## 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:

$\array { gcd(I,J) &=& I+J \\ lcm(I,J) &=& I \cap J }$

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

$\beta \in \cap_{i=1}^r{(P_i^{n_i}-P_i^{n_i+1})} \cap \cap_{j=1}^{s}{(R-Q_j)}$

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:

$\array { \beta &=& \beta_i & (mod P_i^{n_i+1}), & i=1,...,r \\ \beta &=& 1 & (mod Q_j), & j=1,...,s }$

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).

Ben Lynn blynn@cs.stanford.edu 💡