Galois module structure of square power classes for biquadratic extensions

For a Galois extension $K/F$ with $\text{char}(K)\neq 2$ and $\text{Gal}(K/F) \simeq \mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}$, we determine the $\mathbb{F}_2[\text{Gal}(K/F)]$-module structure of $K^\times/K^{\times 2}$. Although there are an infinite number of (pairwise non-isomorphic) indecomposable $\mathbb{F}_2[\mathbb{Z}/2\mathbb{Z}\oplus\mathbb{Z}/2\mathbb{Z}]$-modules, our decomposition includes at most $9$ indecomposable types. This paper marks the first time that the Galois module structure of power classes of a field has been fully determined when the modular representation theory allows for an infinite number of indecomposable types.

1. Introduction 1.1. Background and Motivation. Let K be a field, and write ξ p for a primitive pth root of unity. We write K sep for a separable closure of K, and K(p) for the maximal p-extension within K sep . Each of these extensions is Galois. The absolute Galois group of K is the group G K := Gal(K sep /K). The group G K (p) := Gal(K(p)/K) is the maximal pro-p quotient of G K . For convenience, we will call G K (p) the absolute p-Galois group of K. One of the major open problems in Galois theory is to determine those profinite groups G for which there exists some field K with G K ≃ G; i.e., to distinguish absolute Galois groups within the class of profinite groups. This problem is very difficult. The analogous question for pro-p groupsto distinguish absolute p-Galois groups within the class of pro-p groups -is also unsolved and extremely difficult.
How does one look for those properties that distinguish absolute p-Galois groups from the broader class of pro-p groups? To motivate the perspective pursued in this paper, note that since G K (p) is a pro-p group, it is natural to study it recursively through its Frattini subgroup and its quotient. This quotient is the maximal elementary p-abelian quotient of G K (p), which by Kummer theory (assuming ξ p ∈ K) corresponds to J(K) := K × /K ×p . In the case that K is itself a Galois extension of a field F , one then has a natural action of Gal(K/F ) on J(K).
(Throughout the remainder of this discussion we will assume that Gal(K/F ) is a p-group, just to stay firmly planted in the context of p-groups.) One field-theoretic lens for studying and each of them is cyclic. In contrast, if G is a non-cyclic elementary p-abelian group then there are infinitely many isomorphism classes of indecomposable F p [G]-modules (and often it is impossible to give a full classification of indecomposables). There has been some work which provides partial information about Galois modules in these more complicated settings, recovering information about the Socle series or arguing that the modules are constant Jordan type in special cases ( [1,11,26]). However, these modules were not determined completely.
In this paper we provide a decomposition for J(K) as an F 2 [Gal(K/F )]-module when Gal(K/F ) ≃ Z/2Z ⊕ Z/2Z, without any restriction on K other than char(K) = 2. The decomposition follows the two themes that have arisen in the context of cyclic Galois groups: the module structure is far more stratified than one would expect for a general module (across all fields K, the summands are drawn from at most 9 indecomposable types), and the decomposition can be determined using relatively concrete techniques and the assistance of Hilbert 90 (see [10] for a discussion on how one interprets Hilbert 90 for biquadratic extensions). Undoubtedly this stratified decomposition -both the appearance of some summand types and the exclusion of others -can be translated into new and exciting group-theoretic properties of absolute 2-Galois groups. The authors are currently looking into such results.
A decomposition of J(K) when Gal(K/F ) ≃ Z/2Z ⊕ Z/2Z was completed by the first, second, and fourth authors in 2005 using more technical machinery. A deeper dive into the module from this perspective was explored in [13] under the joint supervision of Minač and Swallow. The impetus for revisiting this problem using more ubiquitous tools was to give greater insight into how decompositions for J(K) (and its ilk) could be carried out when Gal(K/F ) is some other elementary p-abelian group. This approach has already met with success: it has allowed us to exclude one summand type that appeared in the original decomposition from 2005, and it inspired the recent decomposition of the parameterizing space of elementary p-abelian extensions of K as a module over F p [Gal(K/F )] whenever G F (p) is a free pro-p group and Gal(K/F ) is any finite p-group (see the remark after Theorem 1). We are hopeful that the techniques we develop here can inspire the next steps towards investigations of a broader class of elementary p-abelian Galois modules.

