I am currently searching for an integral domain $R$ which allows lcm, but where the intersection of two principal ideals is not necessarily a principal ideal (or at least an example with two certain elements).

As converse example (where the sum of two ideals is not principal but $R$ allows gcd) is $\mathbb Z[X]$. It allows gcd, but obviously $\left<X+1,2\right>$ is not principal.

An example of an integral domain where the intersection of two principal ideals is not principal would be $\mathbb Z[\sqrt{-5}]$ with the ideal $\left<2\right> \cap \left<1+\sqrt{-5}\right>$, but this doesn't allow lcm.

Since I can't find a nice example, I put the question here. Maybe someone has already seen this? (I am not completely sure, one even exists, but pretty certain there is an example. In the case my hunch is wrong, I would be happy, if someone could proof me wrong.)


You will not find an example:

If $u := \mathrm{lcm}(s,t)$ exists (i.e. $s|u$ and $t|u$ holds and its minimal in the following sense: $s|v$ and $t|v$ implies $u|v$), we have $Rs \cap Rt = Ru$.

Proof. $Rs \cap Rt \supset Ru$ is clear from $s|u$ and $t|u$. We will show the other inclusion. Let $x \in Rs \cap Rt$, i.e. $x=rs=r^\prime t$ for some $r,r^\prime \in R$. In particular $s|x$ and $t|x$, hence $u|x$ by the definition of the lcm. We have shown $x \in Ru$ and are done.

As a consequence, one can define the lcm in any integral domain as follows:

If $Rs \cap Rt$ is principal, we call the generator (unique up to units) the lcm of $s$ and $t$.

As you have pointed out, a similar definition of the gcd is not possible in general.

| cite | improve this answer | |

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.