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.
Best Answer
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.