Statement of Main Result.
We first set terminology that will hold for the balance of the paper. Suppose that F is a field with char(F ) = 2 and that K/F is an extension with G := Gal(K/F ) ≃ Z/2Z ⊕ Z/2Z. Let a 1 , a 2 ∈ F be given so that K = F ( √ a 1 , √ a 2 ), and let σ 1 , σ 2 ∈ Gal(K/F ) be their duals; that is, we have σ i ( √ a j ) = (−1) δ ij √ a j . For i ∈ {1, 2} we define K i = F ( √ a i ). Write H i for the subgroup of Gal(K/F ) which fixes elements in K i , and G i for the corresponding quotient group: G i := Gal(K i /F ) = {id, σ i }. In the same spirit, write K 3 = F ( √ a 1 a 2 ), denote the subgroup of Gal(K/F ) which fixes K 3 as H 3 , and use G 3 for the corresponding quotient G/H 3 = Gal(K 3 /F ). To round out the notation, let H 0 = {id} (the elements which fix the extension K/F ) and H 4 = Gal(K/F ) (the elements which fix the extension F/F ), and use G 0 and G 4 for their quotients. (See Figure 1.) In our result below, we use Ω −n and Ω n to denote certain indecomposable modules of dimension 2n + 1; more information on these modules can be found in Section 2. Figure 1. The lattice of fields for K/F , with corresponding Galois groups Theorem 1. Suppose that char(K) = 2 and that Gal( where • X is isomorphic to one of the following: Remark. When char(K) = 2, elementary 2-abelian extensions of K are parameterized by F 2 -subspaces of K/℘K, where ℘K = {k 2 − k : k ∈ K}. It is therefore natural to ask whether K/℘(K) can be decomposed as an F 2 [Gal(K/F )]-module as well. The answer is a resounding "yes." In fact, this is a special case of a far more general theorem from the forthcoming paper [14]: for any prime p and any Galois extension K/F so that Gal(K/F ) is a finite p-group, if G F (p) is a free pro-p group, then the parameterizing space of elementary p-abelian extensions of K decomposes into a free summand and a single summand isomorphic to Ω −2 Gal(K/F ) .
1.3. Outline of paper. In Section 2 we review some basic facts concerning modules over Section 3 is devoted to producing a "large" module whose fixed part is the "obvious" componenent [F × ] within J(K) G ; the key is to give a filtration of [F × ] that is sensitive to image subspaces coming from particular elements of F 2 [G]. Section 4 aims to find a module whose fixed part spans a complement to [F × ] in J(K) G . This requires a deeper understanding of how J(K) G behaves under the norm maps associated to the intermediate extensions K/K i (for i ∈ {1, 2, 3}). The proof of Lemma 4.5 gives our first appearance of a Hilbert 90 result for biquadratic extensions. Section 5 has another result related to Hilbert 90 for biquadratic extensions (Lemma 5.1), as well as the proof of Theorem 1. In Section 6 we discuss the realizability of some of the possibilities for the X summand in terms of the solvability (or non-solvability) of particular embedding problems.

A Primer in Diagramatic Thinking in Module Theory
We will use G to denote the Klein 4-group with generators σ 1 and σ 2 . When M is an F 2 [G]-module, we assume that M's structure is multiplicative, so that the module action is written exponentially. Despite this, if U, V are submodules of a larger F 2 [G]-module W , we will still write U + V for the set {uv : u ∈ U, v ∈ V }, and we will use U ⊕ V to indicate this set when U ∩ V is trivial.
