Non-unique factorization and principalization in number fields

We give a precise description of how the class group of a number field measures the failure of unique factorization in its ring of integers. Specifically, following ideas of Kummer, we determine the structure of all irreducible factorizations of an element in the ring of integers of a number field, and give a combinatorial description for the number of such factorizations. In certain cases, we show how quadratic forms can explicitly provide all such factorizations.


Introduction
One of the basic issues in algebraic number theory is the fact that for a number field K, and an integer n ∈ O K , an irreducible factorization of n may not be unique (up to ordering and units). Historically, there were three major attempts to deal with this: Gauss's theory of binary quadratic forms for quadratic fields K, Kummer's theory of ideal numbers, and Dedekind's theory of ideals. Kummer's approach was largely abandoned in favor of Dedekind's powerful theory. Ideals lead naturally to the general notion of the class group Cl K , which provides a way of measuring the failure of unique factorization in O K . In fact there are precise ways in which we can characterize the failure of unique factorization of O K in terms of the class group. For instance, Carlitz [3] showed that every irreducible factorization of n over O K has the same length for all n ∈ O K if and only if the class number h K of K is 1 or 2.
Nevertheless, the precise way in which the class group determines the irreducible factorizations of n is still not completely understood. Much of the research on nonunique factorizations to date has been devoted to the study of lengths of factorizations, motivated by [3], determining which elements have unique factorization, and related questions about asymptotic behavior. For an introduction to this subject, see [18], [11] or [7]. See [6] for a more comprehensive reference.
In this note we will give an explicit description of the structure of the irreducible factorizations of n in O K in terms of the prime ideal factorization of (n) which depends only upon the class group Cl K . Thus we have a very precise description of how the class group measures the failure of unique factorization in O K . For example, Carlitz's result is an immediate corollary. To do this, we use the idea of principalization, which is essentially Kummer's theory of ideal numbers, framed in the language of Dedekind's ideals. Specifically, we pass to an extension L which principalizes K, i.e., every ideal of K becomes principal in L. This means that all irreducible factorizations of n ∈ O K come from different groupings of a single factorization in O L .
To understand the irreducible factorizations of n in a more quantitative way, one natural question is, what is the number η K (n) of (nonassociate) irreducible factorizations of n in O K ? Of course if h K = 1, then η K (n) = 1 for all n ∈ O K . If h K = 2, then Chapman, Herr and Rooney [4] established a formula for η K (n) in terms of the prime ideal factorization of (n) in O K . However, it is a rather complicated recursive formula on the number of prime ideal factors of (n). (In fact the work [4] treats the more general case of Krull domains.) For work on determining for which n satisfy η K (n) = 1 see [18]; for asymptotic results on η K (n) see [9], [10] or [6].
We obtain, for an arbitrary number field K and n ∈ O K , a relatively simple combinatorial expression for η K (n), which appears to be about as simple as one could hope for. This formula is particularly simple in the case h K = 2, and we begin in Section 2 by explaining how to treat the standard class number 2 example of K = Q( √ −5). The expression we get for η K (n) is valid for any K with class number 2, and is considerably nicer than the formula in [4]. One can in fact treat the case of K = Q( √ −5) with quadratic forms, and this is what we do in Section 2. This yields, more than just the structure of the factorizations of n in O K , the explicit irreducible factorizations of n in terms of the representations of the primes dividing n (say if n ∈ Z) by certain quadratic forms.
In Section 3, we treat the case of an arbitrary number field K, and discuss some simple consequences. We remark that these results should in fact apply to more general Krull domains than O K by the theory of type monoids [6, Section 3.5], but this is not our focus here. In Section 4, we revisit the approach using quadratic forms presented in Section 2 for quadratic fields K.
We would like to thank Daniel Katz for helpful comments, and suggesting the use of formal power series. We also appreciate the comments provided by the referee after a careful reading of the manuscript.
2. An example: class number 2 . This field has discriminant ∆ = −20 and class group Cl K ≃ Z/2Z. Denote by C 1 the set of principal ideals in O K and C 2 the set of nonprinicpal ideals of O K . Now the reduced (positive binary quadratic) forms of discriminant ∆ are Q 1 (x, y) = x 2 + 5y 2 and Q 2 (x, y) = 2x 2 + 2xy + 3y 2 .
