Rings of small rank over a Dedekind domain and their ideals
 Evan M. O’Dorney^{1}Email authorView ORCID ID profile
https://doi.org/10.1186/s4068701600540
© O’Dorney 2016
Received: 14 August 2015
Accepted: 12 January 2016
Published: 14 April 2016
Abstract
The aim of this paper is to find and prove generalizations of some of the beautiful integral parametrizations in Bhargava’s theory of higher composition laws to the case where the base ring \(\mathbb {Z}\) is replaced by an arbitrary Dedekind domain R. Specifically, we parametrize quadratic, cubic, and quartic algebras over R as well as ideal classes in quadratic algebras, getting a description of the multiplication law on ideals that extends Bhargava’s famous reinterpretation of Gauss composition of binary quadratic forms. We expect that our results will shed light on the statistical properties of number field extensions of degrees 2, 3, and 4.
Keywords
Dedekind domain Ring extension Gauss composition BhargavologyMathematics Subject Classification
Primary 13F05 11E20 11R11 11R16 Secondary 11E16 13B02 13A15 11E761 Background

Dirichlet, who discovered an algorithm simplifying the understanding and computation of the product of two forms, which we will touch on in greater detail (see Example 5.9).

Dedekind, who by introducing the nowstandard notion of an ideal, transformed Gauss composition into the simple operation of multiplying two ideals in a quadratic ring of the form \(\mathbb {Z}[(D + \sqrt{D})/2]\);

Bhargava, who in 2004 astounded the mathematical community by deriving Gauss composition from simple operations on a \(2\times 2\times 2\) cube [3].
A second thread that will be woven into this thesis is the study of finite ring extensions of \(\mathbb {Z}\), often with a view toward finite field extensions of \(\mathbb {Q}\). Quadratic rings (that is, those having a \(\mathbb {Z}\)basis with just two elements) are simply and classically parametrized by a single integer invariant, the discriminant. For cubic rings, Delone and Faddeev prove a simple lemma (as one of many tools for studying irrationalities of degree 3 and 4 over \(\mathbb {Q}\)) parametrizing them by binary cubic forms ([9], pp. 101ff). A similar classification for quartic and higher rings proved elusive until Bhargava, using techniques inspired by representation theory, was able to parametrize quartic and quintic rings together with their cubic and sextic resolvent rings, respectively, and thereby compute the asymptotic number of quartic and quintic rings and fields with bounded discriminant [4, 5, 6, 7]. The analytic virtue of Bhargava’s method is to map algebraic objects such as rings and ideals to lattice points in bounded regions of \(\mathbb {R}^n\), where asymptotic counting is much easier. (Curiously enough, the ring parametrizations seem to reach a natural barrier at degree 5, in contrast to the classical theory of solving equations by radicals where degree 4 is the limit.)
Bhargava published these results over the integers \(\mathbb {Z}\). Since then, experts have wondered whether his techniques apply over more general classes of rings; by far the most ambitious extensions of this sort are Wood’s classifications of quartic algebras [13] and ideals in certain nic algebras [14] over an arbitrary base scheme S. In this paper, all results are proved over an arbitrary Dedekind domain R. The use of a Dedekind domain has the advantage of remaining relevant to the original application (counting number fields and related structures) while introducing some new generality.
We will focus on two parametrizations that are representative of Bhargava’s algebraic techniques in general. The first is the famous reinterpretation of Gauss composition in terms of \(2 \times 2 \times 2\) boxes. Following [3], call a triple \((I_1,I_2,I_3)\) of ideals of a quadratic ring S balanced if \(I_1I_2I_3 \subseteq S\) and \(N(I_1)N(I_2)N(I_3) = 1\), and call two balanced triples equivalent if \(I_i = \gamma _i I'_i\) for some scalars \(\gamma _i \in S \otimes _\mathbb {Z}\mathbb {Q}\) having product 1. (If S is Dedekind, as is the most common application, then the balanced triples of equivalence classes correspond to triples of ideal classes having product 1.) Then:
Theorem 1.1

pairs \((S,(I_1,I_2,I_3))\) where S is an oriented quadratic ring of nonzero discriminant over \(\mathbb {Z}\) and \((I_1,I_2,I_3)\) is an equivalence class of balanced triples of ideals of S;

trilinear maps \(\beta : \mathbb {Z}^2 \otimes \mathbb {Z}^2 \otimes \mathbb {Z}^2 \rightarrow \mathbb {Z}\), up to \(\mathrm {SL}_2 \mathbb {Z}\)changes of coordinates in each of the three inputs (subject to a certain nondegeneracy condition).
Our parametrization is analogous, with one crucial difference. Whereas over \(\mathbb {Z}\), the only twodimensional lattice is \(\mathbb {Z}^2\), over a Dedekind domain R there are as many as there are ideal classes, and any such lattice can serve as the Rmodule structure of a quadratic algebra or an ideal thereof. Using a definition of balanced and equivalent essentially identical to Bhargava’s (see Definition 5.1), we prove:
Theorem 1.2

pairs \((S,(I_1,I_2,I_3))\) where S is an oriented quadratic algebra over R and \((I_1,I_2,I_3)\) is an equivalence class of balanced triples of ideals of S;

quadruples \((\mathfrak {a}, (M_1,M_2,M_3), \theta , \beta )\) where \(\mathfrak {a}\) is an ideal class of R, \(M_i\) are lattices of rank 2 over R (up to isomorphism), \(\theta : \Lambda ^2 M_1 \otimes \Lambda ^2 M_2 \otimes \Lambda ^2 M_3 \rightarrow \mathfrak {a}^3\) is an isomorphism, and \(\beta : M_1 \otimes M_2 \otimes M_3 \rightarrow \mathfrak {a}\) is a trilinear map whose three partial duals \(\beta _i: M_j \otimes M_k \rightarrow \mathfrak{A} M_i^*\) \((\{i,j,k\} = \{1,2,3\})\) have image a fullrank sublattice.
In particular R may have characteristic 2, the frequent factors of 1 / 2 in Bhargava’s exposition notwithstanding, and by weakening the nondegeneracy condition, we are able to include balanced triples in degenerate rings.
The second main result of our paper is the parametrization of quartic rings (with the quadratic and cubic parametrizations as preliminary cases). A key insight is to parametrize not merely the quartic rings themselves, but the quartic rings together with their cubic resolvent rings, a notion arising from the resolvent cubic used in the classical solution of the quartic by radicals.
Theorem 1.3

isomorphism classes of pairs (Q, C) where Q is a quartic ring (over \(\mathbb {Z}\)) and C is a cubic resolvent ring of Q;

quadratic maps \(\phi : \mathbb {Z}^3 \rightarrow \mathbb {Z}^2\), up to linear changes of coordinates on both the input and the output.
Our analog is as follows:
Theorem 1.4

isomorphism classes of pairs (Q, C) where Q is a quartic ring (over \(\mathbb {Z}\)) and C is a cubic resolvent ring of Q;

