# Is it possible for an integral domain to have an ideal that cannot be generated by a countable set?

This question came up when I was working on the following problem

Let $$R$$ be an integral domain. Prove that if the following two conditions hold then $$R$$ is a Principal Ideal Domain:

i) any two nonzero elements $$a$$ and $$b$$ in $$R$$ have a greatest common divisor which can be written in the form $$ra+sb$$ for some $$r,s \in R$$.

ii) if $$a_1, a_2, a_3,\ldots$$ are nonzero elements in $$R$$ such that $$a_{i+1}|a_i$$ for all $$i$$, then there is a positive integer $$N$$ such that $$a_n$$ is a unit times $$a_N$$ for all $$n \geq N$$.

I have managed to prove this by assuming that an arbitrary ideal in an integral domain $$R$$ can be written as $$I = (x_1, x_2, \ldots)$$ for some $$x_i \in R$$, if however there existed an ideal that could only be generated by an uncountable set in $$R$$ then my proof would not work.

The obvious counterexample seems to work: that is, $$F[\{x_i\mid i\in I\}]$$ for an uncountable index set $$I$$ and field $$F$$. The ideal generated by the $$x_i$$ is not countable generated.
• I start with a field of the form $I=(x_1,x_2,…)$ then define the sequence $a_1 = x_1$, $a_i = \gcd(a_{i-1},x_i)$ for $i \geq 2$ and then show that $(x_1,\ldots,x_i) = (a_i)$ for all $i \in \mathbb{N}$ using property i). Then I use property ii) to show that there exists an $N$ such that $(a_i) = (a_N)$ for all $i \geq N$. Then this shows that $(x_1, x_2, \ldots) = (a_N)$. Meaning the ideal $I$ is principal. – Daniel Dec 3 '18 at 12:14