# Algebraic Number Theory – IV. Factorization into prime ideals

The arithmetic analogy between ${\mathbb{Z}}$ and ${\mathcal{O}_K}$ is not perfect, as ${\mathcal{O}_K}$ is not generally a unique factorization domain. For instance, note

$\displaystyle 2 \cdot 3 = 6 = (1+\sqrt{-5})(1-\sqrt{-5})$

in ${\mathbb{Z}[\sqrt{-5}] = \mathcal{O}_{\mathbb{Q}(\sqrt{-5})}}$ displays two essentially distinct ways to write ${6}$ as the product of irreducible elements. Nevertheless, one can recover in ${\mathcal{O}_K}$ a different sort of factorization: every ideal may be uniquely written as the product of prime ideals. Any integral domain in which this is possible is called a Dedekind domain. We will see by the end of this post that working in the world of ideals rather than numbers does not preclude us from extracting useful number-theoretic information.

Our goal for this post is to prove some of the important results on Dedekind domains (in particular, that ${\mathcal{O}_K}$ is a Dedekind domain), and it will not hurt to work in greater generality than the number field case. In particular, the setup is as follows: let ${R}$ be a Noetherian integral domain, and let ${F}$ denote its field of fractions.

4.1. “Modular arithmetic”

From the module-theoretic perspective, ideals of ${R}$ are just finitely generated (as ${R}$ is Noetherian) ${R}$-submodules of ${R}$. This point of view is useful because it allows us to naturally generalize the notion of an ideal to include larger subsets of the entire fraction field ${F}$. In particular, a fractional ideal is a finitely generated ${R}$-submodule of ${F}$. Since ideals in ${R}$ are also fractional ideals of ${F}$, for clarity we refer to ordinary ideals in ${R}$ asintegral ideals. We denote by ${J_F}$ the collection of fractional ideals in ${F}$.

Given ${A}$, ${B \in J_F}$, recall one can define their product and sum

$\displaystyle AB := \left \{\sum_{i=1}^{n} a_i b_i \, : \, a_i \in A, b_i \in B, n \in \mathbb{N} \right \}$

$\displaystyle A+ B := \{ a + b \, : \, a\in A, b\in B \}$

which are themselves fractional ideals. This allows one to introduce a notion of invertibility: an ideal ${A \in J_F}$ is invertible if there exists ${B \in J_F}$ such that ${AB = R}$.

We will have to do some hard work now to demonstrate the following proposition.

Proposition 4.1 Let ${R}$ be a Noetherian integral domain. Then, the following are equivalent:

1. ${R}$ is integrally closed (if ${x \in F}$ satisfies a monic polynomial in ${R[t]}$, then ${x\in R}$) and every prime ideal in ${R}$ is maximal.
2. Every integral ideal in ${R}$ can be written uniquely as the product of prime ideals.
3. Every fractional ideal of ${F}$ is invertible. In particular, ${J_F}$ is an abelian group under the module product.

If ${R}$ satisfies any of these conditions, one christens ${R}$ a Dedekind domain.

Before descending into the land of commutative algebra, let us note the corollary:

Corollary 4.2 ${\mathcal{O}_K}$ is a Dedekind domain.

Proof: In the previous post, we showed ${\mathcal{O}_K}$ is Noetherian and that every prime ideal in ${\mathcal{O}_K}$ is maximal. Suppose ${x\in K}$ satisfies a monic polynomial ${p \in \mathcal{O}_K [t]}$, given by

$\displaystyle p(t) = t^n + a_{n-1} t^{n-1} + \cdots + a_1 t + a_0$

Since ${\mathcal{O}_K}$ is a finitely generated ${\mathbb{Z}}$-module, it follows from the fundamental theorem on modules over a principal ideal domain that