quadruples \((L,M,\theta ,\phi )\) where L and M are lattices of ranks 3 and 2 over R, respectively, \(\theta : \Lambda ^2 M \rightarrow \Lambda ^3 L\) is an isomorphism, and \(\phi : L \rightarrow M\) is a quadratic map.
Any quartic ring Q has a cubic resolvent, and if Q is Dedekind, the resolvent is unique.
1.1 Outline
The remainder of this paper is structured as follows. In Sect. 2, we set up basic definitions concerning projective modules over a Dedekind domain. In Sect. 3 and 4, respectively, we generalize to Dedekind base rings two classical parametrizations, namely of quadratic algebras over \(\mathbb {Z}\) and of their ideals. In Sect. 5, we prove Bhargava’s parametrization of balanced ideal triples (itself a generalization of Gauss composition) over a Dedekind domain. In Sect. 6, we work out in detail a specific example—unramified extensions of \(\mathbb {Z}_p\)—that allows us to explore the notion of balanced ideal triple in depth. In Sect. 7 and 8, we tackle cubic and quartic algebras, respectively, and in Sect. 9, we discuss results that would be useful when using the preceding theory to parametrize and count quartic field extensions.
2 Modules and algebras over a Dedekind domain
A Dedekind domain is an integral domain that is Noetherian, integrally closed, and has the property that every nonzero prime ideal is maximal. The standard examples of Dedekind domains are the ring of algebraic integers \(\mathcal {O}_K\) in any finite extension K of \(\mathbb {Q}\); in addition, any field and any principal ideal domain (PID), such as the ring \(\mathbb {C}[x]\) of polynomials in one variable, is Dedekind. In this section, we summarize the properties of Dedekind domains that we will find useful; for more details, see [10], pp. 9–18.
The salient properties of Dedekind domains were discovered through efforts to generalize prime factorization to rings beyond \(\mathbb {Z}\); in particular, every nonzero ideal \(\mathfrak {a}\) in a Dedekind domain R is expressible as a product \(\mathfrak {p}_1^{a_1}\cdots \mathfrak {p}_n^{a_n}\) of primes, unique up to ordering. Our motivation for using Dedekind domains stems from two other related properties. Recall that a fractional ideal or simply an ideal of R is a finitely generated nonzero Rsubmodule of the fraction field K of R, or equivalently, a set of the form \(a\mathfrak {a}\) where \(\mathfrak {a}\subseteq R\) is a nonzero ideal and \(a \in K^\times \). (The term “ideal” will from now on mean “(nonzero) fractional ideal”; if we wish to speak of ideals in the ringtheoretic sense, we will use a phrasing such as “ideal \(\mathfrak{A} \subseteq R\).”) The first useful property is that any fractional ideal \(\mathfrak{A} \subseteq K\) has an inverse \(\mathfrak {a}^{1}\) such that \(\mathfrak {a}\mathfrak {a}^{1} = R\). This allows us to form the group I(R) of nonzero fractional ideals and quotient by the group \(K^\times /R^\times \) of principal ideals to obtain the familiar ideal class group, traditionally denoted \({{\mathrm{Pic}}}R\). (For the ring of integers in a number field, the class group is always finite; for a general Dedekind domain this may fail, e.g. for the ring \(\mathbb {C}[x,y]/(y^2  (xa_1)(xa_2)(xa_3))\) of functions on a punctured elliptic curve.)
The second property that we will find very useful is that finitely generated modules over a Dedekind domain are classified by a simple theorem generalizing the classification of finitely generated abelian groups. For our purposes, it suffices to discuss torsionfree modules, which we will call lattices.
Definition 2.1
Let R be a Dedekind domain and K its field of fractions. A lattice over R is a finitely generated, torsionfree Rmodule M. If M is a lattice, we will denote by the subscript \(M_K\) its Kspan \(M \otimes _R K\) (except when M is denoted by a symbol containing a subscript, in which case a superscript will be used). The dimension of \(M_K\) over K is called the rank of the lattice M.
A lattice of rank 1 is a nonzero finitely generated submodule of K, i.e. an ideal; thus isomorphism classes of rank1 lattices are parametrized by the class group \({{\mathrm{Pic}}}R\). The situation for general lattices is not too different.
Theorem 2.2
(see [10], Lemma 1.5, Theorem 1.6, and the intervening Remark) A lattice M over R is classified up to isomorphism by two invariants: its rank m and its top exterior power \(\Lambda ^m M\). Equivalently, every lattice is a direct sum \(\mathfrak {a}_1 \oplus \cdots \oplus \mathfrak {a}_m\) of nonzero ideals, and two such direct sums \(\mathfrak {a}_1 \oplus \cdots \oplus \mathfrak {a}_m\), \(\mathfrak {b}_1 \oplus \cdots \oplus \mathfrak {b}_n\) are isomorphic if and only if \(m = n\) and the products \(\mathfrak {a}_1\cdots \mathfrak {a}_m\) and \(\mathfrak {b}_1\cdots \mathfrak {b}_n\) belong to the same ideal class.

the tensor product$$\begin{aligned} M \otimes N = \bigoplus _{\begin{array}{c} 1 \le i \le m \\ 1 \le j \le n \end{array}} \mathfrak {a}_i\mathfrak {b}_j (u_i \otimes v_j); \end{aligned}$$

the symmetric powersand the exterior powers$$\begin{aligned} {{\mathrm{Sym}}}^k M = \bigoplus _{1 \le i_1 \le \cdots \le i_k \le m} \mathfrak {a}_{i_1} \cdots \mathfrak {a}_{i_k}(u_{i_1} \otimes \cdots \otimes u_{i_k}) \end{aligned}$$of ranks \(\left( {\begin{array}{c}n+k1\\ k\end{array}}\right) \) and \(\left( {\begin{array}{c}n\\ k\end{array}}\right) \), respectively;$$\begin{aligned} \Lambda ^k M = \bigoplus _{1 \le i_1 < \cdots < i_k \le m} \mathfrak {a}_{i_1} \cdots \mathfrak {a}_{i_k}(u_{i_1} \wedge \cdots \wedge u_{i_k}) \end{aligned}$$

the dual lattice$$\begin{aligned} M^* = {{\mathrm{Hom}}}(M, R) = \bigoplus _{1 \le i \le m} \mathfrak {a}_i^{1} u_i^*; \end{aligned}$$

and the space of homomorphisms$$\begin{aligned} {{\mathrm{Hom}}}(M, N) \cong M^* \otimes N = \bigoplus _{\begin{array}{c} 1 \le i \le m \\ 1 \le j \le n \end{array}} \mathfrak {a}_i^{1}\mathfrak {b}_j (u_i^* \otimes v_j). \end{aligned}$$
Definition 2.3
If M and N are lattices (or, more generally, M is a lattice and N is any Rmodule), then a degreek map \(\phi : M \rightarrow N\) is an element of \(({{\mathrm{Sym}}}^k M^*) \otimes N\). A map to a lattice N of rank 1 is called a form.

Although such a degreek map indeed yields a function from M to N (evaluated by replacing every functional in \(M^*\) appearing in the map by its value on the given element of M), it need not be unambiguously determined by this function if R is finite. For instance, if \(R = \mathbb {F}_2\) is the field with two elements, the cubic map from \(\mathbb {F}_2^2\) to \(\mathbb {F}_2\) defined by \(\phi (x,y) = xy(x+y)\) vanishes on each of the four elements of \(\mathbb {F}_2^2\), though it is not the zero map.

