The lattice of primary ideals of orders in quadratic number fields

Let $O$ be an order in a quadratic number field $K$ with ring of integers $D$, such that the conductor $\mathfrak F = f D$ is a prime ideal of $O$, where $f\in\mathbb Z$ is a prime. We give a complete description of the $\mathfrak F$-primary ideals of $O$. They form a lattice with a particular structure by layers; the first layer, which is the core of the lattice, consists of those $\mathfrak F$-primary ideals not contained in $\mathfrak F^2$. We get three different cases, according to whether the prime number $f$ is split, inert or ramified in $D$.


Introduction
One of the equivalent definition of a Dedekind domain is that every ideal is equal to a product of prime ideals; the fact that this factorization into prime ideals is unique is a consequence of the definition (see for example [9, §6, p. 270]). We are interested here in quadratic orders, that is, integral domains O whose integral closure is the ring of integers D of a quadratic number field K = Q[ √ d], d a square-free integer. We say that an order is proper if it is not integrally closed, that is, O D. Since a Dedekind domain is integrally closed ([9,Theorem 13,p. 275]), a proper order does not enjoy the aforementioned property of unique factorization into prime ideals.
However, since an order is a one-dimensional Noetherian domain, each ideal of O can be written uniquely as a product of primary ideals (see for example [9,Theorem 9,Chapt. IV,§. 5,p. 213]). An order O is determined by its conductor F, defined as the largest ideal of D contained in O; equivalently, F = {x ∈ O : xD ⊆ O}. Since D is a finitely generated O-module, F is always non-zero and it is a proper ideal of O if and only if the order is proper. Each ideal coprime to the conductor has a unique factorization into prime ideals of O [6,Lemma 2.26,p. 389]. These ideals are said to be regular. In particular, each regular primary ideal is equal to a power of its radical. Actually, this condition characterizes the regular primary ideals (see [5,Lemma 2.3]). More interesting is the situation for primary ideals that are non-regular. Here we consider the case where the conductor F of a quadratic order O is a prime ideal of O, so that F = f D, for some prime number f ∈ Z.
Among many others, we recall that Butts and Pall, in [1] and [2], have investigated ideals in quadratic orders. They addressed the problem of determining the number of factorizations of an ideal as a product of two ideals with given norms, without the assumption that the ideal is regular. The main results they gave are focused on invertible ideals. From a different point of view, Zanardo and Zannier [8] studied the class semigroup of quadratic orders, showing that it has a Clifford semigroup structure.
The purpose of the present paper is to give a detailed description of the structure of the lattice of F-primary ideals of a quadratic order O. We also provide generating sets for each F-primary ideal, in a somehow canonical way. By means of such generators, we immediately see the relations of containment between these ideals; in particular, we find at once the minimal power of F contained in a F-primary ideal.
We get three completely different lattices of F-primary ideals, according to whether f D is a prime ideal in D (inert case), or it is the product of two distinct prime ideals of D (split case), or it is equal to the square of a prime ideal of D (ramified case). However, these lattices have a crucial property in common, namely, a structure by layers. This means that the structure of the lattice is determined by its first layer, namely the set of F-primary ideals not contained in F 2 , which we call basic F-primary ideals. The remaining part of the lattice is formed by the n-th layers (n > 1) of the ideals contained in F n and not contained in F n+1 , and all these layers reproduce the same pattern of the first layer.
Quadratic orders in number fields is a natural and important class of integral domains whose lattice of primary ideals is a disjoint union of lattices structured by layers. From the previous discussion it follows that when N = F is a prime ideal of O, the lattice of N-primary ideals is just a chain.
In Sections 1 and 2 we give the definition of F-basic ideals, which are F-primary ideals not contained in F 2 , and prove their main properties. The F-basic ideals constitute the first layer of the lattice of F-primary ideals, hence it suffices to describe them to determine the whole structure. We firstly characterize the F-basic ideals which are also D-modules (that is, ideals of D). This is a crucial step to get a complete description of the first layer, since every F-basic ideal lies between a suitable F-basic D-module Q and f Q. Moreover, for each F-basic ideal (either D-module or not), we define a special set of generators that make the mutual relations of containment evident. We also identify the F-basic ideals that are principal. We show that there are exactly f + 1 pairwise distinct intermediate ideals properly lying between F and F 2 . Moreover, the quotient group of the units of D/F modulo the units of O/F acts freely on the set of intermediate ideals which are not D-modules (Proposition 2.10).
