This question might be a bit simple for many of those reading this text, but I was wondering why, given the fact that $\mathbb{Z}_8$ is not even an integral domain (and therefore not a PID), all the ideals of this ring are generated by one element. Is there any ideal in this ring generated by two elements? If the answer is negative, doesn't it imply that $\mathbb{Z}_8$ is a PID? Can a ring have only principal ideals but not be a domain?
Abstract Algebra – Can a Ring Have Only Principal Ideals but Not Be a Domain?
abstract-algebraring-theory
Related Solutions
Suppose $\def\ZZ{\mathbb Z}I\subseteq\ZZ\times\ZZ$ is an ideal.
If $(x,y)\in I$, then $(x,0)=(1,0)(x,y)$ and $(0,y)=(0,1)(x,y)$ are also in $I$. If we write $I_1=\{x\in\ZZ:(x,0)\in I\}$ and $I_2=\{y\in\ZZ:(0,y)\in I\}$. then it follows easily from this that $I=I_1\times I_2$. Now $I_1$ and $I_2$ are ideals of $\ZZ$, so that there are $a$, $b\in\ZZ$ such that $I_1=(a)$ and $I_2=(b)$, and then $I$ is generated by $(a,b)$.
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.
Best Answer
In the definition of a principal ideal domain you assume the ring is already an integral domain, i.e. a ring $A$ is a principal ideal domain if it is a domain in which every ideal is principal. So yes, it is possible for a ring to have only principal ideals but not to be a domain.
For any $n\in \mathbb{N}_{>0}$ all the ideals of $\mathbb{Z}/n\mathbb{Z}$ are of the form $d\mathbb{Z}/n\mathbb{Z}$ for $d|n$ so they are principal but $\mathbb{Z}/n\mathbb{Z}$ is not a domain if $n$ is not prime.