Also, one must not confuse \(({{\mathrm{Sym}}}^k M^*) \otimes N\) with the space \(({{\mathrm{Sym}}}^k M)^* \otimes N\) of symmetric kary multilinear functions from M to N. Although both lattices have rank \(n\left( {\begin{array}{c}m+k1\\ k\end{array}}\right) \) and there is a natural map from one to the other (defined by evaluating a multilinear function on the diagonal), this map is not in general an isomorphism. For instance, the quadratic forms \(\phi : \mathbb {Z}^2 \rightarrow \mathbb {Z}\) arising from a symmetric bilinear form \(\lambda ((x_1,y_1),(x_2,y_2)) = ax_1x_2 + b(x_1y_2 + x_2y_1) + cy_1y_2\) are exactly those of the form \(\phi (x,y) = ax^2 + 2bxy + cy^2\), with middle coefficient even.
2.1 Algebras
2.2 Orientations
When learning about Gauss composition over \(\mathbb {Z}\), one must sooner or later come to a problem that vexed Legendre (see [8], p. 42): If one considers quadratic forms up to \(\mathrm {GL}_2 \mathbb {Z}\)changes of variables, then a group structure does not emerge because the conjugate forms \(ax^2 \pm bxy + cy^2\), which ought to be inverses, have been identified. Gauss’s insight was to consider forms only up to “proper equivalence,” i.e. \(\mathrm {SL}_2 \mathbb {Z}\) coordinate changes. This is tantamount to considering quadratic forms not simply on a rank2 \(\mathbb {Z}\)lattice M, but on a rank2 \(\mathbb {Z}\)lattice equipped with a distinguished generator of its top exterior power \(\Lambda ^2 M\). For general lattices over Dedekind domains, whose top exterior powers need not belong to the principal ideal class, we make the following definitions.
Definition 2.4
3 Quadratic algebras
Before proceeding to Bhargava’s results, we lay down as groundwork two parametrizations that, over \(\mathbb {Z}\), were known classically. These are the parametrizations of quadratic algebras and of ideal classes in quadratic algebras. The extension of these to other base rings has been thought about extensively, with many different kinds of results produced (see [12] and the references therein). Here, we prove versions over a Dedekind domain that parallel our cubic and quartic results.
Lemma 3.1
Quadratic algebras over R are in canonical bijection with rank2 Rlattices M equipped with a distinguished copy of R and a quadratic form \(\phi : M \rightarrow R\) that acts as squaring on the distinguished copy of R.
Proof
If there is a second copy of R on which \(N_{S/R}\) restricts to the squaring map, it must be generated by a unit of S with norm 1, multiplication by which induces an automorphism of the lattice with norm form. Hence we can eliminate the distinguished copy of R and arrive at the following arguably prettier parametrization:
Theorem 3.2
Quadratic algebras over R are in canonical bijection with rank2 Rlattices M equipped with a quadratic form \(\phi : M \rightarrow R\) attaining the value 1.
For our applications to Gauss composition it will also be helpful to have a parametrization of oriented quadratic algebras. An orientation \(\alpha : \Lambda ^2 S \rightarrow \mathfrak {a}\) can be specified by choosing an element \(\xi \) with \(\alpha (1 \wedge \xi ) = 1\). Since \(\xi \) is unique up to translation by \(\mathfrak {a}^{1}\), the parametrization is exceedingly simple.
Theorem 3.3
Example 3.4
When \(R = \mathbb {Q}\) (or more generally any Dedekind domain in which 2 is a unit), then completing the square shows that oriented quadratic algebras are in bijection with the forms \(x^2  ky^2\), \(k \in \mathbb {Q}\), each of which yields an algebra \(S = \mathbb {Q}[\sqrt{k}]\) oriented by \(\alpha (1 \wedge \sqrt{k}) = 1\).
Example 3.5
Example 3.6
4 Ideal classes of quadratic algebras
We can now parametrize ideal classes of quadratic algebras, in a way that partially overlaps [12]. To be absolutely unambiguous, we make the following definition for quadratic algebras that need not be domains:
Definition 4.1
Let S be a quadratic algebra over R. A fractional ideal (or just an ideal) of S is a finitely generated Ssubmodule of \(S_K\) that spans \(S_K\) over K. Two fractional ideals are considered to belong to the same ideal class if one is a scaling of the other by a scalar \(\gamma \in S_K^\times \). (This is clearly an equivalence relation.) The ideal classes together with the operation induced by ideal multiplication form the ideal class semigroup, and the invertible ideal classes form the ideal class group \({{\mathrm{Pic}}}S\).
The condition in bold means that, for instance, the submodule \(R \oplus \{0\} \subseteq R \oplus R\) is not a fractional ideal. Of course, any ideal that is invertible automatically satisfies it.
Theorem 4.2

ideal classes of oriented quadratic rings of type \(\mathfrak {a}\), and

rank2 lattices M equipped with a nonzero quadratic map \(\phi : M \rightarrow \mathfrak {a}^{1} \cdot \Lambda ^2 M\).
Proof

If f is irreducible, then \(S_K\) is a field, and \(M_K\) is an \(S_K\)vector space of dimension 1, isomorphic to \(S_K\).

If f has two distinct roots, then \(S_K \cong K \oplus K\). There are three different \(S_K\)modules having dimension 2 as Kvector spaces: writing \(I_1\) and \(I_2\) for the two copies of K within \(S_K\), we can describe them as \(I_1 \oplus I_1\), \(I_2 \oplus I_2\), and \(I_1 \oplus I_2\). But on the first two, every element of \(S_K\) acts as a scalar. If \(M_K\) were one of these, then the quadratic form \(\phi (\omega ) = \omega \wedge \xi \omega \) would be identically 0, which is not allowed. So \(M_K \cong I_1 \oplus I_2 \cong S_K\).