In Section 3 we examine separately the three cases mentioned above, namely, f inert, split or ramified in D.
When f is inert, F itself is the only F-basic ideal that is a D-module. As a consequence of this fact all the basic ideals lie between F and F 2 , and, in particular, their number is finite, equal to f + 2. Note that, under the present circumstances, the local ring O F is F-chained, in the sense of Salce's paper [7].
The split case is the most interesting one. In fact, we have infinitely many Fbasic ideals, each of them lying between a special F-basic ideal Q k (k > 0) that is also a D-module, and f Q k . The Q k 's are generated by f k and a naturally defined element t k ∈ F which generates a basic F-primary ideal. There are exactly f − 1 ideals properly between Q k and f Q k that are not D-modules, and they are described in terms of the generators of Q k .
Finally, in the case f is ramified, we again get finitely many F-basic ideals, but not all of them are between F and F 2 . Indeed, if F = f D = P 2 , where P is a prime ideal of D, the F-basic ideals all lie either between F = P 2 and F 2 = P 4 , or between P 3 and P 5 . Under the present circumstances, in order to obtain the generators of the F-basic ideals, we must take care of the exceptional case when f = 2 and d ≡ 3 modulo 4.

General definitions and results
In what follows, we will freely use the standard results on rings of integers in number fields, as can be found, for instance, in [9,Chapter V] and [6, §6.2]. As usual, for both elements z ∈ D and ideals I, the symbolsz,Ī and N(z), N(I) denote the conjugates and the norms, respectively; D * , O * denote the multiplicative groups of the units of D and O. If I is an ideal of O, ID denotes the extended ideal in D, i.e., the ideal of D generated by I. Moreover, in order to simplify the notation, the symbol A ⊂ B will denote proper containment. We fix some standing notation. Let d be a square-free integer. The ring of integers In the latter case, we get ω 2 = ω − (1 + d)/4. Let us fix f > 0 and consider the order Since O/F ∼ = Z/f Z, we immediately see that F is a prime ideal of O if and only if f is a prime number. We are interested in studying the F-primary ideals of O, hence, in what follows, F will be a prime ideal in O; equivalently, f will always denote an assigned prime number. Lemma 1.1. In the above notation, take any α ∈ F \ f O. Then F = (f, α).
Proof. It suffices to show that f ω ∈ (f, α). Say α = f a + f ωb, where a, b ∈ Z and f does not divide b, since α / ∈ f O. Take c, k ∈ Z such that cb = 1 + f k. We get We give a definition which is crucial for our discussion.
If Q/f m is F-basic for some m > 0, then m is unique and coincides with k − 1.
Let us note that a F-basic ideal is a primitive ideal of O, that is, an ideal that cannot be written as mJ, where m ∈ N, m ≥ 2 and J ⊂ O is an ideal (the definition of primitive ideal is quite standard, see for example [2]).
Given a F-primary ideal Q, the F-basic ideal Q ′ contained in Q, as defined in (ii) above, is called the basic component of Q. Roughly speaking, the basic component of a F-primary ideal Q is obtained by extracting from Q the highest power of f . It is uniquely determined by (ii) and (iii). It follows that the lattice L of all the F-primary ideals will be determined as soon as we know the lattice L 1 of the F-basic ideals. In fact, L 1 will be the first layer of L, and the other layers of the lattice will be the L n (n > 0), consisting of those F-primary ideals contained in F n but not in F n+1 . By Lemma 1.3, the elements of L n are obtained by those of L 1 , just multiplying by f n−1 . Therefore, the L n are reproductions of L 1 , and the whole lattice L is structured by layers.
The preceding remarks show that if suffices to focus our attention on L 1 . Hence, in what follows, we will investigate the F-basic ideals of O and their mutual relations of containment.