Throughout this paper we will be considering the solvability of certain systems of equations within various F 2 [G]-modules. Though one could of course write these systems out, it will often be convenient to have graphical representations for the equations. We adopt the convention that an arrow between elements denotes that one is the image of another through some given element of F 2 [G], with the direction of the arrow indicating the acting element from F 2 [G]. If the arrow points down and to the left, this indicates that the bottom element is the image of the upper element under 1 + σ 2 , and likewise if the arrow points down and to the right this means the lower element is the image of the upper element under 1 + σ 1 . In the event that the action of 1 + σ 1 and 1 + σ 2 is the same on a given element, then we write the image immediately below, and use two bent arrows to signify the equality of the two actions. Figure 2 gives some basic examples. Figure 2. A sampling of linear equations. On the left we have the relation α 1+σ 2 = α 1 ; in the middle we have the simultaneous equations in β and β 1 given by β 1+σ 1 = β 1+σ 2 = β 1 ; and on the right we have the simultaneous equations in γ, γ 1 , γ 2 given by γ 1+σ 2 = γ 1 and γ 1+σ 1 = γ 2 .
Since these diagrams represent simultaneous linear equations in the module, we will say that a solution to a system of equations is a solution to the corresponding diagram; if we have some fixed values for particular parameters in a system of equations, and there exist values for the remaining parameters so that the underlying system is solved, then we will say that the diagram is solvable for those (original) fixed values. For example, to say that the diagram on the left side of Figure 2 is solvable for some particular α 1 is equivalent to saying that α 1 is in the image of 1 + σ 1 .
Our decomposition will not require us to have a classification of indecomposable F 2 [G]modules, but for the reader's benefit we review some basic information about these modules. For a full treatment, the reader can consult [3,Theorem 4.3.3]. There are seven ideals in the ring F 2 [Z/2Z ⊕ Z/2Z], and hence six cyclic, non-trivial indecomposable submodule classes. Aside from the even-dimensional cyclic modules, there are also families of indecomposable even-dimensional F 2 [G]-modules that correspond to certain rational canonical form matrices. These will not appear in our decomposition. There are also odd-dimensional indecomposable F 2 [G]-modules: for each odd number 2n + 1 with n > 1, there are two irreducible F 2 [G]modules of dimension n, denoted Ω n and Ω −n . As it happens, our decomposition of J(K) will only require the cyclic modules we have already introduced together with Ω 1 , Ω 2 , Ω −1 , and Ω −2 . We will need formal definitions for these latter modules, but there is no additional cost to define Ω n and Ω −n in general. Using our depiction scheme, these modules are shown in Figure 3. One key fact we'll use about F 2 [G]-module is that we can detect independence of two Proof. Of course if M ∩ N = {1}, then M G ∩ N G = {1} as well. Suppose, then, that M G ∩ N G = {1}, and let w ∈ M ∩ N be given. If w is nontrivial, then w is isomorphic to precisely one of the following: , or w ≃ Ω −1 , then w 1+σ 1 is a nontrivial element in W G ; but this again leads to a contradiction, since then we again have w 1+σ , then w 1+σ 2 is the nontrivial element in W G which leads to a contradiction, and if w ≃ F 2 [G] then the contradictory nontrivial element of W G is w (1+σ 1 )(1+σ 2 ) .

A maximal submodule with fixed part [F × ]
In Section 2 we saw that fixed submodules play an important role in determining independence amongst F 2 [G]-modules. Of course the most natural fixed submodule of J(K) is [F × ].
Our objective in this section will be to find a "sufficiently large" submoduleĴ of J(K) for whichĴ G = [F × ]. For the purposes of the decomposition that we are building, being "sufficiently large" will mean thatĴ contains solutions to certain systems of equations, assuming such equations have solutions within the full module J(K).
In a certain sense, we are most interested in finding free summands -by which we mean free over F 2 [G i ] for some i ∈ {0, 1, 2, 3, 4}) -with the general philosophy that larger submodules are preferrable. Hence primary preference goes to free (cyclic) F 2 [G]-modules, and secondary preference goes to free (cyclic) F 2 [G i ]-modules for i ∈ {1, 2, 3}; for concreteness, we give preference to i = 1 over i = 2, and i = 2 over i = 3. We finish with free F 2 [G 4 ]-modules (i.e., trivial modules).
The issue in pursuing this agenda is that there are potential interrelations between these free modules. For example, suppose a free cyclic Hence in our pursuit of free submodules, we are obliged to look for solutions to this type of system and ensure our decomposition of [F × ] captures these elements.