$\displaystyle M_1 := \mathbb{Z}[a_{n-1}, a_{n-2}, \cdots, a_0] \subset \mathcal{O}_K$

is finitely generated as a ${\mathbb{Z}}$-module. Since ${x^n}$ is a ${M_1}$-linear combination of ${1}$, ${x}$, ${\cdots}$, ${x^{n-1}}$, it follows

$\displaystyle M_2 := \mathbb{Z}[a_{n-1}, a_{n-2}, \cdots, a_0,x]$

is also finitely generated as a ${\mathbb{Z}}$-module. Applying the fundamental theorem one last time, we have

$\displaystyle M_3 := \mathbb{Z}[x] \subset \mathbb{Z}[a_{n-1}, a_{n-2}, \cdots, a_0, x]$

is itself a finitely generated ${\mathbb{Z}}$-module. Since ${xM_3 \subset M_3}$, we conclude ${x}$ is an algebraic integer in ${K}$, i.e. ${x\in \mathcal{O}_K}$. $\Box$

4.2. Proof of Proposition 4.1

As with many arguments in commutative algebra, we piece together the proof from a series of fairly sophisticated lemmas. We shall first show the direction ${(a) \implies (b)}$ in Prop. 4.1, so assume ${R}$ is Noetherian, integrally closed, and every prime ideal in ${R}$ is maximal.

Lemma 4.3 Assume ${(a)}$. Let ${I}$ be an integral ideal of ${R}$. Then, there exist prime ideals ${\mathfrak{p}_1}$, ${\cdots}$, ${\mathfrak{p}_r}$ such that

$\displaystyle \mathfrak{p}_1 \mathfrak{p}_2 \cdots \mathfrak{p}_r \subseteq I$

Proof 1: (General). Denote by ${\mathcal{C}}$ the collection of integral ideals which fail to satisfy the given condition. Suppose ${\mathcal{C}}$ is nonempty. Since ${R}$ is Noetherian, an appeal to Zorn’s lemma admits a maximal element ${I \in \mathcal{C}}$. Of course ${I}$ cannot be prime, so there exist ${a}$, ${b\in R}$ such that ${ab \in I}$ but ${a}$, ${b\not\in I}$. There are the proper inclusions ${I \subset I + (a) := I_1}$ and ${I \subset I + (b) := I_2}$, so each of ${I_1}$ and ${I_2}$ contains a product of prime ideals. However, ${I_1 I_2 \subseteq I}$, which is a contradiction. $\Box$

Proof 2: (${\mathcal{O}_K}$-specific). For the axiom of choice truthers among us, one can show this result in the specific case ${R = \mathcal{O}_K}$ without invoking Zorn. Use that every quotient ring ${\mathcal{O}_K / I}$ is finite, and choose an ideal ${I}$ for which ${[\mathcal{O}_K : I ]}$ is minimal. Then argument proceeds exactly as in the first proof. $\Box$

For any ${A \in J_F}$, define the fractional ideal

$\displaystyle \tilde{A} := \{x \in F \, : x A \subseteq R \}$

By construction, ${A \tilde{A} \subseteq R}$. In fact, ${\tilde{A}}$ is the only candidate for the inverse of a fractional ideal: one can easily show if ${A}$ is invertible, then ${A^{-1} = \tilde{A}}$ (exercise!).

Lemma 4.4 Assume ${(a)}$. Let ${\mathfrak{p}}$ be a prime ideal of ${R}$. Then, ${\mathfrak{p} \tilde{\mathfrak{p}} = R}$.

Proof: The argument proceeds in three steps.