The next notion is related to the basic ideals that are principal.
We say that t is F-primary if tO is an F-primary ideal. We say that t is F-basic (or simply basic) if tO is a F-basic ideal.
Obviously, f is a basic element of O. An element t in O which is F-primary lies in F = f Z + f ωZ and therefore has the form t = f x + f ωy, for some x, y ∈ Z. The next proposition characterizes primary ideals and primary elements in terms of their norms. The proof is straightforward, using the properties of the norm.
∈ f Z (note also that in this case x, y are coprime, since f is the only common prime factor of x and y). If t is a basic element and t ∈ f O, then tO = f O, that is, t and f are associated in O.
However, it is possible that t / ∈ f O, but F 2 ⊂ tO ⊂ F. We will see in the next section that this happens precisely when t and f are associated in D but not in O (Lemma 2.7).
Proof. Let us assume, for a contradiction, that t = rs, where r, s ∈ O, and neither r nor s is a unit in O. Since the norm is a multiplicative function on O, r, s are F-primary elements. In particular, r, s ∈ F. But then t = rs ∈ F 2 , contradiction.
Moreover, tO is not a prime ideal, since it is strictly contained in the conductor F (the only prime ideal containing t), which is not principal.

Intermediate F-primary ideals
Let Q be a F-primary ideal not contained in f O; then, clearly, Q is F-basic. Using, for instance, the Claim in [8], page 358, it is readily seen that there exists The following easy lemma will be useful to determine whether or not an ideal of O is a D-module.
Proof. To get I = ID, it suffices to show that ωI ⊆ I.
We conclude that ωI ⊆ I.
The next result characterizes the F-basic ideals of O that are also D-modules. These kind of ideals will be crucial in the description of the lattice of F-basic ideals. The equivalence (ii) ⇔ (iii) of Theorem 2.2 also follows from [1, p. 34]. We give a direct proof for the sake of completeness.
without any assumption on N(f α).
Let us verify that ωf α ∈ Q. We have for a suitable x 0 to be chosen below. Then the above equality is equivalent to If I is an ideal of O and ID is the extended ideal in D, [ID : I] denotes the index of I in ID as Abelian groups. Note that I = JD, for some ideal J of O, if and only if I is also a D-module. Proof. We firstly show that f α ∈ I for every α ∈ ID. In fact, by definition of extended ideal, α = i a i β i , for suitable a i ∈ I and β i ∈ D. Then f α = i a i f β i is an element of I, since each f β i is in O. In particular, we have f ID ⊂ I ⊂ ID, where the inclusions are strict, since I is not a D-module. Since f is a prime number and the index of f ID in ID is f 2 , it follows that the index of I in ID is f . Note that the above proof works for any order of index f in the ring of integers of any number field, not necessarily quadratic.
We firstly assume that Now we assume that I ⊂ ID, so f = [ID : I] by Lemma 2.3. If ID ⊂ Q, then by the first part of the proof it follows that ID ⊆ f QD, hence I ⊂ f QD, as well. If ID ⊂ Q, then ID∩Q = ID. We get [ID : An easy computation shows that the f +1 subgroups of Z/f Z⊕Z/f Z generated by the elements (1, a)+f Z⊕f Z, a ∈ A, and (0, 1)+f Z⊕f Z are distinct and cover all the proper nonzero subgroups of Z/f Z ⊕ Z/f Z.
between Q and f Q, namely the pairwise distinct ideals Since Q/f Q ∼ = Z/f Z ⊕ Z/f Z (as Abelian groups) and Z/f Z ⊕ Z/f Z has exactly f + 1 proper non-zero subgroups, it suffices to show that the ideals J, J a (a = 0, . . . , f − 1) are pairwise distinct and lie properly between Q and f Q.
It is clear that the ideals J, J a , 0 ≤ a ≤ f −1 lie between Q and f Q = (f k+1 , f 2 α). We firstly verify that these ideals are pairwise distinct.