Finally, if f has a double root, then \(S_K \equiv K[\epsilon ]/\epsilon ^2\). There are two \(S_K\)modules having dimension 2 as a Kvector space: \(K\epsilon \oplus K\epsilon \) and \(S_K\). On \(K\epsilon \oplus K\epsilon \), \(S_K\) acts by scalars and we get a contradiction as before. So \(M_K \cong S_K\).
We now come to the equivalence between invertibility of ideals and primitivity of forms. Suppose first that \(\phi : M \rightarrow \mathfrak {a}^{1} \cdot \Lambda ^2 M\) is imprimitive, that is, there is an ideal \(\mathfrak {a}'\) strictly containing \(\mathfrak {a}\) such that \(\phi \) actually arises from a quadratic map \(\phi ' : M \rightarrow \mathfrak {a}'^{1} \cdot \Lambda ^2 M\). Following through the (first) construction, we see that \(\phi \) and \(\phi '\) give the same \(\xi \)action on \(I = M\) but embed it as a fractional ideal in two different rings, \(S = R \oplus \mathfrak {a}\xi \) and \(S' = R \oplus \mathfrak {a}'\xi \). We naturally have \(S_K \cong S'_K \cong K[\xi ]/(\xi ^2  q\xi + pr)\), and S is a subring of \(S'\). Suppose I had an inverse J as an Sideal. Then since I is an \(S'\)ideal, the product \(IJ = S\) must be an \(S'\)ideal, which is a contradiction.
Note that our proof of the invertibilityprimitivity equivalence shows something more: that any fractional ideal I of a quadratic algebra S is invertible when considered as an ideal of a certain larger ring \(S'\), found by “canceling common factors” in its associated quadratic form. The following relation is worth noting:
Corollary 4.3
Proof
The ring \(S'\) is the one occurring in the proof that imprimitivity implies noninvertibility, provided that the ideal \(\mathfrak {a}'\) is chosen to be as large as possible (i.e. equal to \((p \mathfrak {b}_1^2 + q \mathfrak {b}_1\mathfrak {b}_2 + r \mathfrak {b}_2^2)^{1}\)), so that I is actually invertible with respect to \(S'\). This \(S'\) must be the endomorphism ring \({{\mathrm{End}}}I\), or else I would be an ideal of an even larger quadratic ring. (We here need that \({{\mathrm{End}}}I\) is finitely generated and hence a quadratic ring. This is obvious, as it is contained in \(x^{1}I\) for any \(x \in S_K^\times \cap I\).)
Viewing \(\alpha \), by restriction, as an orientation on \(S'\), we have \(\alpha (\Lambda ^2 S') = \mathfrak {a}'\) and the formula is reduced to that for \(I^{1}\) above. \(\square \)
Example 4.4
5 Ideal triples
 (a)
\(I_1 I_2 I_3 \subseteq S\);
 (b)
\(N(I_1)N(I_2)N(I_3) = 1\). Here N(I) is the norm of the ideal I, defined by the formula \(N(I) = [A:I]/[A:S]\) for any \(\mathbb {Z}\)lattice A containing both S and I. (This should not be confused with the ideal generated by the norms of the elements of I. Even over \(\mathbb {Z}\), the two notions differ: \(2\cdot \mathbb {Z}[i]\) is an ideal of norm 2 in the ring \(\mathbb {Z}[2i]\), but every element of \(2\cdot \mathbb {Z}[i]\) has norm divisible by 4.)
Our task will be to generalize this result to an arbitrary Dedekind domain. First, the definition of balanced extends straightforwardly, provided that we define the norm of a fractional ideal I properly, as the index of I in S as an Rlattice. The resulting notion of balanced is a special case of the definition used in [14]:
Definition 5.1
 (a)
\(I_1 I_2 I_3 \subseteq S\);
 (b)
the image of \(\Lambda ^2 I_1 \otimes \Lambda ^2 I_2 \otimes \Lambda ^2 I_3\) in \((\Lambda ^2 S_K)^{\otimes 3}\) is precisely \((\Lambda ^2 S)^{\otimes 3}\).
The objects that we will use on the other side of the bijection are, as one might expect, not merely 8tuples of elements from R, because the class group intrudes. The appropriate notion is as follows:
Definition 5.2

three rank2 lattices \(M_1\), \(M_2\), \(M_3\);

an orientation isomorphism \(\theta : \Lambda ^2 M_1 \otimes \Lambda ^2 M_2 \otimes \Lambda ^2 M_3 \rightarrow \mathfrak {a}^3\);

a trilinear map \(\beta : M_1 \otimes M_2 \otimes M_3 \rightarrow \mathfrak {a}\) satisfying the following nondegeneracy condition: each of the three partial duals \(\beta _i : M_j \otimes M_k \rightarrow \mathfrak{A} M_i^*\) \((\{i,j,k\} = \{1,2,3\})\) has image a fullrank sublattice.
Theorem 5.3

balanced triples \((I_1,I_2,I_3)\) of ideals in an oriented quadratic ring S of type \(\mathfrak {a}\), up to scaling by factors \(\gamma _1\), \(\gamma _2\), \(\gamma _3 \in S_K^\times \) with product 1;

Bhargava boxes of type \(\mathfrak {a}\).
Remark
Two balanced ideal triples may be inequivalent for the purposes of this bijection even if corresponding ideals belong to the same class (see Example 5.9(d)). Consequently a Bhargava box cannot be described as corresponding to a balanced triple of ideal classes.
Proof

That \(I_1 I_2 I_3 \subseteq S\). Since \(I_1I_2I_3\) is the Rspan of the eight \(\mathfrak {b}_{1i}\mathfrak {b}_{2j}\mathfrak {b}_{3k}\tau _{ijk}\), this is evident from the construction of the \(\tau _{ijk}\).

That \(\prod _i \Lambda ^2(I_i) = \prod _i \Lambda ^2(S)\), and more strongly that the diagram commutes. This is a verification similar to that which showed the correspondence of the forms \(\phi _i\). Indeed, if we had recovered a triple of ideals that produced the correct \(\beta \) but the wrong \(\theta \), then the \(\phi \)’s as computed from \(\beta \) and the two \(\theta \)’s would have to mismatch.
5.1 Relation with the class group
Just as in the case \(R = \mathbb {Z}\), we can restrict to invertible ideals and get a new description of the class group.
Theorem 5.4

\(\phi _1 *\phi _2 *\phi _3 = 1\) for all \((\phi _1,\phi _2,\phi _3)\) arising from a Bhargava box;

