Find all maximal ideals of $\mathbb{Z}_{540}$ Find all maximal ideals of $\mathbb{Z}_{540}$
By using the following statement, 
'$f:R \rightarrow S$ be a surjective ring homomorphism and let $K=ker(f)$.
Observe that there is a one-to-one correspondence between primes ideals of R containing K and primes ideals of S.'
we obtain that prime ideals of $\mathbb{Z}_{540}$ is of the form $P/(540)$ where $P$ is the prime ideal of $\mathbb{Z}$
My question is how do we know the prime ideals of $\mathbb{Z}_{540}$ is of that form? Or maybe someone can explain to me how to use the statement to obtain the prime ideals.
 A: The ideals of $\mathbb{Z}_n$ are, first of all, additive subgroups of $\mathbb{Z}_n$. These we know to all have the form $\langle d\rangle$, where $d$ divides $n$. But, as we know, the set $\langle d\rangle$ is the ideal generated by $d$. So we have just proven that
the ideals in $\mathbb{Z}_n$  are precisely the sets of the form $\langle d\rangle$ where $d$ divides $n$.
Since we are interested in maximal ideals, and this concept is defined in terms of containment
of ideals in one another, we now need to determine when we can have $\langle d_1\rangle\subset \langle d_2\rangle$. This is the case if and only if $d_1 \in <d_2>$.
Here is the main result that you are seeking for:An ideal $I$ in $\mathbb{Z}_n$ is maximal if and only if
$I = \langle p \rangle$ where p is a prime dividing n.
A: The statement you are using is the correspondence theorem for ideals in a quotient ring.  Recall that if $f\colon R \to S$ is surjective with kernel $K$ then $S$ is isomorphic to $R/K$ by the first isomorphism theorem.
Now the correspondence theorem says:
Theorem: There is a one to one correspondence between the ideals of $R$ that contain $K$ and the ideals of $R/K$, given by sending any ideal $I \subseteq R$ containing $K$ to the ideal $I/K$ of $R/K$.  Moreover, this correspondence sends prime ideals to prime ideals and is inclusion preserving (so it sends maximal ideals to maximal ideals).
So the standard quotient map $\mathbb Z \to \mathbb Z_{540}$ has kernel $K = (540)$.  To find the maximal ideals of $\mathbb Z_{540}$ we use this theorem, which says that they are in one to one correspondence with the maximal ideals of $\mathbb Z$ that contain $540$.
Note that in $\mathbb Z$ all ideals are of the form $(a)$ for some $a \in \mathbb Z$ and $(a)$ is maximal if and only if $a$ is prime.  So you want all primes $p$ such that $540 \in (p)$, i.e., you want to find all prime $p$ such that $p$ divides $540$.