Let us suppose that J a = J b . Then we get the equality for suitable x 0 , x 1 , y 0 , y 1 ∈ Z. It follows that where ωQ ⊆ Q since Q is a D-module. The above relation yields (1 − y 0 )α ∈ O, so 1 − y 0 ∈ f Z, since a 1 / ∈ f Z. Then we get af k−1 − y 0 bf k−1 ∈ Q, hence a − y 0 b ∈ f Z, by the minimality of k. We conclude that 1 ≡ y 0 , a ≡ y 0 b mod f, so a ≡ b modulo f , and therefore a = b, since these integers both lie in {0, 1, . . . , f − 1}. We remark that we have actually proved that J a ⊆ J b whenever a = b. Since It remains to show that J a = Q, for a = 0, . . . , f −1. Assume, for a contradiction, that J b = Q for some b ≤ f − 1. Then we get J a ⊆ Q = J b for every a = b, which is impossible, as remarked above. (iii) Under the present circumstances, we get Q ⊃ f QD ⊃ f Q, since Q is not a D-module. Take any primary ideal, say I, properly lying between Q and f Q. Then Lemma 2.4 shows that I ⊆ f QD, hence we actually get the equality The preceding theorem allows us to determine the ideals lying between F and F 2 . Indeed, the following corollary derives immediately from (ii) of Theorem 2.5, since F is a D-module and F 2 = f F. Corollary 2.6. There are exactly f + 1 ideals of O that lie properly between F and F 2 , namely the pairwise distinct ideals J = (f, f 2 ω) = f O, J a = (f 2 , f (a + ω)), a = 0, 1, . . . , f − 1.
Next, we determine the intermediate ideals that are principal, or, equivalently, the basic elements t ∈ O such that F 2 ⊂ tO ⊂ F.
The last statement is immediate.
In particular, Lemma 2.7 implies that the number of principal F-primary ideals between F and F 2 is equal to |D * /O * |.
Since F = f D, it is well-known that there are three possibilities for the factorization into prime ideals of F in D, namely: iii) P ⊂ D a prime ideal (ramified case).
Proof. As we have said at the beginning of Section 1, since f is prime, F is a prime ideal in O and O/F ∼ = F f , the finite field with f elements. In particular, the group of units of O/F has cardinality f − 1. Since F is also an ideal of D, we have three possibilities: In the ramified case, D/P 2 is a local ring with maximal ideal P/P 2 . In each of the three cases, the group of units of D/F has cardinaly equal to f 2 − 1, f 2 − f and (f − 1) 2 , respectively. The canonical ring homomorphism π : D ։ D/F induces a group homomorphism π * : D * → (D/F) * (which is not necessarily surjective). We have the following group homomorphism: We claim that the latter group homomorphism is injective. In fact, if π * (u) ∈ (O/F) * , then π(u) ∈ O/F, so we get u ∈ O * , since π −1 (O/F) = O. It follows that τ = |D * /O * | divides the cardinality of (D/F) * /(O/F) * , which in the three cases is equal to: Remark 2.9. We note that the same conclusion of Proposition 2.8 can be obtained by means of a well-known formula that gives the class number of O in terms of the class number of D (see [3, p. 146-148]). By Corollary 2.6, there are f + 1 ideals properly between F and F 2 . In each of the three cases mentioned above, the number of these intermediate ideals of O that are also D-modules is: i) inert case: there is no intermediate D-module, since there are no D-modules between F = P and F 2 = P 2 .
ii) split case: 2; the only D-modules between F = P P and F 2 = P 2 P 2 are P 2 P and P P 2 .
iii) ramified case: 1; the only D-module between F = P 2 and F 2 = P 4 is P 3 .
Hence, τ = |D * /O * | divides the number of ideals properly between F and F 2 that are not D-modules (f + 1, f − 1 and f , resp.), and this last number is equal to the cardinality of (D/F) * /(O/F) * .
This last fact is an evidence of the following general result. We recall that an action of a group G on a set S is free if the stabilizer of each element s ∈ S is trivial, that is, Stab(s) = {g ∈ G | gs = s} = {1}.