Step 1: (${\tilde{\mathfrak{p}} \neq R}$). Choose a nonzero ${x\in \mathfrak{p}}$ for which

$\displaystyle \mathfrak{p}_1 \mathfrak{p}_2 \cdots \mathfrak{p}_r \subseteq (x) \subseteq \mathfrak{p}$

such that ${r}$ is as small as possible. By the prime avoidance lemma, one of the ${\mathfrak{p}_i}$ (without loss of generality, assume ${\mathfrak{p}_1}$), is contained in ${\mathfrak{p}}$. Then ${\mathfrak{p} = \mathfrak{p}_1}$ since ${\mathfrak{p}_1}$ is maximal. By the minimality of ${r}$, one knows

$\displaystyle \mathfrak{p}_2 \mathfrak{p}_3 \cdots \mathfrak{p}_r \not\subseteq (x)$

so there exists ${y \in \mathfrak{p}_2 \mathfrak{p}_3 \cdots \mathfrak{p}_r }$ such that ${y \not \in (x)}$. In particular, ${y^{-1} x \not \in R}$. However, by the choice of ${y}$,

$\displaystyle (b) \mathfrak{p} = (b) \mathfrak{p}_1 \subseteq \mathfrak{p}_1 \mathfrak{p}_2 \cdots \mathfrak{p}_r \subseteq (x) \implies (y^{-1} x) \mathfrak{p} \subseteq R$

Hence, ${y^{-1} x \in \tilde{\mathfrak{p}}}$. It follows ${\tilde{\mathfrak{p}} \neq R}$.

Step 2: (${I\tilde{\mathfrak{p}} \neq I}$ for any nonzero ideal ${I \subset R}$). Let ${b_1}$, ${\cdots}$, ${b_n}$ be a basis for ${I}$ as an ${R}$-module. Suppose ${I\tilde{\mathfrak{p}} = I}$ and choose ${x\in \tilde{\mathfrak{p}}}$. Then there is a linear map ${I \rightarrow I}$ given by ${r \mapsto xr}$, with matrix ${M_x}$. Then ${x}$ is an eigenvalue of ${M_x}$, so that ${\det(M_x - x \cdot \text{Id}) = 0}$. But the left-hand side is a polynomial in ${R[t]}$. By integral closure of ${R}$, it follows ${x\in R}$. Thus, ${\tilde{\mathfrak{p}} = R}$, contradicting step 1.

Step 3: (${\mathfrak{p} \tilde{\mathfrak{p}} = R}$). By step 2, we see ${\mathfrak{p} \tilde{\mathfrak{p}} \neq \mathfrak{p}}$, so the inclusion ${\mathfrak{p} \subset \mathfrak{p} \tilde{\mathfrak{p}}}$ is proper. Since ${ \mathfrak{p} \tilde{\mathfrak{p}} \subseteq R}$ is an ${R}$-submodule of ${R}$, it is an integral ideal. But ${\mathfrak{p}}$ is a maximal ideal, so ${\mathfrak{p} \tilde{\mathfrak{p}} = R}$ as desired. $\Box$

Using these lemmas, it is actually not so difficult to finally furnish the proof of ${(a) \implies (b)}$. We demonstrate existence of the prime ideal factorization, and leave uniqueness as an (easy) exercise.

Lemma 4.5 Let ${R}$ be an integral domain and ${I \subset R}$ an integral ideal. If ${I}$ can be factored as the product of invertible prime ideals, then this factorization is unique.

Proof: Exercise! $\Box$

Proof 1 of ${(a) \implies (b)}$: (General). Let ${\mathcal{C}}$ denote the collection of integral ideals that do not permit a prime ideal factorization. Suppose ${\mathcal{C}}$ is nonempty. As earlier, Zorn’s Lemma gives a maximal element ${I \in \mathcal{C}}$. Since every ideal is contained in a maximal ideal, there is some prime ideal ${\mathfrak{p}}$ such that ${I \subset \mathfrak{p}}$. Then, one may factor ${I = \mathfrak{p} (\mathfrak{p}^{-1} I)}$, where

$\displaystyle \mathfrak{p}^{-1} I \subset \mathfrak{p}^{-1} \mathfrak{p} = R$

is an integral ideal. Since ${I = \mathfrak{p} (\mathfrak{p}^{-1} I) \subset \mathfrak{p}^{-1} I}$ is maximal in ${\mathcal{C}}$, it follows ${\mathfrak{p}^{-1} I}$ has a prime ideal factorization. But this gives ${\mathfrak{p}}$ a prime ideal factorization, which is a contradiction. $\Box$