\(\phi = 1\) if \(\mathfrak {a}^{1} \cdot \Lambda ^2 M\) is principal and \(\phi \) attains a generator of it
Proof
It is easy to see that a triple \((I_1,I_2,I_3)\) of invertible ideals in a ring S is balanced if and only if \(I_1I_2I_3 = S\). Each \(\sim \)equivalence class in the theorem is the family of forms corresponding to the ideals in a single ring, since we showed that the three forms arising from one Bhargava box belong to the same ring, and conversely if \(I_1\) and \(I_2\) belong to the same ring then \((I_1,I_2,I_1^{1}I_2^{1})\) is balanced (which also shows that \(\sim \) is truly an equivalence relation).
The condition that \(\phi \) attains a generator of \(\mathfrak {a}^{1} \cdot \Lambda ^2 M\) simply says that \(\phi \) matches the form corresponding to the entire ring S itself in Theorem 3.2, which is also the form corresponding to the principal class in Theorem 4.2. Now the theorem is reduced to the elementary fact that the structure of an abelian group is determined by the triples of elements that sum to 0, together with the identification of that 0element (without which any 3torsion element could take its place). \(\square \)
After establishing the corresponding theorem in [3] establishing a group law on quadratic forms, Bhargava proceeds to Theorem 2, which establishes a group law on the \(2\times 2\times 2\) cubes themselves, or rather on the subset of those that are “projective,” i.e. correspond to triples of invertible ideals. This structure is easily replicated in our situation: it is only necessary to note that the product of two balanced triples of invertible ideals is balanced. In fact, a stronger result holds.
Lemma 5.5
Let \((I_1,I_2,I_3)\) and \((J_1,J_2,J_3)\) be balanced triples of ideals of a quadratic ring S, with each \(I_i\) invertible. Then the ideal triple \((I_1J_1,I_2J_2,I_3J_3)\) is also balanced.
Proof
Corollary 5.6
(cf. [3], Theorems 2 and 12) The Bhargava boxes which belong to a fixed ring S (determined by the quadratic form (7)) and which are primitive (in the sense of having all three associated quadratic forms primitive) naturally form a group isomorphic to \(({{\mathrm{Pic}}}S)^2\).
Corollary 5.7
The Bhargava boxes which belong to a fixed ring S naturally have an action by \(({{\mathrm{Pic}}}S)^2\).
It is natural to think about what happens when the datum \(\theta \) is removed from the Bhargava box. As one easily verifies, multiplying \(\theta \) by a unit \(u \in R^\times \) is equivalent to multiplying the orientation \(\alpha \) of S by \(u^{1}\) while keeping the same ideals \(I_i\). Accordingly, we have the following corollary, which we have chosen to state with a representationtheoretic flavor:
Corollary 5.8
These orbits do not have a group structure. Indeed, the identifications cause a box and its inverse, under the group law of Corollary 5.6, to become identified.
Example 5.9
 (a)The boxes (for D even and odd, respectively), have as all three of their associated quadratic forms \(x^2  (D/4)y^2\) and \(x^2 + xy  (D1)/4\cdot y^2\), respectively, the defining form of the ring S of discriminant D. They correspond to the balanced triple (S, S, S). These are the “identity cubes” of [3], Eq. (3).
 (b)The boxes (for b even and odd, respectively), have as two of their associated quadratic forms the conjugatesand as the third associated form the form \(x^2  (D/4)y^2\) or \(x^2 + xy  (D3)/4\cdot y^2\) defining the ring S of discriminant \(D = b^2  4ac\). These boxes express the fact that the triple$$\begin{aligned} ax^2 + bxy + cy^2 \quad \text {and} \quad ax^2  bxy + cy^2 \end{aligned}$$is always balanced (compare Corollary 4.3). If \(\gcd (a,b,c) = 1\), we also get that I and \(\bar{I}\) represent inverse classes in the class group and that, correspondingly, \(ax^2 + bxy + cy^2\) and \(ax^2  bxy + cy^2\) are inverse under Gauss’s composition law on binary quadratic forms.$$\begin{aligned} (S, I, \alpha (\Lambda ^2 I)^{1} \bar{I}) \end{aligned}$$
 (c)The box has as associated quadratic formsAs Bhargava notes ([3], p. 249), Dirichlet’s simplification of Gauss’s composition law was essentially to prove that any pair of primitive binary quadratic forms of the same discriminant can be put in the form \((\phi _1,\phi _2)\), so that the multiplication relation that we derive from this box,$$\begin{aligned} \phi _1(x,y)= & {} dx^2 + hxy + fgy^2 \\ \phi _2(x,y)= & {} gx^2 + hxy + dfy^2 \\ \phi _3(x,y)= & {} fx^2 + hxy + dgy^2. \end{aligned}$$encapsulates the entire multiplication table for the class group.$$\begin{aligned} \phi _1 *\phi _2 = fx^2  hxy + dgy^2 \text { (or, equivalently, }dgx^2 + hxy  fy^2), \end{aligned}$$
 (d)For some examples not found in the classical theory of primitive forms, we consider the nonDedekind domain \(S = \mathbb {Z}[5i]\), whose ideal class semigroup was computed above (Example 4.4). Let us find all balanced triples that may be formed from the idealsof S. We compute$$\begin{aligned} S = \mathbb {Z}[5i], \quad A = {\mathbb {Z}}{\left\langle 5, 1+i\right\rangle }, \quad B = \mathbb {Z}[i] \end{aligned}$$For each triple \((I_1,I_2,I_3)\) of ideal class representatives, finding all balanced triples of ideals in these classes is equivalent to searching for all \(\gamma \in S_K^\times \) satisfying \(\gamma \cdot I_1I_2I_3 \subseteq S\) which have the correct norm$$\begin{aligned} \alpha (\Lambda ^2 S) = \mathbb {Z}, \quad \alpha (\Lambda ^2 A) = \mathbb {Z}, \quad \alpha (\Lambda ^2 B) = \frac{1}{5}\mathbb {Z}. \end{aligned}$$(the right side is an ideal of \(\mathbb {Z}\), so \(N(\gamma )\) is hereby determined up to sign, and as we are in a purely imaginary field, \(N(\gamma ) > 0\)). Using the class B zero or two times, we get four balanced triples$$\begin{aligned} \left\langle N(\gamma )\right\rangle = \frac{1}{\alpha (\Lambda ^2 I_1) \cdot \alpha (\Lambda ^2 I_2) \cdot \alpha (\Lambda ^2 I_3)} \end{aligned}$$each of which yields one Bhargava box. We get no balanced triples involving the ideal class B just once; indeed, it is not hard to show in general that if two ideals of a balanced triple are invertible, so is the third. The most striking case is \(I_1 = I_2 = I_3 = B\), for here there are two multipliers \(\gamma \) of norm 125 that send \(B^3 = \mathbb {Z}[i]\) into \(\mathbb {Z}[5i]\), namely \(10 + 5i\) and \(10  5i\) (we could also multiply these by powers of i, but this does not change the ideal B). The balanced triples \((B,B,(10+5i)B)\) and \((B,B,(105i)B\) are inequivalent under scaling, although corresponding ideals belong to the same classes. Thus we get two inequivalent Bhargava boxes with the same three associated forms, namely$$\begin{aligned} (S,S,S), \quad (S,A,iA), \quad (S,B,5B), \quad \text {and} \quad (A,B,5B), \end{aligned}$$
 (e)The triply symmetric boxes correspond to balanced triples of ideals that all lie in the same class; those that are projective—that is, whose associated forms are primitive—correspond to invertible ideal classes whose third power is the trivial class. This correspondence was used to prove estimates for the average size of the 3torsion of class groups in [1]. Our work suggests that similar methods may work for quadratic extensions of rings besides \(\mathbb {Z}\).
6 Another example: padic rings
Example 6.1
It is instructive to look at the local rings \(R = \mathbb {Z}_p\), where for simplicity we assume \(p \ge 3\). Thanks to the large supply of squares, the corresponding field \(K = \mathbb {Q}_p\) has but five (unoriented) quadratic extensions, namely those obtained by adjoining a square root of 0, 1, p, u, and pu where u is an arbitrary nonsquare modulo p. The quadratic ring extensions S of R then break up into five classes according to the corresponding extension \(S_K\) of K. We will work out one representative case, namely the oriented ring extensions \(S_n = \mathbb {Z}_p[p^n \sqrt{u}]\) corresponding to the unique unramified extension \(L = K[\sqrt{u}]\) of degree 2.
For any fractional ideal I of \(S_n\), we can pick an element of I of minimal valuation (recalling that L possesses a unique extension of the valuation on K) and scale it to be 1. Then \(S_n \subseteq I \subseteq S_0\), since \(S_0 = \mathbb {Z}_p[\sqrt{u}]\) is the valuation ring, and it is easy to see that the only possible ideals are the subrings \(S_0,S_1,\ldots , S_n\). In particular \(S_n\) is the only invertible ideal class, and the class group \({{\mathrm{Pic}}}S\) is trivial.
7 Cubic algebras
The second main division of our paper has as its goal the parametrization of quartic algebras. We begin with cubic algebras, for there the parametrization is relatively simple and will also furnish the desired ring structure on the cubic resolvents of our quartic rings. The parametrization was done by Delone and Faddeev for cubic domains over \(\mathbb {Z}\), by Gan, Gross, and Savin for cubic rings over \(\mathbb {Z}\), and by Deligne over an arbitrary scheme ([13], p. 1074 and the references therein). Here we simply state and prove the result over a Dedekind domain, taking advantage of the construction in [4], section 3.9.
Theorem 7.1
(cf. [2], Theorem 1; [13], Theorem 2.1; [11], Proposition 5.1 and the references therein) Let R be a Dedekind domain. There is a canonical bijection between cubic algebras over R and pairs consisting of a rank2 Rlattice M and a cubic map \(\phi : M \rightarrow \Lambda ^2 M\).
Proof
Given the cubic ring C, we let \(M = C/R\) so \(\mathfrak{A} = \Lambda ^2 M \cong \Lambda ^3 C\) is an ideal class. Consider the map \(\tilde{\phi } : C \rightarrow \mathfrak {a}\) given by \(x \mapsto 1 \wedge x \wedge x^2\). This is a cubic map, and if x is translated by an element \(a \in R\), the map does not change. Hence it descends to a cubic map \(\phi : M \rightarrow \mathfrak {a}\). We will show that each possible \(\phi \) corresponds to exactly one ring C.
Note that the basis change \(\xi _1 \mapsto \xi _1 + t_1\), \(\xi _2 \mapsto \xi _2 + t_2\) (\(t_i \in \mathfrak {a}_i^{1}\)) diminishes c and d by \(t_2\) and \(t_1\), respectively (as well as wreaking greater changes on the rest of the multiplication table). Hence, there is a unique choice of the lifts \(\xi _1\) and \(\xi _2\) such that \(c=d=0\).
Example 7.2