Let P 0 denote the primes p ∈ N which are not represented by Q 1 or Q 2 and P i denote the primes p ∈ N which are represented by Q i for i = 1, 2. Then P 0 is the set of inert primes in K/Q, P 1 is the set of primes p such that the ideal pO K factors into two principal ideals in O K , and P 2 is the set of primes p such that pO K factors into two nonprincipal ideals of O K . Set If p ∈ P 0 ∪ P 1 then any prime ideal p of O K lying above p is in C 1 , and if q ∈ P 2 , then any prime ideal of O K lying above q is in C 2 . Specifically, if q = 2 ∈ P ram 2 , then qO K = r 2 where r is the prime ideal (2, 1 + √ −5) of O K , and if q ∈ P unr 2 then q = qq where q andq are distinct prime ideals of O K . Hereq denotes the conjugate ideal of q in K/Q. Now let n ∈ O K be a nonzero nonunit, and write the prime ideal factorization of (n) as sq e2s s r f , where each p i ∈ C 1 , q j ∈ C 2 with conjugateq j , and the p i 's, q j 's,q j 's and r are all distinct. Since each p i = (π i ) for some irreducible π i of O K , any irreducible factorization of n must contain (up to units) π d1 1 · · · π dr r . Thus it suffices to consider irreducible factorizations of n ′ = n/(π d1 1 · · · π dr r ). Let q j be the prime in N such that q j lies above q j . Since q j is nonprincipal, we must have that q j ∈ P 2 , i.e., q j is represented by Q 2 . Note that we can factor the quadratic form into linear factors over the field L = K( √ 2). Hence, while q j is irreducible over O K (otherwise the prime ideal factors of q j O K would be principal), the fact that q j = Q 2 (x, y) for some x, y gives us a factorization Observing that α 1j and α 2j are conjugate with respect to the nontrivial element of Gal(K/Q), the ideals (α 1j ) ∩ O K and (α 2j ) ∩ O K must be conjugate ideals of O K which divide q j , and hence in some order equal q j andq j . Thus, up to a possible relabeling, we can write q j O L = (α 1j ) andq j O L = (α 2j ). Similarly rO L = ( √ 2). To simplify notation below, we set α 00 = √ 2 and e 00 = f . This means the following. If n ′ = β k is any irreducible factorization of n ′ in O K , we have (n ′ ) = q e11 1q e12 1 · · · q e1s sq e2s s r e00 = (β k ) as ideals of O K . By unique factorization of prime ideals, any (β k ) = q g11q g12 1 · · · q g1s sq g2s s r g00 where 0 ≤ g ij ≤ e ij . Passing to ideals of O L , we see each β k is a product of the α ij . In other words, all the irreducible factorizations of n ′ in O K , up to units, come from different groupings of the factorization n ′ = u α eij ij in O L , where u is a unit of O K . Thus to determine the factorizations of of n ′ in O K , it suffices to determine when a product of the α ij is an irreducible element of O K . But this is simple! Note from the factorization of Q 2 (x, y) in (1), we see that each α ij ∈ √ 2K. Hence the product of any two α ij lies in K, and therefore O K , and must be irreducible since no individual α ij ∈ O K . What we have done, together with the fact that the α ij 's are all nonassociate (they generate different ideals), proves the following.
If {a i } is a collection of distinct objects, denote the multiset containing each a i with cardinality m i by {a and n ∈ O K be a nonzero nonunit. Write the prime ideal factorization of (n) in O K as (n) = p di i q ej j , where each p i ∈ C 1 , q j ∈ C 2 and the p i 's and q j 's are all distinct. Let π i ∈ O K and α j ∈ O L such that p i = (π i ) and q j O L = (α j ). Then the irreducible nonassociate factorizations of n are precisely n = u π di i β k where u is a unit, each β k is a product of two (not necessarily distinct) α j 's and β k = α ej j .
