# Prove Principal Ideal Domain from Bezout's condition, and terminating divisibility chain

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.

1. Any two non-zero elements $a$ and $b$ in $R$ have a greatest common divisor that can be written as $d=ra+sb$, $r,s \in R$

2. 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.

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 $a:=a_N$.

• Hi. I know this is late, but I'm working with the same problem. I've noticed in your answer there will be at most a countable number of steps. But what if $I$ is generated by uncountablly many elements? Then we'd be missing generators in our process. I am probably missing something simple, but this is confusing me. – Freddie Feb 7 '17 at 1:37

Other answers have already shown why you need the second condition. As a counterexample, consider the ring of all algebraic integers, that is the set of complex numbers which satisfy a monic polynomial over $\mathbb{Z}$. This ring satisfies your first condition, but isn't even a unique factorisation domain, let alone a principal ideal domain (this ring has no irreducible elements, since the square root of an algebraic integer is itself an algebraic integer).

Conceptually, we can view this as a generalization of the proof that ideals in Euclidean domains are generated by any element of minimal value. The Bezout condition $$(1)\Rightarrow$$ ideals are closed under gcd, since $$\rm\:a,b \in I \:\Rightarrow\: gcd(a,b) = ra+si \in I.\:$$ The chain condition $$(2)$$ says that the divisor relation is well-founded, i.e. that there are no infinite descending chains of proper divisors $$\rm\ \cdots\ a_3 \mid a_2 \mid a_1,\:$$ which implies ideals can be generated by elements least/minimal w.r.t. divisibility (proof below). Combining $$(1)$$ and $$(2)$$ we infer that $$\rm\,I\ne 0\,$$ is principal, since least generators $$\rm\,g_i$$ must be associate, else $$\rm\:gcd(g_i,g_j) \in I\:$$ and it is a proper divisor of $$\rm\:g_i,\:$$ contra leastness of $$\rm\,g_i$$ w.r.t. divisibility.

That $$\rm\,I\ne 0\,$$ can be generated by such minimal generators is provable as follows. First, there exists a set of generators for $$\rm\,I,\,$$ e.g. the set of nonzero elements of $$\rm\, I.\:$$ By the chain condition, each generator can be replaced by some divisor $$\rm\in I\,$$ that is least w.r.t. divisibility, since repeatedly taking proper divisors $$\rm\in I\,$$ of a generator yields a proper divisor chain $$\rm\,\cdots\, a_3 \mid a_2\mid a_1.\:$$ This chain must terminate with some $$\rm\:a_n\in I\:$$ having no proper divisors $$\rm\in I\,$$ (else it would yield an infinite descending chain of proper divisors, contra the hypothesized chain condition).

Your argument is quite right if all ideals are finitely generated. For example, consider $R=\{\frac {a}{2^b} : a,b\in \mathbb{Z}, b\ge 0\}$. $R$ is itself an ideal, that is not finitely generated (hence $R$ is not a PID). It passes your first condition but fails the second one.

• I don't get it: isn't $R$ always a principal ideal (generated by $1$)? Moreover, your example is nothing but a ring of fractions of $\mathbb Z$, so it is a PID. – user26857 Apr 16 '16 at 15:51