With all this in mind, let us move toward statements that are more precise. In Figure 4 we introduce 5 subspaces of [F × ] that capture the ideas we alluded to in the previous paragraphs. We denote these spaces A, V, B, C, and D.
It is readily apparent that A ⊆ V, and furthermore that V is a subspace of both B and C. We just observed that B ∩ C = V. Continuing in the theme of being careful about interrelations that exist between these subspaces, the following lemma considers how elements of D are related to elements from B + C.
Lemma 3.1. Let B, C, and D be defined as in Figure 4.
is solvable for some Diagram for B [γ] [f ] [1] Diagram for C [γ] [f ] Diagram for D Proof. Suppose first that Equation (1) holds. From this we see that [1].
Hence we will be particularly interested in understanding solutions to Equation (1) that come from outside V. The following lemma characterizes such solutions.
Then there exists an Proof. That B W and C W are subspaces follows since Equation (1) and so we see The same argument, of course, shows that for a given [c] ∈ C W there exists a unique [b] ∈ B W for which Equation (1) is solvable. We define φ W as the function which associates to each [b] ∈ B W its corresponding [c] ∈ C W . The fact that the equations represented by Equation (1) are linear implies that φ W is linear as well, and hence an isomorphism of F 2spaces.
We are now prepared to state and prove the main result in this section. Theorem 2. There exists a submoduleĴ so thatĴ G = [F × ], and for whicĥ Proof. Choose A to be an F 2 -basis for A. By the definition of A, for each Certainly the appropriate Ω 2 relations hold by construction, so we simply need to ensure that there are no additional relations.
. As before, we in fact have Let B 0 be a basis for a complement to We have already detailed the fixed parts of each submodule, and we will use this to show that the sum is direct.   We have left to settle the statement for M = D. This will take a bit more work. Since D 0 is a basis for a complement to (B + C) ∩ D within D, we can write From our previous equations we get This means that [b W ] ∈ B W , and by Lemma 3

4.
A module whose fixed part complements [F × ] in J(K) G Lemma 2.1 tells us that independent summands of J(K) have independent fixed parts. Since we've already constructed a module whose fixed part is [F × ], we now are interested in finding a complementary module whose fixed part spans a complement to [F × ] in J(K) G -at least to the degree that such a goal is achievable at all. Ultimately this search will culminate in Theorem 3 at the end of this section, but to work towards this result we must first determine precisely which elements from J(K) G come from [F × ].
Kummer theory tells us that we can determine whether an element [γ] ∈ J(K) G comes from [F × ] by examining the Galois group of the extension it generates over F : The following result gives a slightly more nuanced view of this phenomenon. Note that in this result -and hence for much of the duration of this section -we use the notation [γ] i to indicate the class of an element γ ∈ K × i ∩ K ×2 considered in the set (K × i ∩ K ×2 )/K ×2 i for i ∈ {1, 2, 3}. represent lifts of σ 1 , σ 2 ∈ Gal(K/F ) to the group Gal(K( √ γ)/F ), then we have Proof. We consider the first statement first. Observe that we already know thatσ 2 2 acts trivially on √ a 1 and √ a 2 , so we only need to determine the action ofσ 2 2 on √ γ. For this, note that √ γσ , and by Kummer theory this means that N K/K 1 (γ) = a ε 2 k 2 1 for some ε ∈ {0, 1} and k 1 ∈ K × 1 . Note that ε = 0 if and only if N K/K 1 (γ) ∈ K ×2 1 , which is equivalent to [N K/K 1 (γ)] 1 = [1] 1 . Hence our previous calculation continues √ γσ This gives the desired result.
Similar calculations give the other two results.

Then ker(T ) = [F × ].
Remark. Note that Kummer theory tells us that each (K × i ∩ K ×2 )/K ×2 i consists of only two distinct classes, with representatives drawn from {1, a 1 , a 2 , a 1 a 2 }. For example (K × 3 ∩ K ×2 )/K ×2 3 has [1] 3 = [a 1 a 2 ] 3 and [a 1 ] 3 = [a 2 ] 3 as its elements. For the sake of lightening what would otherwise be fairly weighty notation, when considering elements in the image of T we will suppress the bracket notation in its coordinates; that is to say, if T ( Our goal, then, is to build a module whose fixed part spans the image of T , ideally while avoiding [F × ] as much as possible. The first question we consider when looking for such a module is to determine when elements with a nontrivial image under T are themselves in the image of either 1 + σ 1 or 1 + σ 2 . We start with

Proof. Observe first that since [N K/F (γ)] = [1], Kummer theory tells us that [N
The result then follows from the following calculations: ) as well, a contradiction. The same argument gives the result for i = 2.