In particular η K (n) is the number of ways we can arrange the multiset {α (ej ) j } in pairs, i.e., the number of partitions of this multiset into sub-multisets of size 2. In other words, if the number of distinct q j 's is m, then η K (n) is the coefficient of x ej j in the formal power series expansion of i≤j Note the final sentence is essentially a tautology.
Thus, in addition to η K (n), we have provided an explicit determination of the irreducible factorizations of an arbitrary n ∈ O K , provided one knows the (irreducible) factorization in O L . (The point is that there is a choice of which of α 1j and α 2j is q j O L and which isq j O L in the above argument.) However if n ∈ Z, then it suffices to know the prime factorization n = 2 f p and they are all distinct. For then each p ′ k is irreducible in O K , the irreducible factorization each of p i in O K is given by solving , and the factorization q j = α 1j α 2j in O L above is given by solving q j = 2x 2 + 2xy + 3y 2 . Here there is no need to worry which of α 1j and α 2j correspond to which of q j andq j since both q j andq j both occur to the same exponent e j .
Except for the explicit factorization we get from the quadratic forms Q 1 and Q 2 above, all of this is true for arbitrary number fields with class number 2.

Proposition 2.
If K is a number field with class number 2, and n ∈ O K is a nonzero nonunit, write the prime ideal factorization of (n) as (n) = p di One can either conclude this result from the work of [4] together with our example of K = Q( √ −5) or remark it is a special case of our main result, Theorem 1, below. In [4], the authors prove a recursive formula for η K (n), being recursive on the number of nonprincipal prime ideal factors of (n), which is independent of field K (in fact it is valid more generally for Krull domains also, but we will not stress this). As the formula in [4] is rather complicated, we will not state their complete formula here, but just give the first two cases to give the reader an idea of form of their expressions. In the notation of the corollary above, they show, assuming and the expression for η Xm+1 involves an m-fold summation over η Xm . Hence our approach of principalization and factoring the form Q 2 (x, y) in K( √ 2) provides a much nicer combinatorial answer to the question of what is η K (n). We now proceed to see what our result says in some simple cases. For the rest of this section, we maintain the notation of Proposition 2.
The first thing we observe is that e i must be even.
Corollary 1. An irreducible factorization of n is unique, i.e., η K (n) = 1, if and only if (i) there is at most one nonprincipal prime ideal dividing (n), or (ii) (n) = p di i · q e 1 q 2 where the p i 's are principal, the q i 's are nonprincipal, and e is odd. In particular, to return to the original example of K = Q( √ −5), if n ∈ Z, then η K (n) = 1 if and only if (i) it is not divisible by any primes in P unr 2 , i.e., any primes of the form q ≡ 3, 7 mod 20, or (ii) n = q p di i where the p i 's and q are primes with q ≡ 3, 7 mod 20 and each p i ∈ P 0 ∪ P 1 , i.e., p i ≡ 3, 7, mod 20 and p i odd. This classification of n ∈ N with η Q( √ −5) (n) = 1 was previously established by Fogels using an approach similar in spirit to ours in [5], where he used this to show that "almost all" n ∈ N do not have unique factorization.
Note this matches with the formula for η X2 (x 1 , x 2 ) in [4]. This was observed earlier in the case of elementary abelian 2-class groups ([10, Example 1]).
Proof. We want to count the number of ways we can pair e 1 x 1 's and e 2 x 2 's. This is simply determined by the number of x 1 's which are paired up with x 2 's. This can be any number k between 0 and min(e 1 , e 2 ) such that e i − k is even.
Proof. This is just the number of ways in which we can arrange the set When K = Q( √ −5) and q 1 , . . . , q k are distinct primes in P unr 2 , this means h K (q 1 · · · q k ) = (2k − 1)!!.

General Number Fields
Let K be an arbitrary number field and let Cl K = {C i } be the ideal class group of K. Denote the class of principal ideals in O K by I.
