4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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? $\endgroup$ – Derek Holt Jan 22 '14 at 8:46
  • $\begingroup$ @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. $\endgroup$ – JN. Jan 22 '14 at 8:58
  • $\begingroup$ 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 $I$. $\endgroup$ – Derek Holt Jan 22 '14 at 9:08
  • $\begingroup$ 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! $\endgroup$ – JN. Jan 24 '14 at 8:17
1
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.