(ii) Lemma 2.7 shows that every principal F-basic intermediate ideal has the form f u n O, for some n ≥ 0. Moreover, if n = qτ + j, with 0 ≤ j < τ , we clearly get f u n O = f u j O. An easy exercise gives our assertion in the case when u ∈ O, i.e., τ = 1. So we assume that τ > 1. The principal ideals f u j O, j = 0, . . . , τ − 1, are F-basic intermediate, and are pairwise distinct, since u i−j / ∈ O whenever i = j and 0 ≤ i, j < τ . Recall that y j / ∈ f Z, since u j / ∈ O for j = 1, . . . , τ − 1. By Corollary 2.6, f u j O = (f 2 , f (a + ω)), for some 0 ≤ a < f , and this equality of ideals holds exactly if f (a + ω) ∈ f u j O. Equivalently, we get the following equality in D j , modulo f . An easy computation allows us to find the mutual conjugates of the F-basic intermediate ideals.
Proposition 2.12. In the above notation, the conjugate ideal of (

The lattice of basic ideals.
In the present section we analyze separately the lattice L 1 of F-basic ideals, in each of the three cases that may appear, namely f inert, split or ramified in D.

Inert case.
The next theorem gives a complete description of the lattice of the F-basic ideals of O = Z[f ω], in the case when f is a prime element of D = Z[ω]. Recall that infinitely many prime numbers remain prime elements in D; actually the well-known Tchebotarev Density Theorem (see [3, §8]) says that the density of such primes is 1/2.
Proof. Let Q be a basic ideal. The extended ideal QD is equal to F, since Q is F-primary (F is the only prime ideal of O that contains Q, hence the only prime ideal of D that contains Q) and Q is not contained in F 2 by definition. It follows by Lemma 2.3 Finally, Q lies in the set J by Corollary 2.6.
The principal basic F-primary ideals are described in Theorem 2.11. Their number is exactly equal to the number of distinct non-associated basic elements of O, which is equal to |D * /O * |, see Lemma 2.7. Moreover, their norm is equal to f 2 , since the ideal they generate lies in between F and F 2 .
The following diagram represents the lattice of F-primary ideals in the inert case. Note that, in the diagram, only the powers of F are D-modules.

Split case.
Throughout this section, we assume that F = f D splits as an ideal of D, say f D = PP , where P =P are prime ideals of D of norm f , both of which lie above F, considered as an ideal of O. Note that P is principal if and only if f is not irreducible in D (recall that f is always irreducible in O, by Proposition 1.6). However, some power of P is a principal ideal of D, since the class group of D is finite. For the remainder of this section, we will denote by m the order of P in the class group of D (i.e., the minimum power m of P such that P m is principal), and by β a fixed generator of P m .
Similarly to the inert case, there exist infinitely many prime numbers that split in D.
Lemma 3.2. In the above notation, β n / ∈ O for every n > 0.
Proof. Assume, for a contradiction, that β n ∈ O. Then β n ∈ O ∩ P = F. It follows that β n D = P mn ⊆ F = PP ⊂P , whence P ⊆P , impossible.
The next proposition gives a way to construct F-basic principal ideals, not containing F 2 .
Proposition 3.3. The elements t n = f β n , for n ∈ N, are F-basic, and the principal ideals t n O, for n > 0, are pairwise incomparable and do not contain F 2 .
Proof. Recall that N(β) = f m . Then N(t n ) = f 2 N(β n ) = f mn+2 , so the principal ideal t n O is F-primary, for every n > 0, in view of Proposition 1.5. Now note that t n / ∈ F 2 = f F, for every n ≥ 0, since t n /f = β n / ∈ F ⊂ O. We conclude that t n is F-basic for every n ≥ 0. Pick now two distinct non-negative integers n, m, with n = m + h, h > 0. Then t n O and t m O are not comparable, since t n /t m = β h / ∈ O and t m /t n = β −h / ∈ O. It follows that the ideals t n O (n ≥ 0) are pairwise incomparable. Finally, since t n has norm strictly greater than f 2 , for n > 0, F 2 is not contained in t n O.
In the notation of Proposition 3.3, we define t 0 = f = f β 0 . Differently from the t n with n > 0, t 0 O ⊃ F 2 . The Proposition also shows that in the split case, unlike the inert case, there are basic elements of arbitrary large norm, so, they are infinitely many.