We say K i is a principalization field for C i if K i is an extension of K such that every ideal in C i becomes principal in O Ki . Such a field always exists. For example if C i has order m, then for any ideal a ∈ C i , we have a m is principal. Say a m = (a). Consequently a, and therefore every ideal in C i , becomes principal in the field K i = K( m √ a). We say L is a prinicpalization field for K if every ideal in O K becomes principal in O L . For instance if K i is a principalization field for C i for each C i ∈ Cl K , then the compositum L = K i is a principalization field for K. By the principal ideal theorem of class field theory, the Hilbert class field of K is a principalization field for K.
If α, β ∈ O K and α = uβ for a unit u ∈ O K , i.e., if α and β are associates, write α ∼ β. Theorem 1. Let K be a number field and Cl K = {C i }. Let n ∈ O K be a nonzero nonunit. Suppose the prime ideal factorization of nO K is (n) = (i,j)∈T p ij where the p ij 's are (not necessarily distinct) prime ideals such that p ij ∈ C i , and T is some finite index set. Let K i be a principalization field for C i , so We also remark that one could take each K i = L for any principalization field L of K, but we will see in the next section reasons why one may not always want to do this. In fact, for specific n, L need not be a principalization field for K, but just for the ideal classes containing ideals dividing nO K .
Proof. Suppose n = β l is an irreducible factorization of n in O K , i.e., each β l is a (nonunit) irreducible. By unique factorization of prime ideals, each (β l ) is a subproduct of p ij . Write (β l ) = (i,j)∈S p ij where S ⊆ T . Since (β l ) is principal, the subproduct of prime ideals yielding (β l ) must be trivial in the class group, i.e., (i,j)∈S C i = I. Further, S must be minimal such that the corresponding product in the class group is trivial, otherwise we would be able to write (β l ) as a product of two principal ideals, contradicting irreducibility.
Write S = {(i 1 , j 1 ), (i 2 , j 2 ), . . . , (i r , j r )}, so that where P iiji = p iiji O Ki 1 . Passing to O Ki 1 Ki 2 and using the fact that p i2j2 = (α i2j2 ) in O Ki 2 , we see that Proceeding inductively, we obtain (β l ) = (α i1j1 )(α i2j2 ) · · · (α ir jr ) as ideals in O L , yielding (ii) as asserted in the theorem. This proves that any irreducible factorization of n in O K is of the form stated above, namely that any irreducible factorization of n is obtained from a grouping of the terms in the (not necessarily irreducible) factorization n = u π i α ij in O L such that each group of terms is minimal so that the corresponding product in the class group Cl K is trivial. (Here u is some unit.) It remains to show that any such grouping gives an irreducible factorization of O K .
It suffices to show that if S is a minimal subset of T such that (i,j)∈S C i = I, then u (i,j)∈S α ij is an irreducible element of O K for some unit u ∈ O L . Suppose S is such a subset. Then (i,j)∈S q ij = (β) for some β ∈ O K . As above, looking at ideals in O L , we see β ∼ (i,j)∈S α ij , hence the product on the right is, up to a unit of O L , an element of O K . If β were reducible, say β = γγ ′ where γ, γ ′ ∈ O K are nonunits, then by unique factorization into prime ideals, we would have (γ) = (i,j)∈S ′ q ij where S ′ is a proper subset of S, i.e., (i,j)∈S ′ C i = I, contradicting the minimality of S.
where S runs over all minimal sub-multisets of U such that the product (i,j)∈S C i = I. Combinatorially, η K (n) is the number of ways one can partition the multiset {x eij ij } into minimal subsets V such that xij ∈V C i = I. Proof. Let K i be a principalization field for C i , and write p ij O Ki = (α ij ). Set L = K i . Then we have n ∼ T α eij ij = U α ij over O L . By the theorem, the irreducible factorizations of n in O K correspond to the partitions of U into minimal sub-multisets S such that S C i = I. Hence it remains to show that any two distinct partitions give nonassociate factorizations of n.
It suffices to prove that if S α ij ∼ S ′ α ij over O L for two sub-multisets S, S ′ of T , then S = S ′ . But this hypothesis means that Intersecting our ideals with O K gives S p ij = S ′ p ij , which means S = S ′ by unique factorization into prime ideals.
