The following is a problem from Dummit & Foote.
Let $R$ be an integral domain. Prove that if the following two conditions are true, then $R$ is a principal ideal domain.
-
Any two non-zero elements $a$ and $b$ in $R$ have a greatest common divisor that can be written as $d=ra+sb$, $r,s \in R$
-
If $a_1,a_2,\dots$ are non-zero elements of $R$ such that $a_{i+1}\mid a_i$ for all $i$, then there exists a positive integer $N$ such that $a_n$ is a unit times $a_N$ for all $n \ge N$.
I am a bit confused really. Isn't the first condition sufficient? For e.g. an ideal generated by two elements $a,b$ would be a principal ideal, as any two elements have a gcd. This can then be extended to ideals generated by an arbitrary number of elements by induction.
Obviously, there is something wrong in what I am doing. What is the second condition for? I can only think of defining principal ideals generated by the elements in a chain, and then finding a maximal ideal with respect to divisibility. I have no idea.
Best Answer
The second condition is to assure that every ideal has a finite set of generators. It is an ascending chain condition for principal ideals. Let $I\subseteq R$ be any ideal and start with $a_0:= 0$. In each step, pick $a'_i\in I\setminus \langle a_{i-1}\rangle$ (an element which can not be generated by $a_i$) and set $a_i:=\gcd(a_{i-1},a'_i)$. By the first assumption, $a_i\in I$ and clearly $\langle a_{i-1}\rangle \subseteq \langle a_i \rangle$. Since the $a_i$ form a chain as in (2), we may now conclude that this process actually ends at some point $a_N$, at which we have found the generator $a:=a_N$.