The following theorem describes all the F-basic elements of O which are associated to the elements t n = f β n , as introduced in Proposition 3.3. Note that the only elements of norm f 2 are f w = t 0 w, where w is a unit of D (see Lemma 2.7).
Theorem 3.4. Let t be a basic element of O of norm f s+2 , s ≥ 0. Then s = mn, for some n ∈ N, and t ∈ {t n w,t n w : w ∈ D * }. Moreover, t h wO =t k w ′ O for any positive integers h, k and w, w ′ ∈ D * , and t h wO = t k w ′ O if and only if h = k and w/w ′ ∈ O.
Proof. Since t is F-basic, P,P are the only prime ideals of D above tD. Then we get tD = P kP h , h, k > 0.
Moreover, since t / ∈ F 2 = P 2P 2 , the integers h, k are not both > 1. Let us assume that h = 1, whence tD = f P k−1 . Then P k−1 is principal, hence k − 1 = mn, for some positive integer n. It follows that N(t) = f s+2 = f 2 N(P mn ) = f mn+2 , hence we get the desired equality s = mn. Now, we have tD = f P mn = f β n D = t n D, which is possible only if t = t n w, for some w ∈ D * .
In the case k = 1 we will symmetrically get t =t n w for some w ∈ D * . Finally, if t h wO =t k w ′ O, then h = k otherwise t h , t k have different norms and we get that some power of β is in O, which is impossible by Lemma 3.2. Moreover, t h wO = t k w ′ O implies h = k as before, hence we also get w/w ′ ∈ O.
Our next step is to classify the non-principal basic F-primary ideals. We recall that a Special PIR (Special Principal Ideal Ring) R is a principal ideal ring with a unique prime ideal M, such that M is nilpotent (see [9, p. 245]). So, in the case when M = pR, for some p ∈ R, we get p n = 0 for some n > 1. Proof. The claim is immediate when t Thus we may assume that n ≥ 1.
Note that, if I is an ideal of O containing t n , then I is basic F-primary, since any prime ideal containing I must contain the F-basic element t n . The ideal I is basic since t n / ∈ F 2 . In particular, O/t n O has a unique maximal ideal, equal to F/t n O. Since F = (f, t n ) by Lemma 1.1, it follows that F/t n O is a principal ideal of O/t n O, generated by f + t n O. From this fact, it is not difficult to see that every nonzero ideal of O/t n O is principal, generated by some f i + t n O, for some 1 ≤ i ≤ mn + 1 (see [4,Proposition 4], for example). Indeed, f h ∈ t n O if and only if h ≥ mn + 2, since N(t n ) = f mn+2 . The next corollary follows at once from the previous lemma and gives all the basic F-primary ideals that contain some F-basic element.
Corollary 3.6. The ideals (necessarily F-primary) that contain t n O are the (f i , t n ), for i = 1, . . . , mn + 2. Moreover, the norm of (f i , t n ) is f i .
Proof. The first claim follows from Lemma 3.5. The second claim follows from Theorem 2.2, since f i is the least power of f contained in (f i , t n ). Proposition 3.7. Let i ∈ N and t ∈ O be a basic F-primary element of norm f m , m > i. The ideal I = (f i , t) of O is a D-module, equal either to I = P iP or I = PP i . In particular, we get (f i , t i ) = (f i , t n ), for every n ≥ i.
Proof. Since f i+1 | N(t), we get I = ID, by Theorem 2.2. Without loss of generality, we suppose that tD = P m−1P . Since D is a Dedekind domain, f i D + tD is the greatest common divisor of f i D and tD, so it is equal to P iP , since f i D = (PP ) i . Hence, I = ID = P iP . The last claim follows immediately, since f i divides N(t n ) = nm + 2 for every n ≥ i.
For every k ≥ 1, we define Q k = (f k , t k ) = P kP ; in this notation, Q 1 = F.
