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This is a problem from Borevich and Shafarevich (p.180).

First, a definition:
A theory of divisors on a domain $\mathfrak{O}$ consists of a commutative semigroup $D$ (semigroup = associative multiplication with identity) having the property of unique factorization into irreducibles along with a homomorphism $\alpha \rightarrow (\alpha)$ of the semigroup $\mathfrak{O}^{\times}$ into $D$ satisfying
(i) An element $\alpha \in \mathfrak{O}^{\times}$ is divisible by $\beta \in \mathfrak{O}^{\times}$ in the ring $\mathfrak{O}$ if and only if $(\alpha)$ is divisible by $(\beta)$ in $D$.
(ii) If $\alpha$ and $\beta$ of $\mathfrak{O}$ are divisible by $a \in D$, then $\alpha \pm \beta$ are also divisible by $a$.
(iii) If $a$ and $b$ are two elements of $D$ and the set of all elements $\alpha \in \mathfrak{O}$ which are divisible by $a$ coincides with the set of all elements $\beta \in \mathfrak{O}$ which are divisible by $b$, then $a = b$.
The elements of $D$ are called divisors and elements of the form $(\alpha)$ for $\alpha \in \mathfrak{O^{\times}}$ are called principal divisors.

The problem is to show that every element $a$ of $D$ is the greatest common divisor of two principal divisors $(\alpha),(\beta)$ [which means that $a$ is a common divisor of $(\alpha)$ and $(\beta)$ that is divisible by every other common divisor of these two elements].

My first thought was to mimic the proof that every ideal $a$ (in a Dedekind domain) is generated by two elements $\alpha, \beta \in a$, which means exactly that $a$ is the gcd of the principal ideals $(\alpha),(\beta)$. However, this proof seems to rely critically on the addition of ideals and the notion of the inverse of an ideal, neither of which seem to transfer over to the setting of divisors. Most basic books in algebraic number theory have this proof; see for instance Stewart and Tall.

I have been told that (iii) implies that $a$ divides at least one principal divisor $(\alpha)$ [though I don't see how].

Since $D$ has unique factorization, I know that $(\alpha) = \prod_{i}^{n} p_{i}^{r_i}$ for irreducibles $p_i$ and positive integers $r_i$, and that $a = \prod_{i}^{n} p_{i}^{s_i}$ where $s_i \leq r_i$. Hence, if $(\beta)$ exists it must have the form $(\beta) = \prod_{i}^{n} p_{i}^{t_i} \cdot q_1 \cdots q_m$ such that $r_i = \min(s_i,t_i)$ and the $q_i$'s are irreducibles not equal to the $p_i$'s. This means that the $t_i$ are restricted as $t_i = r_i$ if $s_i < r_i$ and $t_i =$ any integer $\geq s_i$ if $s_i = r_i$.

If I can make a valid choice of the $t_i$'s and $q_i$'s that makes $\prod_{i}^{n} p_{i}^{t_i} \cdot q_1 \cdots q_m $ principal, then I will be done. However, I am not sure how to do this. I am not even sure if this is the right approach.

Can someone please help me?

EDIT: What I call a semigroup above is actually a monoid.

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up vote 1 down vote accepted

If $\rm\:A|a\:\Rightarrow\: a=0\ $ then ditto for $\rm\:A^2\:,\:$ since $\rm\:A|A^2|a\:.\:$ Thus by (iii) $\rm\ A^2 = A\ $ so $\rm\: A = (1)\:\Rightarrow\: A|1\ \Rightarrow\Leftarrow\:$ Therefore there exists an $\rm\ a\ne 0\ $ such that $\rm\: A\:|\:a\:,\:$ say $\rm\ AB = (a)\:.\:$ Applying the lemma below yields $\rm\ (b) = AC,\ (B,C) = 1\:.$ So $\rm\ (a,b) = (AB,AC) = A\:(B,C) = A\:.$

LEMMA $\ $ For any $\rm\:A,B\:$ there exists $\rm\:b,C\:$ with $\rm (b) = AC,\ (B,C)=1\:.$

Proof $\ $ Let $\rm\: B = P_1^{e_1}\cdots P_n^{e_n},\ P_i\:$ prime, $\rm\ Q = P_1\cdots P_n,\ \:Q_i = Q/P_i\:.\ Q_i|Q\ $ proper $\rm\:\Rightarrow\ AQ_i|AQ$ proper, so there exists $\rm\ b_i:\ AQ_i|b_i,\ AQ\nmid b_i\:.\:$ Put $\rm\ b = b_1+\cdots+b_n\:.\:$ Note $\rm\ A|b\ $ by $\rm\ A|AQ_i|b_i\:.\:$ $\rm\ i\ne j\ \Rightarrow\ P_i|Q_j\ \Rightarrow\ AP_i|AQ_j|b_j\:,\:$ but $\rm\ AP_i\nmid b_i\ $ (else $\rm\ AP_i,AQ_i|b_i $ $\rm \Rightarrow$ $\rm\: lcm(AP_i,AQ_i)=AQ|b_i\:$). Thus $\rm\ AP_i\nmid b\ $ so $\rm\ P_i\nmid (b)/A =: C\:,\: $ hence $\rm\ (B,C) = 1,\ \ (b) = AC\:.\ \ $ QED

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What do you mean when you say $Q_i | Q$ proper? – admchrch Feb 25 '11 at 1:45
@adm: $\rm\ a|b\ $ proper(ly) means $\rm\:a\:$ is a proper divisor of $\rm\:b\:,\:$ i.e. $\rm\ a|b,\ b\nmid a\:.$ The deduction in that sentence implicitly uses (iii) to deduce the existence of the $\rm\:b_i\:$. In fact it's easy to show that the axioms imply "divides = contains" i.e. $\rm\ A|B \iff \bar A \supseteq \bar B\ $ where $\rm\ \bar A = \{ a\in \mathfrak{O}\ :\ A|a\}\ $ is the ideal of elements divisble by $\rm\:A\:$. So if $\rm\ A|B\ $ *properly* then then $\rm\ \bar A\supsetneq \bar B\ $ *properly* so there is some $\rm\ a \in \bar A,\ a\not\in \bar B\ $ i.e. $\rm\ A|a,\ B\nmid a$ – Bill Dubuque Feb 25 '11 at 3:41
Thanks for the clarification (and the whole answer). That bit of terminology slipped my mind. – admchrch Feb 25 '11 at 4:37
@adm: If you're familiar with valuations you might find it more enlightening to express the proof in that language. It may make clearer the similarity with CRT (Chinese remainder theorem) and related results. – Bill Dubuque Feb 25 '11 at 5:05
Yes, I found such a proof. It uses a result in Borevich and Shafarevich that gives, for any finite set of valuations $v_i$ and set of integers $k_i$, a principal divisor $(\alpha)$ such that $v_i(\alpha)=k_i$. Using the result twice with appropriate choices for the $k_i$'s gives the desired $(\alpha)$ and $(\beta)$. I also see now that the proof of this result contains essentially your solution to the problem. – admchrch Feb 25 '11 at 7:34

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