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.

It’s hard to imagine how your argument relied on countable generation. Could you explain more clearly what you argued? Otherwise it is hard to help.

  • $\begingroup$ 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. $\endgroup$ – Daniel Dec 3 '18 at 12:14
  • $\begingroup$ Sorry I meant Ideal* at the beginning $\endgroup$ – Daniel Dec 3 '18 at 12:24

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