TITLE: If $I$ is a maximal ideal in $R$, $(I,x)$ is a maximal ideal in $R[x]$
QUESTION [5 upvotes]: Click Link to Original Text
Let $R$ be a commutative ring with $1$, and $I$ is an ideal of $R$.  Then, $(I) = I[x]$ is an ideal in $R[x]$.  I was able to prove, via first isomorphism, that $\frac{R[x]}{(I)}$ is isomomorphic to $\frac{R}{I}[x]$.  It follows that if $I$ is a prime ideal, then $(I)$ is a prime ideal in $R[x]$.
Then the author noted that, if $I$ is maximal, it does not mean that $(I) = I[x]$ is maximal.  If my understanding is correct, it is because while $R/I$ is a field, $\frac{R}{I}[x]$ may still only be an integral domain.  Finally, the author added that $(I, x)$ is a maximal ideal in $R[x]$.  What does the notation $(I, x)$ stand for?

REPLY [6 votes]: The (sloppy) notation $(I,x)$ should mean the ideal generated by $I$ and $x$, that is
$$
(I,x)=IR[x]+xR[x]
$$
The quotient
$$
R[x]/(I,x)
$$
is isomorphic to $R/I$. Just consider the homomorphism
$$
R[x]\to R/I
$$
sending $r\in R$ to $r+I$ and $x$ to $0+I$.
Your intuition is correct: if $(I)=IR[x]=I[x]$, the quotient $R[x]/I[x]$ is isomorphic to $(R/I)[x]$ which is never a field.