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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.

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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 $a:=a_N$.

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  • $\begingroup$ 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. $\endgroup$ – Freddie Feb 7 '17 at 1:37
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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).

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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).

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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.

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    $\begingroup$ 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. $\endgroup$ – user26857 Apr 16 '16 at 15:51
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The second condition, I think, proves that division terminates after a finite number of steps. Hence, once you've found the gcd, you can only divide it a finite number of times before you find you can't divide it any further. You can then form a principal ideal generated by one of the factors of the gcd.

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