The current approach to investigating lengths and number of factorizations has primarily been through block and type monoids [6,Chapter 3]. Our theorem essentially gives the theory of block and type monoids in the case of rings of integers of number fields. In particular it can be used to provide new proofs of many known results in the theory of non-unique factorizations. Here we just illustrate the most basic example of Carlitz's result.
If n is a nonzero nonunit in O K and n = α i where each α i (not necessarily distinct) is a (nonunit) irreducible of O K , we say the number of α i 's occurring in this product (with multiplicity) is the length of this factorization. Proof. It is immediate from the theorem (or Proposition 2) that if h K ≤ 2, then every irreducible factorization of an element must have the same length. Suppose h K > 2.
First suppose Cl K has an element C of order e > 2. Then let p ∈ C and q ∈ C −1 be prime ideals of O K . Let n ∈ O K such that (n) = p e q e . Then one irreducible factorization of n is corresponds to the grouping (n) = (pq)(pq) · · · (pq) which has length e > 2. Another irreducible factorization of n corresponds to the grouping (n) = (p e )(q e ), which has length 2.
Otherwise Cl K has at least three elements C 1 , C 2 and C 3 = C 1 C 2 of order 2. Let p i ∈ C i be a prime ideal of O K for each i = 1, 2, 3. Let n ∈ O K such that (n) = p 2 1 p 2 2 p 2 3 . The two different groupings (p 1 p 2 p 3 )(p 1 p 2 p 3 ) and (p 2 1 )(p 2 2 )(p 2 3 ) give irreducible factorizations of n of lengths 2 and 3.
This proof might be considered a slight simplification, but it does not differ in any essential way from Carlitz's original proof. However, looking at this proof suggests that if Cl K ≃ Z/hZ then the ratio of the maximal length of an irreducible factorization of n to a minimal length is bounded by h 2 for any nonzero nonunit n ∈ O K . In fact this is true, and the maximum value of this ratio is called the elasticity ρ K of K. More generally, the Davenport constant D(Cl K ) of Cl K is defined to be the maximal m such that there is a product of length m which is trivial in Cl K but no proper subproduct is. Then the above theorem can be used to provide a new proof the known result (e.g., see [18]) that ρ K = D(Cl K )/2.
Specializing to certain cases, we can obtain simple formulas for η K (n) or criteria on when η K (n) = 1. A few examples were given in the case of class number 2 in the previous section. Here we give two more simple examples for arbitrary class number. Corollary 6. Let K be a quadratic field and p ∈ Z a rational prime. Let p be a prime of O K above p, and let m be the order of p in Cl K . If m = 1 or p is ramified in K/Q then η K (p n ) = 1 for all n ∈ N. Otherwise, η K (p n ) = ⌊ n m ⌋ + 1. Proof. If m = 1, the statement is obvious. If p is ramified, then pO K = p 2 , and again the result is immediate from our main result. Otherwise pO K = pp wherē p = p andp is the inverse of p in Cl K . Then any irreducible of O K dividing p corresponds to one of the groupings pp, p m orp m . The number of times p m appears in an irreducible grouping of p npn is the same as the number of timesp m will appear. Hence the irreducible factorizations of p n in O K are determined by the number of p m 's which appear in an irreducible groupings of p npn .
We remark that in [10], Halter-Koch showed for any number field K and x ∈ O K (or more generally a Krull monoid), η K (x n ) = An d + O(n d−1 ) for some A ∈ Q and d ∈ Z. Corollary 7. Let K be a number field and C ∈ Cl K be an ideal class of order m. Suppose n ∈ O K such that (n) = p 1 p 2 · · · p k where the p i 's are distinct prime ideals in C. Then η K (n) is the number of partitions of {1, 2, . . . , k} into subsets of size m, i.e., η K (n) = k! (m!) k/m (k/m)! . This is immediate from our main result, and a generalization of Corollary 3.

Explicit factorizations in quadratic fields
As we pointed out earlier, the approach via quadratic forms in Section 2 in some sense gives the irreducible factorizations of an element of O K in a more explicit fashion. Specifically, one does not know a priori the elements α ij occurring in Theorem 1 explicitly. Therefore one might ask in what generality can one apply the prinicipalization argument from Section 2 using quadratic forms. First we must restrict to the case of quadratic fields.
