
prime but not maximal?
Then it would be easy to show that they contain $I[x]$ since they are maximal unlike $I[x]$ and contain all other ideals.
I know that $[]$ is generated by $$ generator and by $$
No. The ideal $I[x]$ does not contain $x$. It is the set of polynomials in the variable $x$ whose coefficients all lie in $I$.
You should show that $R[x]/I[x]\cong (R/I)[x]$. The latter is the polynomial ring over the field $R/I$ in the variable $x$. Since it is a polynomial ring over a field, it is an integral domain but not a field, equivalently $I[x]$ is prime but not maximal in $R[x]$.
You will be led to find two maximal ideals of $R[x]$ containing $I[x]$ if you can find two maximal ideals of $(R/I)[x]$. I will give a hint for one of them: since the polynomial $x-1$ is irreducible over the field $R/I$, the ideal $\langle x-1\rangle$ is maximal in $(R/I)[x]$, hence the ideal $\langle I[x], x-1\rangle$ is maximal in $R[x]$.