Unique Factorization of Ideals

Theorem: Let $I$ be an ideal of a Dedekind domain $R$. Then there exists an ideal $J$ with $I J$ principal.

Proof: Take any nonzero $\alpha \in I$, and let $J = \{\beta\in R: \beta I\subset (\alpha)\}$. Then $J$ is a nonzero ideal, and we have $I J \subset (\alpha)$. We wish to show that this is in fact an equality. We require two lemmas:

Lemma: In a Dedekind domain, every ideal contains the product of prime ideals.

Proof: Suppose not. Then consider the set of ideals that do not contain products of prime ideals. Because a Dedekind domain is Noetherian, this set has a maximal member $M$. $M$ cannot be prime (it does not contain the product of primes) so there exist $r,s\in R\setminus M$ with $r s \in M$. Now $M+(r),M+(s)$ are strictly bigger than $M$ so they contain products of primes, but this implies $(M+(r))(M+(s))$ does too. This is a contradiction since this ideal is contained in $M$.

Lemma: Let $A$ be a proper ideal in a Dedekind domain $R$ with field of fractions $K$. Then there exists $\gamma \in K\setminus R$ with $\gamma A \subset R$.

Proof: Fix any nonzero $a \in A$. By the previous lemma the principal ideal $(a)$ contains a product of primes, so let $P_1 P_2 ... P_r \subset (a)$ where $r$ is as small as possible. Every proper ideal is contained in a maximal ideal, and since a maximal ideal is prime, this means $A \subset P$ for some prime ideal $P$. Then $P$ also contains this product of primes. This implies $P$ contains some $P_i$: if not, then if we take an $a_i \in P_i \setminus P$ for all $i$ and consider the product $a_1...a_r$, then since $P$ is prime, it must contain one of the $a_i$, which is a contradiction. Relabel so that $P_1 \subset P$. But since every prime ideal is maximal, we have $P = P_1$. By the minimality of $r$, there exists $b\in(P_2 P_3...P_r) \setminus(a)$. Then take $\gamma = b/a \in K\setminus R$ and we have $\gamma A \subset R$.

Now we can complete the proof of the theorem. Consider the set $A = \frac{1}{\alpha} I J$. Since $I J \subset (\alpha)$ we have $A \subset R$, and in fact $A$ is an ideal. If $A = R$, then $I J = (\alpha)$, otherwise, $A$ is a proper ideal of $R$. Applying the latter of the above lemmas, we have $\gamma A \subset R$ for some $\gamma \in K \setminus R$.

Since $\alpha \in I$ we have $J \subset A$, thus $\gamma J subset \gamma A \subset R$. This implies $\gamma J \subset J$.

Now pick generators $\alpha_1,...,\alpha_m$ for $J$. We have

\[ \gamma { ( \array { \alpha_1 \\ \vdots \\ \alpha_m } ) } = M { ( \array { \alpha_1 \\ \vdots \\ \alpha_m } ) } \]

for some $m\times m$ matrix $M$ over $R$. We can then argue that $det(\gamma I - M) = 0$ and obtain a monic polynomial over $R$ having $\gamma$ as a root. Thus $\gamma \in R$ by integral closure, which is a contradiction.

Corollary: The ideal classes in a Dedekind domain form a group.

Corollary (Cancellation Law): If $A,B,C$ are ideals in a Dedekind domain with $A B = A C$, then $B=C$.

Proof: There exists an ideal $J$ with $A J$ principal. Let $A J = (\alpha)$. Then $\alpha B = \alpha C$ whence $B = C$.

Corollary: If $A,B$ are ideals in a Dedekind domain $R$, then $A|B$ if and only if $A \supset B$.

Proof: The forward direction is trivial. Assuming $A \supset B$, find a $J$ with $A J$ principal. Set $A J = (\alpha)$. Then since $J B \subset A J$ we have that $C = (1/\alpha)J B$ is an ideal in $R$. Lastly, note we have $A C = B$.

Theorem: Every ideal in a Dedekind domain $R$ is uniquely representable as a product of prime ideals.

Proof: First we shall prove that every ideal can be represented as a product of prime ideals. Suppose not. Then consider the set of ideals that cannot be represented as a product of primes. Take any maximal member $M$ (which is guaranteed to exist in a Dedekind domain). By convention, $R$ is the empty product so $M \ne R$ (or instead we may restrict the set to proper ideals of $R$).

As in the proof of one of the above lemmas, this implies $M$ is contained in some prime ideal $P$. Thus by the above corollary, $M = P I$ for some ideal $I$. So $I$ contains $M$. Furthermore, this contaiment is strict, for $I = M$ implies $R M = P M$ which by the cancellation law implies $R = P$, a contradiction. Since $M$ was chosen to be maximal, we must have $I$ a product of prime ideals, which then implies $M$ is also a product of primes, a contradiction.

Now we handle uniqueness. Suppose $P_1 P_2 ... P_r = Q_1 Q_2 ... Q_s$. Then $P_1 \supset Q_1 ... Q_s$. As in the proof of one of the above lemmas, this means $P_1 \supset Q_i$ for some $i$. Relabel so that $P_1 \supset Q_1$. Then we must have $P_1 = Q_1$ because every prime ideal is maximal. By the cancellation law we obtain $P_2 ... P_r = Q_2 ... Q_s$. Repeating this argument shows that $r = s$ and $P_i = Q_i$ for all $i$.

Since every number ring is a Dedekind domain, we immediate obtain the following:

Corollary: The ideals in a number ring factor uniquely into prime ideals.

Example: Consider $(2), (3)$ in the number ring $R = \mathbb{Z}[\sqrt{-5}]$. Then it turns out that

\[ \array { (2) & = & (2, 1+\sqrt{-5})^2 \\ (3) & = & (3, 1+\sqrt{-5})(3,1-\sqrt{-5}) } \]

All the ideals on the right-hand side are primes: observe that $|R/(2)| = 4$, so $|R/(2,1+\sqrt{-5})|$ divides 4. It must be 2, because $(2,1+\sqrt{-5})$ properly contains $(2)$ and cannot be all of $R$. Thus $(2,1+\sqrt{-5})$ is maximal as an additive subgroup. Thus it is maximal and hence prime. Similarly the factors of $(3)$ are primes.

As elements, we have the non-unique factorization into primes $6 = 2 \cdot 3 = (1+\sqrt{-5})(1-\sqrt{-5})$, but as ideals, we find that this becomes a unique factorization once it is verified that $(1+\sqrt{-5})=(2,1+\sqrt{-5})(3,1+\sqrt{-5})$ and $(1-\sqrt{-5})=(2,1+\sqrt{=5})(3,1-\sqrt{-5})$.