The trivial ring \(\mathbb {Z}[\epsilon _1,\epsilon _2]/(\epsilon _1^2,\epsilon _1 \epsilon _2, \epsilon _2^2)\) corresponds to the zero form 0.

Rings which are not domains correspond to reducible forms (e.g. \(\mathbb {Z}\oplus \mathbb {Z}\oplus \mathbb {Z}\) corresponds to \(xy(x+y)\)), and rings which have nontrivial nilpotents correspond to forms with repeated roots.

A monogenic ring \(\mathbb {Z}[\xi ]/(\xi ^3 + a\xi ^2 + b\xi + c)\) corresponds to a form \(x^3 + a x^2 y + b x y^2 + c y^3\) with leading coefficient 1. Accordingly a form which does not represent the value 1 corresponds to a ring that is not monogenic; for instance, the form \(5x^3 + 7y^3\) (which attains only values \(\equiv 0, \pm 2\) mod 7) corresponds to the subring \(\mathbb {Z}[\root 3 \of {5^2\cdot 7}, \root 3 \of {5\cdot 7^2}]\) of the field \(\mathbb {Q}[\root 3 \of {5^2 \cdot 7}] = \mathbb {Q}[\root 3 \of {5\cdot 7^2}]\), proving that this ring (which is easily checked to be the full ring of integers in this field) is not monogenic.

If a form \(\phi \) corresponds to a ring C, then the form \(n \cdot \phi \) corresponds to the ring \(\mathbb {Z}+ nC\) whose generators are n times as large. Hence the content \({{\mathrm{ct}}}(\phi ) = \gcd (a,b,c,d)\) of a form \(\phi (x,y) = ax^3 + bx^2y + cxy^2 + dy^3\) equals the content of the corresponding ring C, which is defined as the largest integer n such that \(C \cong \mathbb {Z}+ nC'\) for some cubic ring \(C'\). The notion of content (which is also not hard to define for cubic algebras over general Dedekind domains) will reappear prominently in our discussion of quartic algebras (see Sect. 8.2).
8 Quartic algebras
Definition 8.1
The resolvent \((M,\theta ,\phi )\) is called minimal if \(\phi \) has full image \(\phi (Q/R) = M\), that is, it is not really a map to any proper sublattice \(M' \subseteq M\) (cf. page 6). The resolvent is called numerical if \(\theta \) is an isomorphism.
Our minimal resolvent corresponds to the ring \(R^{\mathrm {inv}}\) in Bhargava’s treatment ([4], p. 1337), while our numerical resolvents correspond to Bhargava’s resolvent. The numerical resolvents are more suited to analytic applications, while the minimal resolvent has the advantage of being canonical (for nontrivial Q), as we prove below.
Example 8.2
8.1 Resolvent to ring
Our first result is that the resolvent encapsulates the data of the ring:
Theorem 8.3
(cf. [4], Theorem 1 and Proposition 10; [13], Corollary 1.2) Let \(\tilde{Q}\) and M be Rlattices of ranks 3 and 2 respectively. Let \(\theta : \Lambda ^2 M \rightarrow \Lambda ^3 \tilde{Q}\) be a morphism, and let \(\phi : \tilde{Q} \rightarrow M\) be a quadratic map. Then there is a unique quartic ring Q with an isomorphism \(Q/R \cong \tilde{Q}\) such that \((M,\theta ,\phi )\) is a resolvent for Q.
Proof

The notation \([\omega ]_i\) denotes the coefficient of \(\xi _i\) when \(\omega \) is expressed in terms of the basis \(\{1,\xi _1,\xi _2,\xi _3\}\).

Each of the first two equations is a direct calculation. For instance:$$\begin{aligned}&{[(\xi _i\xi _i)\xi _j  (\xi _i\xi _j)\xi _i]_k}\\&\quad = \left[ \left( c_{ii}^0 + c_{ii}^i\xi _i + c_{ii}^j\xi _j + c_{ii}^k\xi _k\right) \xi _j \left( c_{ij}^0 + c_{ij}^i\xi _i + c_{ij}^j\xi _j + c_{ij}^k\xi _k\right) \xi _i\right] _k \\&\quad = c_{ii}^i c_{ij}^k + c_{ii}^j c_{jj}^k + c_{ii}^k c_{jk}^k  c_{ij}^i c_{ii}^k  c_{ij}^j c_{ij}^k  c_{ij}^k c_{ik}^k \\&\quad = \left( c_{ii}^i  c_{ij}^j  c_{ik}^k\right) c_{ij}^k + c_{ii}^k\left( c_{jk}^k  c_{ij}^i\right) + c_{ii}^j c_{jj}^k \\&\quad = \epsilon \left( \lambda ^{ij}_{ik} \lambda ^{jj}_{ii}  \lambda ^{ii}_{ij}\lambda ^{ik}_{jj} + \lambda ^{ii}_{ik} \lambda ^{jj}_{ij}\right) \\&\quad = {{\mathrm{Pl}}}(jk,kk). \end{aligned}$$

