## 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 $$(1-\sqrt{-5})=(2,1+\sqrt{=5})(3,1-\sqrt{-5})$$.

Ben Lynn blynn@cs.stanford.edu 💡