Proof 2 of ${(a) \implies (b)}$:(${\mathcal{O}_K}$-specific). We shall prove by induction (on ${r}$) that if an integral ideal ${I \subset \mathcal{O}_K}$ contains a product of ${r}$ nonzero prime ideals, then ${I}$ is a product of prime ideals. By Lemma 4.3, this is sufficient to prove ${(a) \implies (b)}$. For ${r = 1}$, if ${\mathfrak{p} \subset I}$ for some prime ideal ${\mathfrak{p}}$, then ${I = \mathfrak{p}}$ by maximality. Suppose now that ${\mathfrak{p}_1 \mathfrak{p}_2 \cdots \mathfrak{p}_r \subset I}$. Since ${\mathcal{O}_K / I}$ is finite, there is a maximal ideal ${\mathfrak{p}}$ containing ${I}$. Then the prime avoidance lemma implies, without loss of generality, that ${\mathfrak{p}_1 = \mathfrak{p}}$. Multiply through the inclusion by ${{\mathfrak{p}}^{-1}}$:

$\displaystyle \mathfrak{p}_1 \mathfrak{p}_2 \cdots \mathfrak{p}_r \subset I \subset \mathfrak{p}_1$

$\displaystyle \implies \mathfrak{p}_2 \cdots \mathfrak{p}_r \subset I {\mathfrak{p}_1}^{-1} \subset \mathcal{O}_K$

Induction gives a prime ideal factorization for ${ I {\mathfrak{p}_1}^{-1}}$, and hence one for ${I}$$\Box$

Next on the agenda: we must show ${(b) \implies (c)}$.

Lemma 4.6 Choose any nonzero ${x \in F}$. If ${(x) = I_1 I_2 \cdots I_r}$ for fractional ideals ${I_k \in J_{F}}$, then each ${I_k}$ is invertible.

Proof: Simply set ${{I_k}^{-1} = (x^{-1}) I_1 I_2 \cdots I_{k-1} I_{k+1} \cdots I_r}$. $\Box$

Lemma 4.7 Assume ${(b)}$. Let ${\mathfrak{p} \subset R}$ be a nonzero prime ideal. If ${\mathfrak{p}}$ is invertible, then ${\mathfrak{p}}$ is maximal.

Proof: Choose any ${x\in R \setminus \mathfrak{p}}$. Suppose that ${R \neq \mathfrak{p} + (x)}$. Then we may factorize

$\displaystyle \mathfrak{p} + (x) = \mathfrak{p}_1 \cdots \mathfrak{p}_m$

$\displaystyle \mathfrak{p} + (x^2) = \mathfrak{q}_1 \cdots \mathfrak{q}_n$

where each ${\mathfrak{p}_i}$, ${\mathfrak{q}_j}$ is a prime ideal. Let ${y = x + \mathfrak{p} \in R/\mathfrak{p}}$. Then in ${R/\mathfrak{p}}$, we have the prime ideal factorizations

$\displaystyle (y) = (\mathfrak{p}_1 / \mathfrak{p}) \cdots (\mathfrak{p}_m / \mathfrak{p})$

$\displaystyle (y^2) = (\mathfrak{q}_1 / \mathfrak{p}) \cdots (\mathfrak{q}_n / \mathfrak{p})$

so Lemma 4.6 implies each ${\mathfrak{p}_i / \mathfrak{p}}$, ${\mathfrak{q_j}/\mathfrak{p}}$ is invertible. Equating these factorizations as