The two lower diagrams show the instances of the associative law that produce a summand of \(c_{ii}^0\) or \(c_{ij}^0\), respectively. Each node in the diagrams yields a formula for \(c_{ii}^0\) or \(c_{ij}^0\) (having no denominator, and consequently belonging to the correct ideal \(\mathfrak {a}_i^{2}\) resp. \(\mathfrak {a}_i^{1}\mathfrak {a}_j^{1}\)); and where two nodes are joined by a line, the difference between the two corresponding formulas is expressible as a Plücker relation.
8.2 Ring to resolvent
Conversely, we will now study all possible resolvents of a given quartic ring Q. There is one case in which this problem takes a striking turn: the trivial ring \(Q = R[\mathfrak {a}_1\epsilon _1,\mathfrak {a}_2\epsilon _2,\mathfrak {a}_3\epsilon _3]/\sum _{i,j}(\mathfrak {a}_i\mathfrak {a}_j\epsilon _i\epsilon _j)\) where all entries of the multiplication table are zero. Here \(\phi \) can be an arbitrary map to a 1dimensional sublattice of M, or alternatively M and \(\phi \) can be chosen arbitrarily while \(\theta = 0\). For all other quartic rings, the family of resolvents is much smaller, as we will now prove.
Theorem 8.4
 (a)
Q has a unique minimal resolvent \((M_0, \theta _0, \phi _0)\);
 (b)
we have \(\theta _0(\Lambda ^2 M_0) = \mathfrak {c}\cdot \Lambda ^3 (Q/R)\), where \(\mathfrak {c}\) is the ideal (called the content of Q) characterized by the following property: For each ideal \(\mathfrak {a}\subseteq R\), there exists a quartic Ralgebra \(Q'\) such that \(Q \cong R + \mathfrak{A} Q'\) if and only if \(\mathfrak{A}  \mathfrak {c}\);
 (c)
all other resolvents \((M,\theta ,\phi )\), up to isomorphism, are found by extending \(\theta _0\) linearly to \(\Lambda ^2 M\), where M is a lattice with \([M:M_0]\mid \mathfrak {c}\), and taking \(\phi = \phi _0\);
 (d)
the numerical resolvents arise by taking \([M:M_0]=\mathfrak {c}\) in the preceding.
Proof
Write \(Q = R \oplus \mathfrak {a}_1\xi _1 \oplus \mathfrak {a}_2\xi _2 \oplus \mathfrak {a}_3\xi _3\). The multiplication table can be encoded in a family of \(c_{ij}^k\)’s, from which the fifteen values \(\lambda ^{ij}_{k\ell }\) are determined through (12). These \(\lambda ^{ij}_{k\ell }\) satisfy the fifteen Plücker relations by (13). It then remains to construct the target module M, the map \(\theta \), and the vectors \(\mu _{ij} \in \mathfrak {a}_i^{1}\mathfrak {a}_j^{1}M\) such that their pairwise exterior products \(\mu _{ij} \wedge \mu _{k\ell }\) have, via \(\theta \), the specified value \(\lambda ^{ij}_{k\ell }\).
The six \(\mu _{ij}\) are in complete symmetry at this point, and it will be convenient to denote \(\mu _{ij}\) by \(\mu _x\), where x runs over \(\{11,12,13,22,23,33\}\) or, if you prefer, \(\{1,2,3,4,5,6\}\). Likewise we write each \(\lambda ^{ij}_{k\ell }\) as \(\lambda ^x_y\) or simply \(\lambda _{xy}\).
If \(\mathfrak{A} \supseteq \mathfrak {c}\) for some \(\mathfrak{A} \subseteq R\), we can replace each of the three \(\mathfrak {a}_i\) with \(\mathfrak {a}^{1}\mathfrak {a}_i\) without changing the validity of the \(\lambda \)system. This means there is an extension ring \(Q' = R \oplus \mathfrak {a}^{1} \mathfrak {a}_1\xi _1 \oplus \mathfrak {a}^{1} \mathfrak {a}_2\xi _2 \oplus \mathfrak {a}^{1} \mathfrak {a}_3\xi _3\) with the same multiplication table as Q, and we see that \(Q = R + \mathfrak{A} Q'\). Conversely, given such a \(Q'\), we write its multiplication table with respect to the basis \(Q'/R = \mathfrak {a}^{1} \mathfrak {a}_1\xi _1 \oplus \mathfrak {a}^{1} \mathfrak {a}_2\xi _2 \oplus \mathfrak {a}^{1} \mathfrak {a}_3\xi _3\) and get that \(\lambda ^{ij}_{k\ell } \in \mathfrak {a}\mathfrak {a}_1\mathfrak {a}_2\mathfrak {a}_3\mathfrak {a}_i^{1}\mathfrak {a}_j^{1}\mathfrak {a}_k^{1}\mathfrak {a}_\ell ^{1}\), so \(\mathfrak {c}\subseteq \mathfrak {a}\). This proves (b).
Bhargava proved ([4], Corollary 4) that the number of (numerical) resolvents of a quartic ring over \(\mathbb {Z}\) is the sum of the divisors of its content. Likewise, we now have:
Corollary 8.5
Proof
Here we simply have to count the superlattices M of index \(\mathfrak {c}\) over a fixed lattice \(M_0\). The classical argument over \(\mathbb {Z}\) extends rather readily; for completeness, we give the proof.
In particular, we have the following.
Corollary 8.6
[cf. [4], Corollary 5] Every quartic algebra over a Dedekind domain possesses at least one numerical resolvent.
8.3 The cubic ring structure of the resolvent
In contrast to the classical presentation, the resolvent maps we have constructed take their values in modules, without any explicit connection to a cubic ring. In fact, there is the structure of a cubic ring already latent in a resolvent:
Theorem 8.7
To any quartic ring Q and resolvent \((M,\theta ,\phi )\) thereof, one can canonically associate a cubic ring C with an identification \(C/R \cong M\).
Remark
As stated, this theorem has no content, as one can take the trivial ring structure on \(R \oplus M\). However, we will produce a ring structure generalizing the classical notion of cubic resolvent. This C may be called a cubic resolvent of Q, the maps \(\theta \) and \(\phi \) being suppressed.
Proof
Two theorems concerning this cubic ring structure we will state without proof, since they are mere polynomial identities already implied by Bhargava’s work over \(\mathbb {Z}\). The first may be used as an alternative to Theorem 7.1 to determine the multiplicative structure on C; as Bhargava notes, it uniquely determines the ring C in all cases over \(\mathbb {Z}\) except when Q has nilpotents.
Theorem 8.8
Theorem 8.9
Example 8.10

First note that there is a resolvent map of \(\mathbb {C}\)algebras from \(Q_0 = \mathbb {C}^{\oplus 4}\) to \(C_0 = \mathbb {C}^{\oplus 3}\) given by the roots of the equationsolver’s resolventor, more accurately, by its reduction modulo \(\mathbb {C}\)$$\begin{aligned} (x,y,z,w) \mapsto (xy + zw, xz + yw, xw + yz) \end{aligned}$$supplemented of course by the standard identification$$\begin{aligned} \phi _0 : (x,y,z,0) \mapsto (xy  yz, xz  yz, 0), \end{aligned}$$Accordingly, if we have a quartic \(\mathbb {Z}\)algebra \(Q \subseteq Q_0\) and a cubic \(\mathbb {Z}\)algebra \(C \subseteq C_0\) on which the restrictions of \(\phi _0\) and \(\theta _0\) are well defined, then it automatically follows that \(C/\mathbb {Z}\) is a resolvent for Q with attached cubic ring structure C.$$\begin{aligned} \theta _0 : \Lambda ^2 (C_0/\mathbb {C}) \rightarrow \Lambda ^3 (Q_0/\mathbb {C}). \end{aligned}$$

