All Ideals in integers mod n

Short, sweet, to the point.

We are trying to show how to find all ideals in ${\mathbb{Z}}_n$ and then classify them.

Go through first part of Introductory Ring Theory and also Behaviors of Z modulo

This will help build a lot of intuition. Note, the only ideals in a field are trivial so if it was ${\mathbb{Z}}_p$ for some prime $P$ then the ideals would be 0 or the ring itself and we would be finished.

Lets take ${\mathbb{Z}}_{18}$ and find all ideals. Note, all ideals in ${\mathbb{Z}}_n$ are principal. And we can generate any ideal from all the elements; ie:

are all valid ideals. However, lets first show that all ideals can be reduced down to one generator.

Suppose this is not the case, and we have some ideal $⟨4,6⟩$. Notice, this ideal is defined as such:

Now, I would like to assert that $⟨4,6⟩=⟨GCD(4,6)⟩=⟨2⟩$. Lets prove it.

1. We have to show inclusivity in both directions, clearly 2 can generate 4 and 6 and therefore its multiples are all in $⟨2⟩$, so we have it that $⟨4,6⟩\subseteq ⟨2⟩$.
2. Now to show the other side, notice $gcd(a,b)=ar+bs$ from number theory. This means $2\in ⟨4,6⟩$, and by def of an ideal $rI\subset I$ for all r in R (commutative so no need to prove both directions). that means, $r(2)\in ⟨4,6⟩$ for all $r$ in ${\mathbb{Z}}_{18}$. This is legit the definition of the principal ideal of 2, thus $⟨2⟩\subseteq ⟨4,6⟩$.

So in general $⟨a,b⟩=⟨GCD(a,b)⟩$. By def the GCD can generate all elements in $⟨a,b⟩$ as it is a devisor, and by the Euclidean algorithm the gcd is in the $⟨a,b⟩$, meaning its entire generated principal ideal is too.

You can do this for any $a,b$. Thus all ideals can be simplified down to one generator and we now only need to explore $⟨0⟩,⟨1⟩,⟨2⟩...⟨17⟩$. Now, if $⟨a,b⟩=⟨GCD(a,b)⟩$, then we know

As 18 is 0, the definition does not change (we have $r_{1}a+r_{2}0$). This means, for any ideal in $⟨0⟩,⟨1⟩,⟨2⟩...⟨17⟩$ we have it so $⟨a⟩=⟨GCD(a,18)⟩$ which simplifies down to all the devisors of 18. Ie, $⟨4⟩=⟨GCD(4,18)⟩=⟨2⟩$. Using this, we see that the unique ideals of ${\mathbb{Z}}_{18}$ are generated by divisors of 18:
$⟨0⟩$
$⟨1⟩$
$⟨2⟩$
$⟨3⟩$
$⟨6⟩$
$⟨9⟩$
$⟨18⟩$ = $⟨0⟩$

Prime and Maximal Ideals:

From above, we know $⟨0⟩$ and $⟨1⟩$ are trivial. Notice all elements in $⟨9⟩$ or $⟨6⟩$ are contained in $⟨3⟩$ as 3 divides both. Similarly, all elements in $⟨6⟩$ are contained in $⟨2⟩$. Thus the maximal ideals are $⟨3⟩$ and $⟨2⟩$.

We already know the maximal ideals are prime, and it is easy to check they are the only prime ideals as, of the non proper ideals, $2\ast 3=6$, $3\ast 3=9$ and $2\notin ⟨6⟩$ similarly $3\notin ⟨9⟩$