# 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.

• It is unclear what you are asking. By defining $I = (2,1+\sqrt{-5})$ you have constructed the ideal. Do you mean something like how could I ever have thought of that for myself? – Derek Holt Jan 22 '14 at 8:46
• @DerekHolt yep, that was what I was trying to ask (but phrased it quite badly). I have edited the question above. Thank you for the suggestion. – JN. Jan 22 '14 at 8:58
• It can be helpful to look for instances of non-unique factorization. Here we have $6 = 2 \times 3 = (1 + \sqrt{-5})(1-\sqrt{-5})$ which might help you to think of this ideal $I$. – Derek Holt Jan 22 '14 at 9:08
• I understand that all non-unique factorization domains are not principal ideal domains, but how would I get the ideal based on this non-unique factorization? For example, I see that in our case, $I=(3, 1-\sqrt{-5} )$ is a non-principal ideal. Does that hold in general or is it just for this case? In other words, if I had a non-unique factorization, say $a \times b = c \times d$ , does that mean $I=(a,c)$ is definitely a non-principal ideal? Thanks for your help! – JN. Jan 24 '14 at 8:17

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.
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.