Showing a ring is not a principal ideal ring 
If I have a ring and suppose that I want to show that it is not a principal ideal ring. How can I construct an ideal (that is not a principal ideal) as a counterexample?

For example, I saw this question the other day:
The ring $R = \mathbb Z[\sqrt{-5}] $ is not a principal ideal domain because the ideal $I = (2, 1+\sqrt{-5})$ is not a principal ideal. But how would I think of this ideal by myself?
Thank you.
 A: In the above case $R$ is the ring of integers of $\mathbb{Q}(\sqrt{-5})$, and hence a Dedekind ring. Every ideal in a Dedekind ring is generated by at most $2$ elements, but not every ideal need to be principal (this is the case if and only if $R$ is factorial). Hence it is a good idea to consider an ideal $I=\langle a,b\rangle$ with two elements $a,b\in R$. Now we assume that it can be generated by one element $c\in R$. Taking norms (see the answers in the above post, i.e. choosing $a,b$ such that
$N(c)\mid gcd(N(a),N(b))=2$, so that $N(c)=1,2$ etc.) it is easy to see how to choose $a,b$ to obtain a contradiction.
In other cases, like $R=\mathbb{Z}[x]$, or $\mathbb{Q}[x,y]$ the question has been answered here. 
A: I think it is worth mentioning that in fact, for $R$ a commutative ring, any number of elements in $R$ can be defined similarly as an ideal. Don't forget that $I = (2,1+\sqrt{-5}) $ is really $ 2\mathbb{Z}[\sqrt{-5}] +(1+\sqrt{-5})\mathbb{Z}[\sqrt{-5}]$. So you could technically arbitrarily pick any ideal, and show that it is not generated by a single element.