Then there exists some
Proof. Our approach will be to argue that the appearance of these elements in the image of T guarantees the solvability of certain embedding problems, from which we deduce the solvability of certain equations involving norms.
We are now prepared for the main result of this section.
if dim(im(T )) = 2 and im(T ) is one of the "coordinate planes" if dim(im(T )) = 2 and im(T ) is not one of the "coordinate planes"

In all cases except the last, we have
Proof. We proceed by cases based on dim(im(T )). First, if dim(im(T )) = 1 then let [x] ∈ J(K) G be given so that T ([x]) = (1, 1, 1). Then X := [x] has the desired properties. Now suppose that dim(im(T )) = 2. By Lemma 4.5 we know that if im(T ) is any of (1, a 1 , a 1 ), (1, 1, a 1 1, a 1 ), (a 2 , 1, a 1 or  (1, a 1 , 1), (a 1 , 1, 1 then we can find the some [γ] ∈ J(K) so that T [N K/K 1 (γ)] , T [N K/K 2 (γ)] = im(T ). It is easy to see that X := [γ] ≃ Ω −1 , and since the nontrivial fixed elements in this module have nontrivial images under T , we get On the other hand, if dim(im(T )) = 2 but im(T ) is none of the three subspaces above, then Lemma 4.3 tells us that no element from J(K) G \ ker(T ) is in the image of 1 + σ 1 or 1 + σ 2 . In this case we let [x 1 ], [x 2 ] ∈ J(K) G be given so that {T ([x 1 ]), T ([x 2 ])} forms a basis for im(T ); we then get X :  1, a 1 ), and furthermore that T ([N K/K 1 (γ 1 )]) = (1, a 1 , a 1 ) and T ([N K/K 2 (γ 2 )]) = (a 2 , 1, a 1 ). We claim that X := [γ 1 ], [γ 2 ] ≃ Ω −2 ; certainly the appropriate relations hold, so we only need to check that the module is 5-dimensional. Note that {[N K/K 1 (γ 1 )], [x], [N K/K 2 (γ 2 )]} must be independent since their images under T are independent, and hence any nontrivial dependence must involve [γ 1 ] or [γ 2 ]. But an application of 1 + σ 1 (or 1 + σ 2 ) to such a relation creates a nontrivial relation amongst Alternatively, suppose that dim(im(T )) = 3 but We define X = [γ 1 ], [γ 2 ] . One sees that [γ 1 ] ≃ [γ 2 ] ≃ Ω −1 in the same manner as above (these modules satisfy the appropriate relations by definition, and one can argue they generate a module of the appropriate dimension by leveraging the independence of the image of their fixed components under T ). We claim that X ≃ Ω −1 ⊕ Ω −1 ; for the sake of contradiction, then, assume instead that [ Considering images under T and using Lemma 4.3, we must have [N K/K 2 (γ 1 )] = [x] = [N K/K 1 (γ 2 )], contrary to the assumption in this case that The former follows from the rank-nullity theorem applied to the function T ; in fact we see that  (1, 1, 1)}. This runs contrary to the overriding assumption in this case, that

Proof of Theorem 1
We need one final preparatory result, which is again a manifestation of Hilbert 90 in the biquadratic case.
