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 💡