$\displaystyle (\mathfrak{p}_1 / \mathfrak{p})^2 \cdots (\mathfrak{p}_m / \mathfrak{p})^2 = (\mathfrak{q}_1 / \mathfrak{p}) \cdots (\mathfrak{q}_n / \mathfrak{p})$

we see Lemma 4.5 tells that ${n=2m}$ and each ${\mathfrak{p}_i / \mathfrak{p}}$ appears exactly twice in the collection ${\{\mathfrak{q}_j / \mathfrak{p} \}_{j=1}^{n}}$. Hence ${\{\mathfrak{q}_1, \cdots, \mathfrak{q}_n \}}$ is the multiset ${\{\mathfrak{p}_1, \cdots, \mathfrak{p}_n, \mathfrak{p}_1, \cdots, \mathfrak{p}_n\}}$. This gives the following chain:

$\displaystyle \mathfrak{p} \subset \mathfrak{p} + (x^2) = (\mathfrak{p} + (x))^2 \subset \mathfrak{p}^2 + (x)$

Thus, if ${a \in \mathfrak{p}}$, then ${a = b + rx}$ for some ${b \in \mathfrak{p}^2}$ and ${r \in R}$. Since ${rx \in \mathfrak{p}}$ and ${x\not\in \mathfrak{p}}$ by choice, we must have ${r \in \mathfrak{p}}$. This yields another chain:

$\displaystyle \mathfrak{p} \subset \mathfrak{p}^2 + (x) \mathfrak{p} \subset \mathfrak{p} \implies R = \mathfrak{p} + (x)$

since ${\mathfrak{p}}$ is invertible. But ${R \neq \mathfrak{p} + (x)}$ by assumption. $\Box$

Corollary 4.8 Assume ${(b)}$. Let ${\mathfrak{p} \subset R}$ be a nonzero prime ideal. Then, ${\mathfrak{p}}$ is invertible.

Proof: Choose any ${x\in \mathfrak{p}}$. We may factorize ${(x) = \mathfrak{p}_1 \mathfrak{p}_2 \cdots \mathfrak{p}_r}$, where, by Lemma 4.6, each ${\mathfrak{p}_i}$ is an invertible prime ideal. In fact, each ${\mathfrak{p}_i}$ is also maximal, due to Lemma 4.7. The prime avoidance lemma then implies without loss of generality ${\mathfrak{p} = \mathfrak{p}_1}$, so ${\mathfrak{p}}$ is itself invertible. $\Box$

The meat of this portion is now done with, and we can give the proof of ${(b) \implies (c)}$.

Proof: Let ${I \in J_{F}}$ be a fractional ideal. Since ${I}$ is finitely generated, there exists ${d\in R}$ such that ${(d) I \subset R}$. Factoring into prime ideals, we may write ${(d)I = \mathfrak{p}_1 \mathfrak{p}_2 \cdots \mathfrak{p}_r}$. Then ${(d) {\mathfrak{p}_r}^{-1} \cdots {\mathfrak{p}_2}^{-1} {\mathfrak{p}_1}^{-1} }$ is the inverse of ${I}$. $\Box$

Finally, it remains only to prove ${(c) \implies (a)}$, which is (comparatively) not very tough at all.

Proof: Let ${\mathfrak{p}}$ be a prime ideal in ${R}$, and suppose ${I}$ is a maximal ideal containing ${\mathfrak{p}}$. Then ${\mathfrak{p} = I (I^{-1} \mathfrak{p})}$, so the prime avoidance lemma implies either ${I^{-1} \mathfrak{p} \subseteq \mathfrak{p}}$ or ${I \subseteq \mathfrak{p}}$. Suppose the former is true. Then, multiplying through by ${I}$ gives

$\displaystyle \mathfrak{p} \subset I \mathfrak{p} \subset \mathfrak{p} \implies I = R$

which is impossible. Hence ${\mathfrak{p} = I}$ is maximal.