From now on, unless otherwise stated, let ∆ be a fundamental discriminant and K = Q( √ ∆) be the quadratic field of discriminant ∆. Suppose Q(x, y) = ax 2 + bxy + cy 2 is a primitive quadratic form of discriminant ∆. Then Q(x, y) factors into linear factors Note that K ′ = K if and only if a = m 2 or a = m 2 ∆ for some m ∈ Z. The latter is not possible since Q is primitive. The former implies that β ± ∈ O K ′ = O K if and only if a = 1.
Suppose K ′ = K and write Gal( In other words, we can use the factorization of a quadratic form to principalize the corresponding ideal class if and only if the quadratic form is ambiguous. This makes sense because an ideal class corresponds to an ambiguous form if and only if it has order ≤ 2 in the class group. On the other hand, the linear factorization of a binary quadratic form always happens over a quadratic extension, but one needs to use an extension of degree m to principalize an ideal class of order m in Cl K . To see this last assertion, suppose a is an ideal of order m in Cl K , so that a m = (α). If L principalizes a, say aO L = (β), then β m O L = αO L . Hence m √ uα ∈ O L for some unit u ∈ O K . No k-th root of uα is contained in K for 1 = k|m since a has order m. Therefore m|[L : K].
We now set up our notation for the statement and proof of the main result of this section. Let I be the class of principal ideals in Cl K , and C 1 , . . . , C k be the ideal classes in Cl K of order 2. We assume k ≥ 1.
If Q(x, y) = ax 2 + bxy + cy 2 is primitive of discriminant ∆, we define the ideal in ). We will say two forms are (weakly) equivalent if their corresponding ideals are equivalent, so that the equivalence classes of forms form a group isomorphic to Cl K . It is easy to see that Q and −Q correspond to the same ideal if Q is ambiguous.
Let Q j (x, y) = a j x 2 + b j xy + c j y 2 be an ambiguous form corresponding to an ideal in C j . Set K j = K( √ a j ) and L = K 1 K 2 · · · K k .
Lemma 2. K j is a principalization field for C j . Hence L is a principalization field for C 1 , . . . , C k .
Proof. Let a be the ideal of O K corresponding to Q j , andā be its conjugate. One easily checks thatā = a and a 2 = (a j ). Thus aO Kj = ( √ a j ) 2 . Since a ∈ C j , K j principalizes any ideal in C j .
Though we do not need this for the proposition below, it would be decent of us to determine the structure of L/K. This follows from the following. Proof. Write b = ra and e = sd. Note that b 2 − 4ac = ∆ then implies a|∆. Since ∆ is either squarefree or 4 times a squarefree number, we have that a is either squarefree, 2 times a squarefree number or 4 times a squarefree number. On the other hand, if 4|a, then 16|b 2 −4ac = ∆, which is not possible. Hence a is squarefree. Similarly d is a squarefree divisor of ∆.
For any squarefree n|∆ and m ∈ Z, we have a . First suppose d = a. Note that dividing r 2 a 2 − 4ac = s 2 a 2 − 4af by a implies r 2 a ≡ s 2 a mod 4, which implies r ≡ s mod 2 since 4 ∤ a. But this means the ideals a = (a, ra− a , we may replace R(x, y) with the equivalent form R(y, −x), thus interchanging d and f , and negating e. This means both e and f are now divisible by ∆ ′ a , so d cannot be by primitivity. This means d must be ±a, which we have just dealt with. (If d = −a, we can replace R by −R, which corresponds to the same ideal.) We remark that this lemma gives the following well known result.
Corollary 8. If Cl K contains a subgroup isomorphic to (Z/2Z) r , then ∆ has at least r + 1 distinct prime divisors.
Proof. Since there must be at least 2 r pairwise equivalent ambiguous forms a i x 2 + b i xy + c i y 2 of discriminant ∆, the above lemma and its proof imply that the a i 's and ∆ ai 's are distinct divisors of ∆. Each a i is always squarefree, and if ∆ ai is not squarefree, then ∆ 4ai is, and it is distinct from the other divisors. Thus ∆ has at least 2 r+1 distinct squarefree divisors, so it must have at least r + 1 distinct prime factors.
