## GCDs and LCMs

Since every ideal factors in a Dedekind domain,
we may naturally define the *greatest common divisor*
$gcd(I,J)$ and
*least common multiple*
$lcm(I,J)$ of two ideals $I,J$. Note "greatest" and
"least" take the opposite meaning here. We take
"multiple" to mean subideal, and "divisor" means larger ideal, so
$gcd(I,J)$ is the smallest ideal containing both ideals, whilst $lcm(I,J)$
is the largest ideal contained in both:

**Theorem:** Let $I$ be an ideal of a Dedekind domain $R$ and let $\alpha \in I$.
Then there exists $\beta \in I$ such that $I = \langle\alpha,\beta\rangle$

**Proof:** We shall find $\beta \in R$ with
$I = gcd(\langle\alpha\rangle, \langle\beta\rangle)$.
Note this implies $\beta \in I$.

Let $I = P_1^{n_1}...P_r^{n_r}$ be the prime decomposition of $I$. Then $\langle\alpha\rangle$ is divisible by all the $P_i^{n_i}$. Let $Q_1,...,Q_s$ denote the other primes which divide $\langle\alpha\rangle$. We shall find $\beta$ such that none of the $Q_j$ divide $\langle\beta\rangle$, and $P_i^{n_i}$ is the exact power of $P_i$ dividing $\langle\beta\rangle$ for all $i$. In other words

This can be done using the Chinese Remainder Theorem. Fix $\beta_i \in P_i^{n_i} -P_i^{n_i+1}$ and solve the congruences:

We need show that the powers of the $P_i$ and $Q_j$ are pairwise coprime. But this is true since the sum is the greatest common divisor. Alternatively, for all $P_i, Q_j$ we have $1 = \alpha + \beta$ for some $\alpha \in P_i, \beta \in Q_j$. Then for any two positive integers $m,n$ we have $1^{m+n-1} = (\alpha+\beta)^{m+n-1}\in P_i^m + Q_j^n$ by considering the binomial expansion.

In general every PID is a UFD, but the converse is not always true. However, for a Dedekind domain:

**Theorem:** A Dedekind domain is a UFD if and only if it is a PID

**Proof:** We need only show the converse. Suppose we have an ideal
$P$ that is prime but not principal. Consider the set of ideals $I$ with
$P I$ principal, which must be nonempty in a Dedekind domain. Take a maximal
member $M$ and set $P M = \langle\alpha\rangle$.
Then $\alpha$ is an irreducible element
since if $\alpha = \beta \gamma$ then either $\langle\beta\rangle$ or
$\langle\gamma\rangle$ would
be of the form $P J$ for some $J \subset M$, and since $M$ is maximal this
would imply $J = M$ and hence $\beta$ or $\gamma$ is a unit.

Take any $\delta \in P - \langle\alpha\rangle$, $\epsilon \in M-\langle\alpha\rangle$ and note $\langle\alpha\rangle$ contains $\delta \epsilon$. That is $\alpha | \delta \epsilon$ yet $\alpha$ is irreducible, which is impossible in a UFD (since every irreducible should be prime).