Suppose ${x\in F}$ satisfies a monic polynomial of degree ${n}$, whose coefficients are in ${R}$. Consider the fractional ideal

$\displaystyle I = (1, x, x^2, \cdots, x^{n-1})$

Then ${xI \subset I}$, so multiplying through by ${I^{-1}}$ proves ${x \in R}$. We conclude ${R}$ is integrally closed. $\Box$

Let us conclude this section with a very brief exercise, which illustrates two important properties of Dedekind domains.

Lemma 4.9 Let ${R}$ be a Dedekind domain.

1. “To contain is to divide”: the ideals ${I}$, ${J \subset R}$ satisfy ${I \subset J}$ iff there exists an integral ideal ${K}$ such that ${I = JK}$.
2. Comaximal equals relatively prime: the ideals ${I}$ and ${J}$ have no common prime ideal factors iff ${I+J = R}$.

Proof: Easy exercise. $\Box$

4.3. Introducing the ideal class group

We have strayed a bit far from number theory with all of these maximality arguments. Let us return by introducing the ideal class group of a number field ${K}$, whose relation to number theory is beautiful and important.

Recall ${J_K}$ denotes the collection of fractional ideals of ${K}$, which gains the structure of an abelian group under the module product. There is a subgroup ${P_K}$ which consists precisely of the principal fractional ideals, those generated by exactly one element in ${K}$. The class group of ${K}$ is then defined as the quotient

$\displaystyle \text{Cl}_K := J_K/ P_K$

Equivalently, one may place on the integral ideals of ${\mathcal{O}_K}$ a relation given by ${I \sim J}$ iff there exist ${\alpha}$, ${\beta \in \mathcal{O}_K}$ such that

$\displaystyle (\alpha) I = (\beta)J$

The equivalence classes thus obtained gain a natural group structure under ideal multiplication, and one easily sees this is isomorphic to the structure on ${J_K/ P_K}$. In the next post, we will prove the following extremely important proposition.

Proposition 4.10 For a number field ${K}$, the class group ${\text{Cl}_K}$ is finite.

For now, let us take this fact as given. The order of ${\text{Cl}_K}$ is called the class number of ${K}$, and denoted ${h_K}$. Observe that ${|\text{Cl}_K |= 1}$ iff ${\mathcal{O}_K}$ is a principal ideal domain. Thus, in some sense, the class number measures the extent to which a ring of integers fails to have (or succeeds in having) unique factorization for elements.

Consequently, computation of the class number is significant in number theory, as it detects how well one can treat ideals as elements. This interpretation is scalable: finiteness of the class group implies for any integral ideal ${I \subset \mathcal{O}_K}$, the ideal ${I^{h(K)}}$ is principal. One can use this fact to great effect in the solutions to certain Diophantine equations:

Proposition 4.11 Suppose ${d \in \mathbb{N}}$ is square-free, with ${d\equiv 1, 2 \pmod{4}}$. Let ${K = \mathbb{Q}(\sqrt{-d})}$, and assume furthermore that ${\gcd(3, h_K) = 1}$. Then,

1. If there exists ${a\in \mathbb{N}}$ for which ${d = 3a^2 \pm 1}$, then the Mordell equation ${x^3 = y^2 + d}$ has the unique solutions ${(x,y) = (a^2 + d, \pm a(a^2 - 3d))}$.
2. If there exists no such ${a\in \mathbb{N}}$, then the Mordell equation has no solutions.

The proof of this proposition is actually already well within our grasp (given temporary suspension of disbelief regarding the finiteness of ${h_K}$). Recall that when ${d\equiv 1, 2 \pmod{4}}$, the ring of integers of ${\mathbb{Q}(\sqrt{-d})}$ is exactly ${\mathcal{O}_K = \mathbb{Z}[\sqrt{-d}]}$.