One could refine this had we been using the notion of proper equivalence classes of quadratic forms, which we do not need for our purpose. Precisely, if r is maximal so that Cl K contains a subgroup isomorphic to (Z/2Z) r , then one can show that there are either r + 1 or r + 2 distinct prime divisors of ∆. The first case occurs when the extended genus field of K equals the genus field of K, and the second when they are different. (See, e.g., [15]).
However, our interest in the previous lemma is in the structure of L/K (which is closely related to the genus field and extended genus field of K, but different from both in general). We know {I, C 1 , . . . , C k } is the subgroup of Cl K generated by all elements of order 2. We put r such that 2 r = k + 1 so that this subgroup is isomorphic to (Z/2Z) r . Corollary 9. L is an abelian extension of K of degree 2 r and Gal(L/K) ≃ (Z/2Z) r .
Proof. Clearly [L : K] ≤ 2 r and is a power of 2 by construction. Moreover L/K is Galois and the Galois group is an elementary abelian 2-group because L is obtained from K by adjoining square roots of K. By the previous lemma, L/K has 2 r − 1 subextensions of degree 2 over K, so [L : K] = 2 r .
where the p i 's are primes in N represented by the principal form Q 0 (x, y) = x 2 +b 0 xy +c 0 y 2 of discriminant ∆, the q jk 's are primes in N represented by Q j and the r ℓ 's are primes in N not represented by any form of discriminant ∆. Write each p i = Q 0 (u i , v i ) and q jk = Q j (x jk , y jk ) for u i , v i , x jk , y jk ∈ Z. Let Then the irreducible factorizations of n in O K , up to units, are precisely given by the O K -irreducible groupings of the factorization By an O K -irreducible grouping of a product γ in O L , we of course mean a grouping of the terms such that the product of each group of terms is (up to a unit of O L ) an irreducible in O K . In the above proposition, each α ± i and r ℓ is already an irreducible of O K , and the elements β ± jk correspond to the ideal class C j . A product of these β ± jk 's is, up to a unit of O L , an irreducible in O K if and only if the corresponding product of ideal classes is trivial but no proper subproduct is. In fact, such a product of β ± jk 's must actually be an irreducible of O K , since the fact that β ± jk ∈ √ a j K implies such a product lies in O K .
Proof. It is obvious that any prime p i represented by Q 0 satisfies p i O K = p 1 p 2 for some principal prime ideals p 1 and p 2 of O K , since Q 0 factors over K. Further any prime q jk represented by Q j satisfies q jk O K = q 1 q 2 for some prime ideals q 1 , q 2 ∈ C j (see [1, p. 143]). Lastly each r ℓ is inert in K/Q. Now apply Theorem 1.
The above gives a complete answer for the factorization of rational integers n in O K when Cl K ≃ (Z/2Z) r , i.e., when there is one class per genus in the form class group, and a partial answer for other quadratic fields. We end with two examples and some remarks on principalization fields. Example 1. Let ∆ = −87. Then K = Q( √ −87) has class number h K = 6. The principal form is Q 0 (x, y) = x 2 + xy + 22y 2 and there is one other ambiguous form up to equivalence, Q 1 (x, y) = 3x 2 + 3xy + 8y 2 . Let n = 14145 = 3 · 5 · 23 · 41. We of K = Q( √ 5) shows. (It is of course closely related to H, and more generally to the genus field of K.) In general for a number field K it is natural to ask, what we can say about the minimal abelian extensions L which principalize K? By the remarks after Lemma 1, we know m|[L : K] for every cyclic group of order m contained in Cl K . One might be tempted to posit that [L : K] ≥ h K , or even that Gal(L/K) contains Cl K , but this turns out to be false. For instance, the Hilbert class field H of K is an abelian extension of K with Gal(L/K) ≃ Cl K and always principalizes K, but proper subextensions of H may also principalize K ( [12], [13], [14]). We will not survey the literature on principalization, but refer to the expositions [16], [17] and [19], as well as point out the recent works [8] and [2] which study extensions of K not contained in its Hilbert class field.