# Failure of existence of GCD

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...

• @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.
– Unit
Sep 13 '19 at 22:01

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 $$n is actually freely generated by $$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$$.