Proof. Since Q is basic, exactly by the same proof of Theorem 3.4 given in the case of principal basic ideals, we have QD = P kP = Q k , for some k ≥ 1 (or its conjugate), so Q ⊆ Q k . By Lemma 2.3, either Q = Q k or [Q k : Q] = f . In both cases, we get The next theorem gives a description of the ideals of O that contain a basic element. (iii) if Q contains a basic element and it is not principal, then either Q = Q k or Q =Q k for some k ∈ N.
Proof. (i) By Proposition 3.7, we get Q k = P kP (not the conjugate, since β k ∈ P \P ). Hence the Q k are pairwise distinct, as k ranges. (ii) For 0 ≤ i ≤ k, by Proposition 3.7 we get Conversely, if I ⊇ Q k , then I contains t k , hence, by Corollary 3.6, we get I = (f j , t k ), for some j ∈ {1, . . . , k + 1}, so I = (f j , t k ) = (f j , t j ) = Q j . (iii) This follows from (ii) and its proof, possibly replacing the Q i with their conjugates.
In order to complete the description of the lattice of F-basic ideals, it remains to find the basic ideals of O that do not contain a F-basic element.
Theorem 3.10. Let Q be a basic F-primary ideal not containing any basic element. Then In the case where k = ms + 1, s ∈ N, we know by Proposition 3.7 that Q k = (f k , t s ), and Q k ⊃ t s O ⊃ f Q k . We identify which of the ideals J a (see Theorem 2.5 and Theorem 3.10) coincides with t s O. By Proposition 3.7, β n = b n + c n ω, where c n / ∈ f Z, since β n / ∈ O for every n > 0 (Lemma 3.2). We show that Since the intermediate ideals between Q k and f Q k are pairwise incomparable (Theorem 2.5), it suffices to show that t s O ⊇ (f k+1 , f k a + t k ). Since N(β s ) = f ms = f k−1 we immediately get f k+1 ∈ f β s O, as well as the equality Hence it suffices to show that z = β k−s + aβ The desired conclusion follows.
The diagram below represents the lattice of F-primary ideals in the split case.

Ramified case.
Here we assume that f is ramified. Recall that this condition holds if and only if either f | d, when d ≡ 1 modulo 4, or f | 4d, when d ≡ 2, 3 modulo 4.
(ii) Since D is a Dedekind domain and Q is a basic F-primary ideal, QD is equal either to P 2 or to P 3 . In both cases, by Lemma 2.3, f QD ⊂ Q ⊆ QD, which is the statement.
Besides the basic elements t ∈ F such that F 2 ⊂ tO ⊂ F, which are associated to f by a unit of D (see Lemma 2.7), in the ramified case we may have other basic elements such that P 5 ⊂ tO ⊂ P 3 , according to whether P is a principal ideal of D or not, as the next result shows. Proposition 3.13. There exists a basic element t ∈ O such that P 5 ⊂ tO ⊂ P 3 if and only if P is a principal ideal of D. If this condition holds, say P = βD, for some β ∈ D, then every basic element is associated to f β by a unit of D.
Proof. Let us assume that P = βD, for some β ∈ D. Under the present circumstances we get N(β) = f and f = uβ 2 , for some unit u ∈ D. Clearly, β / ∈ O, otherwise β ∈ P ∩ O = F = P 2 , which is impossible. Hence, t = f β is a basic element, according to Lemma 1.3 and Proposition 1.5, since its norm is f 3 and t / ∈ f O. Since tD = β 3 D = P 3 , we get β 5 D = P 5 ⊂ tO ⊂ P 3 .
Conversely, let t ∈ O be a basic element such that P 5 ⊂ tO ⊂ P 3 . Using Lemma 2.3, we get tD = P 3 = f P , so P = t f D is a principal ideal of D. The last claim follows arguing as in Lemma 2.7.
Summarizing, the set of basic elements in the ramified case is: ii) {f u, f βv : u, v ∈ D * }, if P is a principal ideal of D generated by an element β.
We have N(f u) = f 2 , N(f βv) = f 3 , and the total number of these elements is 2τ (Lemma 2.7 and Corollary 3.13).
The diagram below represents the lattice of F-primary ideals in the ramified case. By Theorem 2.2 and the above description of the basic ideals, all the basic ideals, with the exception of F and P 3 , are invertible.