Lemma 4.12 Assume ${\gcd(n, h_K) = 1}$. Suppose there are ${a}$, ${b\in \mathcal{O}_K}$ such that ${I^n = (ab)}$ for some ideal ${I}$. Furthermore, suppose ${(a)}$ and ${(b)}$ are comaximal ideals. Then ${a = ux^n}$ and ${b = vy^n}$ for some ${x}$, ${y \in \mathcal{O}_K}$ and units ${u}$, ${v \in {\mathcal{O}_K}^{\times}}$.

Proof: Since ${(a)}$ and ${(b)}$ are comaximal, they share no prime ideal factors. Therefore, ${(a) = J^n}$ and ${(b) = K^n}$ for some ideals ${J}$, ${K \subset \mathcal{O}_K}$ with ${I = JK}$. The class of principal ideals is the identity in ${\text{Cl}_K}$. By finite group theory, it follows that the order of ${\bar{J} \in \text{Cl}_K}$ must divide both ${n}$ and ${h_K}$. As it was assumed ${\gcd(n, h_K) = 1}$, it follows ${J}$ is principal. Similarly, ${K}$ is a principal ideal. Writing ${J = (x)}$ and ${K = (y)}$, the result drops out. $\Box$

Proof of Prop. 4.11: Assume there exists a solution ${(x,y) \in \mathbb{Z}^2}$ to Mordell’s equation ${y^2 + d = x^3}$. Due to the choice of ${d}$, modulo ${4}$ considerations tell that ${x}$ is necessarily odd.

Additionally, we find ${\gcd(x,d) = 1}$. Otherwise, suppose there is a prime ${p}$ dividing both ${x}$ and ${d}$. Then ${p}$ also divides ${x^3 - d = y^2}$, so ${p^2}$ divides ${x^3 - y^2 = d}$, which was assumed square-free.

Now, factor the equation in ${\mathbb{Z}[\sqrt{-d}]}$ and pass to ideals:

$\displaystyle (y+\sqrt{-d})(y-\sqrt{-d}) = (x)^3$

Suppose ${(y + \sqrt{-d})}$ and ${(y - \sqrt{-d})}$ are not comaximal, i.e. have a common prime ideal factor ${\mathfrak{p}}$. Then ${2\sqrt{-d} \in \mathfrak{p}}$, and hence ${-(2\sqrt{-d})^2 = 4d \in \mathfrak{p}}$. But ${\mathfrak{p}}$ also contains ${y^2 + d = x^3}$, thus by primality contains ${x}$. Since ${x}$ is odd and ${\gcd(x,d) = 1}$, also ${\gcd(x,4d) = 1}$, so there are integers ${m}$ and ${n}$ such that ${1 = mx+4nd \in \mathfrak{p}}$. This is a contradiction.

Therefore, ${(y+ \sqrt{-d})}$ and ${(y - \sqrt{-d})}$ are relatively prime. By Lemma 4.12, there exists ${\alpha \in \mathbb{Z}[\sqrt{-d}]}$ and a unit ${u \in \mathbb{Z}[\sqrt{-d}]^{\times}}$ such that ${y + \sqrt{-d} = u \alpha^3}$. The only units in ${\mathbb{Z}[\sqrt{-d}]}$ are ${\{\pm 1\}}$ (exercise!). Setting ${\alpha = a + b\sqrt{-d}}$ and assuming ${u = 1}$ without loss of generality, it follows

$\displaystyle y + \sqrt{-d} = (a + b \sqrt{-d})^3 = a(a^2 - 3db^2) + b(3a^2 - db^2) \sqrt{-d}$

Equating coefficients gives ${b = \pm 1}$ and ${d = 3a^2 \pm 1}$ accordingly. A little more computation proves ${y = \pm a(a^2 - 3d)}$ and ${x = a^2 + d}$ as desired. $\Box$

As lovely as this little characterization is, really applying it requires one to know the class number of ${K}$. And in general, calculating ${h_K}$ is quite challenging. There do exist explicit formulas and algorithms designed to compute ${h_K}$ in the case where ${K}$ is a quadratic field, which we will discuss in the subsequent posts.