Number Fields

Theorem: Let $I$ be an ideal of a Dedekind domain $R$. Then there exists an ideal $J$ with $IJ$ 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 $IJ\subset (\alpha )$. 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 $rs\in M$. 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 $A=\frac{1}{\alpha }IJ$. Since $IJ\subset (\alpha )$ we have $A\subset R$, and in fact $A$ is an ideal. If $A=R$, then $IJ=(\alpha )$, 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 $\gamma J\mathrm{subset}\gamma A\subset R$. This implies $\gamma J\subset J$.

Now pick generators ${\alpha }_1,...,{\alpha }_m$ for $J$. We have

$$\gamma (\begin{array}{c}{\alpha }_1\\ ⋮\\ {\alpha }_m\end{array})=M(\begin{array}{c}{\alpha }_1\\ ⋮\\ {\alpha }_m\end{array})$$

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 $AB=AC$, then $B=C$.

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

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

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

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, $M=PI$ for some ideal $I$. So $I$ contains $M$. Furthermore, this contaiment is strict, for $I=M$ implies $RM=PM$ 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

$$\begin{array}{ccc}(2)& =& (2,1+\sqrt{-5}{)}^2\\ (3)& =& (3,1+\sqrt{-5})(\mathrm{3,1}-\sqrt{-5})\end{array}$$

All the ideals on the right-hand side are primes: observe that $\mid R/(2)\mid =4$, so $\mid R/(\mathrm{2,1}+\sqrt{-5})\mid$ divides 4. It must be 2, because $(\mathrm{2,1}+\sqrt{-5})$ properly contains $(2)$ and cannot be all of $R$. Thus $(\mathrm{2,1}+\sqrt{-5})$ 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 $(1+\sqrt{-5})=(\mathrm{2,1}+\sqrt{-5})(\mathrm{3,1}+\sqrt{-5})$ and $(1-\sqrt{-5})=(\mathrm{2,1}+\sqrt{=5})(\mathrm{3,1}-\sqrt{-5})$.