Number Fields

Since every ideal factors in a Dedekind domain, we may naturally define the greatest common divisor $\gcd (I,J)$ and least common multiple $\mathrm{lcm}(I,J)$ 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 $\mathrm{lcm}(I,J)$ is the largest ideal contained in both:

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

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

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

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

$$\begin{array}{ccccc}\beta & =& {\beta }_i& (\mathrm{mod}{P}_i^{n_i+1}),& i=1,...,r\\ \beta & =& 1& (\mathrm{mod}{Q}_j),& j=1,...,s\end{array}$$

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 $PI$ principal, which must be nonempty in a Dedekind domain. Take a maximal member $M$ and set $PM=⟨\alpha ⟩$. Then $\alpha$ is an irreducible element since if $\alpha =\beta \gamma$ then either $⟨\beta ⟩$ or $⟨\gamma ⟩$ would be of the form $PJ$ 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 $\delta \in P-⟨\alpha ⟩$, $\epsilon \in M-⟨\alpha ⟩$ and note $⟨\alpha ⟩$ contains $\delta \epsilon$. That is $\alpha \mid \delta \epsilon$ yet $\alpha$ is irreducible, which is impossible in a UFD (since every irreducible should be prime).