13
$\begingroup$

In all the examples I've seen of $\gcd(a, b)$ not existing (e.g. MO, Wiki), the poset of principal ideals containing $(a, b)$ is finite but has more than one minimal element. Are other modes of failure possible?

Specifically, is there a domain $R$ with two elements $a$ and $b$ such that the ideal $(a, b)$ is not principal and either

  1. there are infinitely many distinct minimal principal ideals containing $(a, b)$, or
  2. there is no minimal principal ideal containing $(a, b)$?

Both problems require finding a sequence $d_1, d_2, d_3, \dots$ of elements of $R$, each dividing both $a$ and $b$. In the first case, they should be pairwise non-associate, and if $(a, b) \subseteq (c) \subseteq (d_i)$ for some $i$ then $(c) = (d_i)$. In the second case, each should properly divide the next, meaning that $d_{i+1}$ is a non-unit multiple of $d_i$.

I've considered various polynomial rings, the ring of entire functions, the ring of algebraic integers...

$\endgroup$
1
  • $\begingroup$ @Mastrem In your example, the poset of principal ideals containing $(3, X)$ is just the singleton $\{(1)\}$, so a smallest element exists, and $\gcd(3, X) = 1$ in $\mathbb{Z}[X]$. The principal ideals under consideration don't have to be proper. $\endgroup$
    – Unit
    Sep 13, 2019 at 22:01

1 Answer 1

7
+100
$\begingroup$

Sure: just build the ring with generators and relations to have the properties you want. For instance, consider the commutative ring $R$ which is generated by elements $a$ and $b$ and infinitely many elements $d_i,s_i,t_i$ with relations that $d_is_i=a$ and $d_it_i=b$ for all $i$. Explicitly, we can realize this $R$ as a subring of the field of rational functions $\mathbb{Q}(a,b,d_i)$ by mapping $s_i$ to $a/d_i$ and $t_i$ to $b/d_i$ (in particular, this shows $R$ is a domain). Even more explicitly, $R$ is the set of rational functions of the form $p/q$ where $q$ is a monomial in the $d_i$ of degree $n$ and $p\in\mathbb{Z}[a,b,d_i]$ is in the ideal $(a,b)^n$. This description makes it easy to verify that all the $d_i$ are maximal common divisors of $a$ and $b$ (since, for instance, if $p/q$ divides $a$ then $p$ can only be $\pm 1$ or $\pm a$ times a monomial in the $d_i$).

You can build a similar example where there are no maximal common divisors: take the subring $R$ of the field of rational functions $\mathbb{Q}(a,b,d_n)$ generated by $a,b,d_n,a/d_n,b/d_n,$ and $d_{n+1}/d_n$ for each $n$. In particular, writing $s_n=a/d_n$, $t_n=b/d_n$, and $u_n=d_{n+1}/d_n$, note that for any $N$, the subring $R_N$ generated by $d_0,s_N,t_N$, and the $u_n$ for $n<N$ is actually freely generated by $d_0, s_N,t_N$, and the $u_n$, and that every finite subset of $R$ is contained in some $R_N$. This makes it easy to verify that $a$ and $b$ have no maximal common divisor in $R$, since their greatest common divisor in $R_N$ is $d_N$ but $d_N$ properly divides $d_{N+1}$ for each $N$.

$\endgroup$
1
  • $\begingroup$ Thank you, Eric! I'll have a closer look at this in a few days. $\endgroup$
    – Unit
    Sep 16, 2019 at 15:26

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .