I am stuck on the proof for this. This is what I manage to conclude so far: we have two trivial ideals, (0) and (1). Since it is not a PID, the ring R cannot be a field, so it must have some element that isn't a unit. Let it be x, so (x) is another ideal. Then, as it is principal, we must have another ideal, say I, which is not principal. (x) is not equal to I, so we have to have some element y that: either belongs to I but not to (x) (if I is not a subset of (x)), or y is in (x). Whatever the case may be, (y) is another ideal. That gives 5 ideals, but I am struggling to find a 6th one. My ideas were (x, y) or (x+y) but I cannot conclude that those ones are different from all the rest.
Ring that is not a principal ideal domain has at least 6 ideals
abstract-algebraidealsprincipal-ideal-domains
Best Answer
Let $R$ be a commutative ring with unity that is not a PID. Let $I$ be a nonprincipal ideal.
If $I$ is not finitely generated, let $X$ be an infinite generating set; we may assume no element of $X$ is zero. Let $x_1\in X$. Then $(x_1)\neq (0)$, $(x_1)\neq R$, and $(x_1)\neq I$. If for every $x\in X$ we have $(x_1,x)=(x_1)$, then $X\subseteq (x_1)$, so $I=(x_1)$, which is not possible. Thus, there exists $x_2\in X$ such that $(x_1,x_2)\neq (x_1)$. But also $(x_1,x_2)\neq I$. Continuing this way we obtain an infinite chain of ideals contained in $I$, so you get at least six different ideals.
So we may assume that all ideals of $R$ are finitely generated. Note that this means that there must be a $2$-generated ideal that is not principal, for if every $2$-generated ideal is principal, then every finitely generated ideal is principal (do induction). Thus, there is an ideal $I$, $I\neq R$, $I\neq (0)$, and elements $a,b\in R$, $a\neq b$, such that $I=(a,b)$, $I\neq (a)$, $I\neq (b)$. In particular we have the ideals $(0)$, $(a)$, $(b)$, $(a,b)$, and $R$, pairwise distinct. We just need one more ideal.
Consider $(a+b)$. Note that because $(a)\neq(b)$, we cannot have $a+b=0$, so $(a+b)\neq 0$. Because $a+b\in I$, we also cannot have $(a+b)=R$. If $a\in (a+b)$, then $a,b\in (a+b)$, so $I=(a,b)\subseteq (a+b)\subseteq I$, and we would have $I$ principal, so $a\notin (a+b)$. Therefore, $(a)\neq(a+b)$. Symmetrically, $(b)\neq (a+b)$. So now we have the ideals $(0)$, $(a)$, $(b)$, $(a+b)$, $(a,b)=I$, and $R$, pairwise distinct. That makes six.
In fact, six is the best you can do, as there are non-PIDs with exactly six ideals.
Note that there are rings in which every finitely generated ideal is principal (they are called Bézout domains, and a standard example is the ring of all algebraic integers), so if you are looking for a nonprincipal ideal you cannot just assume that there is one which is finitely generated. But if you know there are finitely generated non-principal ideals, then there must be a 2-generated non-principal ideal.