I am taking a course on commutative algebra and the instructor used this result without any proof or explanation and he is not very keen to take questions of students.
So, I am asking it here.
How can I prove that maximal ideals of $\mathbb{Z_9}$ are of the form $p\mathbb{Z} / 9 \mathbb{Z}$ where $p\mid 9$?
Definition of maximal ideal and ohter details are clear to me as I have done a course on ring theory but I am still unable to prove it.
So, guidence will be appreciated for a rigorous proof.
Let's try and make the question a bit more general.
Given a proper ideal $I$ in the ring $R$, there is a bijection between the set of ideals in the ring $R/I$ and the set of ideals in $R$ that contain $I$. This bijection preserves the order. It's a consequence of the homomorphism theorems and you should be able to spell out what the bijection is.
Thus, finding the maximal ideals in $R/I$ is the same as finding the maximal ideals in $R$ that contain $I$.
In the case of $\mathbb{Z}/n\mathbb{Z}$, where $n>1$, we know that the maximal ideals in $\mathbb{Z}$ have the form $p\mathbb{Z}$, for $p$ a prime. Moreover $p\mathbb{Z}\supseteq n\mathbb{Z}$ if and only if $p\mid n$.
Therefore the maximal ideals in $\mathbb{Z}/n\mathbb{Z}$ are exactly those of the form $p\mathbb{Z}/n\mathbb{Z}$ for $p$ a prime dividing $n$. If $n=9$ there's only one of them, namely $3\mathbb{Z}/9\mathbb{Z}$.