Proof. We prove the result when ℓ = 3, m = 1 and n = 2; the other results follow by the symmetry of the fields K 1 , K 2 and K 3 .
for somek ∈ K × and ε ∈ {0, 1}. To see this, note that f = k 1+σ 1 σ 1k 2 for some k,k ∈ K. Solving fork 2 and using the fact that F ⊆ K 3 , we then havek 2 ∈ K 3 . But this meanŝ k 2 ∈ K ×2 ∩ K × 3 , so by Kummer theory we getk 2 = k 2 3 a ε 1 , where k 3 ∈ K × 3 and ε ∈ {0, 1}. Naturally we have k 2 3 = N K/K 3 (k 3 ), so that our original expression becomes f = k 1+σ 1 σ 2k 2 = k 1+σ 1 σ 2 k 2 3 a ε 1 = N K/K 3 (kk 3 )a ε 1 . Settingk = kk 3 and dividing through by a ε 1 gives Equation (2). Now we argue that For this, suppose that we have elements g ∈ F × and k ∈ K × so that g = N K/K 3 (k). Now k = f 1 + f 2 √ a 1 + f 3 √ a 2 + f 4 √ a 1 a 2 for some f 1 , f 2 , f 3 , f 4 ∈ F × , and so by assumption we get However since g ∈ F × we must have f 1 f 4 = f 2 f 3 . Our goal is to write g as an element of . In other words, we need to solve We proceed by cases. First, suppose that f 1 = 0. Hence we must have either f 2 = 0 or f 3 = 0. Note if f 2 = 0 then our expression for g becomes A similar computation settles the case where f 3 = 0. So now suppose that f 1 = 0, and observe that since With both (2) and (3) in hand, we can prove the lemma. If we apply (3) to f a ε 1 from (2), then we see that f a ε But since [a 1 ] = [1], we get the desired result.
We are now ready for the proof of the main result of this paper. Our basic strategy is to show that the modulesĴ and X from Theorems 2 and 3 provide the desired decomposition, though in the case where dim(im(T )) = 3 and T ([N K/K 1 (K × )] ∩ [N K/K 2 (K × )]) = {(1, 1, 1)} we will need to make a small adjustment toĴ -removing a single trivial summand -to achieve our result.
Proof of Theorem 1. LetĴ be the module from Theorem 2 and let X be the module from Theorem 3.
If we are not in the case where dim(im(T )) = 3 and T ([ (I.e., the moduleJ is just the result of removing the summand [x 0 ] fromĴ .) In either case we will show that J(K) =J ⊕ X. Of course we haveJ + X ⊆ J(K), and furthermore our construction ofJ gives X G ∩J = {[1]}, so thatJ + X =J ⊕ X. Hence we only need to verify that J(K) ⊆J + X. We do this by examining the possible isomorphism classes for [γ] , where [γ] ∈ J(K).   Figure 5 for a graphical description of these relationships.)   In any of these cases our construction for X (see the second case in Theorem 3) gives an ele-

Some realizability results
Theorem 1 tells us that there are a limited number of summands that could possibly appear in a decomposition of J(K), but is it the case that each of these summand types occurs for at least one biquadratic extension K/F ? In this section we offer some partial results concerning this kind of realizability question, focusing particularly on the possible structures for the X summand from Theorem 1. For a more complete treatment of this problem of realizing the various summands, the reader is encouraged to consult [9] which enhances the current work by exploring its connection to the Brauer group Br(F ).
The X summand takes on one of 6 possible structures, with the various possibilities determined by the image of the function T from section 4 (as detailed in Theorem 3). To determine whether these structures are realizable, we will view the conditions found in Theorem 3 through the lens of Galois embedding problems via Lemma 4.1.