As an example, consider the ringof content p, for each prime p. The minimal resolvent of Q comes out to be \(\phi _0(Q/\mathbb {Z}) = C'/\mathbb {Z}\), where$$\begin{aligned} Q = \mathbb {Z}+ p(\mathbb {Z}\oplus \mathbb {Z}\oplus \mathbb {Z}\oplus \mathbb {Z}) = \{(a,b,c,d) \in \mathbb {Z}^{\oplus 4} : a\equiv b\equiv c\equiv d \,\,\mathrm{mod}\,\,{p}\} \end{aligned}$$But \(C'\) is not a numerical resolvent of Q: it has index \(p^4\) in \(\mathbb {Z}^{\oplus 3}\), while Q has index \(p^3\) in \(\mathbb {Z}^{\oplus 4}\), so the restriction of \(\theta _0\) cannot possibly be an isomorphism. We must enlarge \(C'\) by a factor of p. Note that any subgroup C such that$$\begin{aligned} C' = \mathbb {Z}+ p^2 \cdot \mathbb {Z}^{\oplus 3}. \end{aligned}$$is a ring, since the product of two elements in \(p \cdot \mathbb {Z}^{\oplus 3}\) lies in \(p^2 \cdot \mathbb {Z}^{\oplus 3}\). So any ring of the form$$\begin{aligned} \mathbb {Z}+ p^2 \cdot \mathbb {Z}^{\oplus 3} \subseteq C \subseteq \mathbb {Z}+ p \cdot \mathbb {Z}^{\oplus 3} \end{aligned}$$is a numerical resolvent of Q. Letting [a : b] run over \(\mathbb {P}^1(\mathbb {Z}/p\mathbb {Z})\) yields the \(p+1\) numerical resolvents predicted by Theorem 8.4.$$\begin{aligned} C = \mathbb {Z}+ p^2 \cdot \mathbb {Z}^{\oplus 3} + \left\langle ap,bp,0\right\rangle \end{aligned}$$

Note that some of these resolvents are isomorphic under the automorphism group of Q, which is simply \(S_4\) acting by permuting the coordinates. One verifies that \(S_4\) acts through its quotient \(S_3\), which in turn permutes the three distinguished points \(0,1,\infty \) on \(\mathbb {P}^1(\mathbb {Z}/p\mathbb {Z})\). Accordingly, if we are using Theorem 8.3 to count quartic rings, the ring Q will appear not \(p+1\) times but \(\lceil p/6 \rceil + 1\) times (1 time if \(p = 2\)). This is no contradiction with Theorem 8.4, which gives the number of resolvents as maps out of the given ring Q.
9 Maximality
In order to convert his parametrization of quartic rings into one of quartic fields, Bhargava needed a condition for a ring to be maximal, i.e. to be the full ring of integers in a field. In like manner we discuss how to tell if a quartic ring Q over a Dedekind domain R is maximal in its fraction ring \(Q_K\) using conditions on a numerical resolvent. The first statement to make is that maximality is a local condition, i.e. can be checked at each prime ideal.
Proposition 9.1
Let Q be a ring of finite rank n over R. Q is maximal if and only if \(Q_\mathfrak {p}= Q \otimes _R R_\mathfrak {p}\) is maximal over \(R_\mathfrak {p}\) for all (nonzero) primes \(\mathfrak {p}\subseteq R\).
Remark
Proof
When Q is a domain, one can use the facts that Q is maximal if and only if it is normal (integrally closed in its fraction field) and that normality is a local property. A direct proof is also not difficult.
Suppose that Q is not maximal, so that there is a larger ring \(Q'\) with \(Q_K = Q'_K\). The nonzero Rmodule \(Q'/Q\) is pure torsion, so there is a prime \(\mathfrak {p}\) such that \((Q'/Q)_\mathfrak {p}= Q'_\mathfrak {p}/Q_\mathfrak {p}\) is nonzero, i.e. \(Q_\mathfrak {p}\) embeds into the larger ring \(Q'_\mathfrak {p}\).
Lemma 9.2
Proof
Theorem 9.3
 (a)
\(\mathfrak {p}^2\) divides \(a_{11}\), and \(\mathfrak {p}\) divides \(a_{12}\), \(a_{13}\), and \(b_{11}\).
 (b)
\(\mathfrak {p}\) divides \(a_{11}\), \(a_{12}\), \(a_{22}\), \(b_{11}\), \(b_{12}\), and \(b_{22}\).
 (c)
\(\mathfrak {p}^2\) divides \(a_{11}\), \(a_{12}\), and \(a_{22}\), and \(\mathfrak {p}\) divides \(a_{13}\) and \(a_{23}\).
 (d)
\(\mathfrak {p}\) divides all \(a_{ij}\).
Proof
The basic strategy is to find a suitable extension of the resolvent map to the ring \(Q'\) in Lemma 9.2, examining the cases where k is 1, 2, and 3.
First assume that Q has content 1 (by which we mean that the content ideal \({{\mathrm{ct}}}(Q)\) is the whole of R). Then k is 1 or 2 and \(Q'\) also has content 1. Both Q and \(Q'\) have unique (minimal and numerical) resolvents \((M,\theta ,\phi )\) and \((M',\theta ',\phi ')\), where (since \(Q_K = Q'_K\)) we have \(M \subseteq M' \subseteq M_K\), and \(\theta \) and \(\phi \) are the restrictions of \(\theta '\) and \(\phi '\). Also, since Q has index \(\mathfrak {p}^k\) in \(Q'\), M has index \(\mathfrak {p}^k\) in \(M'\).
If \(k = 2\), then the proof works similarly, except that \(M'\) takes one of the two forms \(\left\langle \pi \eta _1,\pi \eta _2\right\rangle \) and \(\left\langle \eta _1,\pi ^2\eta _2\right\rangle \). We leave it to the reader to write out the corresponding matrices \((A',B')\) and conclude cases (b) and (c) above, respectively.
Conversely, if one of the conditions (a)–(d) holds, the foregoing calculations suggest how to embed \(L = Q/R\) and M into lattices \(L'\) and \(M'\) with \(L \subsetneq L'\), such that the extensions of \(\theta \) and \(\phi \) still form a resolvent, yielding a quartic ring \(Q'\) that contains Q as a proper subring. \(\square \)
10 Conclusion
We have found the Dedekind domain to be a suitable base ring for generalizing the integral parametrizations of algebras and their ideals by Bhargava and his forebears. In each case, ideal decompositions \(\mathfrak {a}_1 \oplus \cdots \oplus \mathfrak {a}_n\) fill the role of \(\mathbb {Z}\)bases, and elements of appropriate fractional ideals take the place of integers in the parameter spaces. We have also shown that the notion of “balanced,” introduced by Bhargava to describe the ideal triples parametrized by general nondegenerate \(2\times 2\times 2\) cubes, has some beautiful properties and is worthy of further study. We expect that the methods herein will extend to replicate the other parametrizations in Bhargava’s “Higher Composition Laws” series and may shed light on the analytic properties of number fields and orders of low degree over base fields other than \(\mathbb {Q}\).
A generalization to quintic algebras over a Dedekind domain, following [6], has been found. Details are to appear in a forthcoming publication (see http://arxiv.org/abs/1511.03162).
Declarations
Acknowledgements
A previous version of this paper served as my senior thesis at Harvard College. I thank my thesis advisor, Benedict Gross, for many helpful discussions and comments. I thank Melanie Wood for clarifications on the relationships between my work and hers. I thank Arul Shankar for useful discussions, especially for informing me that he and Wood had been interested in the question answered by Corollary 8.6. I thank Brian Conrad for the suggestion that I work with Prof. Gross. I thank Noam Elkies and the editors for hunting down some subtle errors. I thank Ken Ono for suggesting RMS as a publication venue.
Competing interests
The author declares that they have no competing interests.
Open AccessThis article is distributed under the terms of the Creative Commons Attribution 4.0 International License (http://creativecommons.org/licenses/by/4.0/), which permits unrestricted use, distribution, and reproduction in any medium, provided you give appropriate credit to the original author(s) and the source, provide a link to the Creative Commons license, and indicate if changes were made.
Authors’ Affiliations
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