First we introduce some terminology. Note that since K/F is a biquadratic extension with intermediate fields K 1 , K 2 , and K 3 , if there exists some extension L/K which is Galois over F with Gal(L/F ) ≃ D 4 , then there is a unique i ∈ {1, 2, 3} so that Gal(L/K i ) ≃ Z/4Z. We will refer to such an extension as a D 4 -extension of type i. Likewise if there is an extension L/K with Gal(L/F ) ≃ Z/4Z ⊕ Z/2Z, then there is a unique i ∈ {1, 2, 3} so that there exists some field L with K i L L and Gal( L/F ) ≃ Z/4Z. We will refer to such an extension as a Z/4Z ⊕ Z/2Z-extension of type i. (1, 1, a 1 ), then K( √ γ)/F is a D 4 -extension of type 2 or 3 (respectively). We also have that Happily, these types of embedding problems have already been studied extensively. For example, in [16] one finds that a quadratric extension E(   {(a 2 , a 1 , 1), (a 2 , 1, a 1 ), (1, a 1 , a 1 )} is in im(T ). We also can clearly see that 7 = −5y 2 − x 2 has no rational solutions, so K/F does not embed in a D 4 -extension of type 1; therefore (a 2 , 1, 1) ∈ im(T ). Likewise −5 = −35y 2 − x 2 and −35 = −5y 2 − x 2 have no rational solutions. For example, a rational solution to −5 = −35y 2 − x 2 would imply an integral solution to u 2 = 7v 2 + 5w 2 for which 5 ∤ u and 5 ∤ v. One sees this is impossible by examining this equation modulo 5. Because these equations have no rational solutions, it follows that K/F does not embed in a D 4 -extension of type 2 or 3 either. Hence {(1, a 1 , 1), (1, 1, a 1 )} ∈ im(T ). Finally, since −5 is conspicuously not a sum of three rational squares, we have that K/F does not embed in a Q 8 -extension, and so (a 2 , a 1 , a 1 ) ∈ im(T ). Hence im(T ) = {(1, 1, 1)}, and by Theorem 3 we have X = {[1]}. Example 6.2. Let F = Q and K = Q( √ 7, √ −1). We see that K/F does not embed in any Z/4Z ⊕ Z/2Z-extension since none of 7, −1, nor −7 is a sum of two rational squares; it does not embed in a D 4 -extension of type 1 or 3 since 7 = −y 2 − x 2 and −7 = −y 2 − x 2 have no rational solutions; and it does not embed in a Q 8 -extension since −1 is not a sum of three rational squares. It does, however, embed in a D 4 -extension of type 2 since −1 = −7y 2 − x 2 has a rational solution. Hence im(T ) = {(1, 1, 1), (1, a 1 , 1)}, and so X ≃ F 2 . Example 6.3. Let F = Q and K = Q( √ 2, √ −1). Since 2 is a sum of two rational squares but −1 and −2 are not, we see that K/F embeds in a Z/4Z ⊕ Z/2Z-extension of type 1, but not of types 2 or 3. It's also the case that 2 = −y 2 − x 2 has no rational solutions, but −1 = 2y 2 − x 2 and −2 = 2y 2 − x 2 do have rational solutions, and hence K/F embeds in D 4 -extensions of types 2 and 3, but not type 1. We also have that −1 is not a sum of three rational cubes, so K/F does not embed in a Q 8 -extension. Taken together, this means that im(T ) = {(1, 1, 1), (1, a 1 , a 1 ), (1, a 1 , 1), (1, 1, a 1 )}, which is one of the coordinate planes (the "yz-plane"). Hence from Theorem 3, we have X ≃ Ω −1 .
Example 6.5. Let F = Q and K = Q( √ 5, √ 41). Since 5, 41, and 205 are all expressible as sums of two rational squares, and since we can write 5 = (2) 2 + (1) 2 + 0 2 and 41 = (−1) 2 + (2) 2 + (6) 2 , we see that {(a 2 , a 1 , 1), (a 2 , 1, a 1 ), (1, a 1 , a 1 ), (a 2 , a 1 , a 1 )} ⊆ im(T ). Hence dim(im(T )) = 3 in this case, and we have either X ≃ Ω −1 ⊕ Ω −1 or X ≃ Ω −2 (depending on Readers who are familiar with Galois embedding problems will recognize that absence of a key player from our discussion above: the Brauer group. Indeed, the solvability of each of the embedding problems we've discussed is encoded in the vanishing of certain element(s) drawn from (a 1 , a 1 ), (a 1 , a 2 ), (a 2 , a 2 ) ⊆ Br(Q). The relationship between embedding problems and Galois cohomology has been studied extensively; for a small sampling, see [15,16,19,20]. The focus of the follow-up paper [9] is to reinterpret the decomposition of J(K) provided by Theorem 1 through the lens of certain equations in Br(F ). In particular, this will allow us to compute the multiplicities of the various summands by analyzing subspaces within Br(F ), and ultimately show that all listed summand types from Theorem 1 are realizable.