Classification of quaternion algebras over the field of rational numbers
Bachelor Thesis
Nadir Bayo October 2017
Supervised by Prof. Dr. Richard Pink
Contents
Introduction 4
1 General results about algebras over a field 6
2 Quadratic forms and quadratic spaces 12
3 Quaternion algebras over the fields Rand Qp 22
4 Quaternion algebras over Q 29
References 36
Introduction
All algebras in this bachelor thesis assumed to be associative and unitary. As the title suggests, our goal is to classify the quaternion algebras, i.e. central simple algebras of dimension 4, over the field of rational numbers. To motivate this question we give an example which the reader may already have encountered: the Hamilton quaternions. They are defined as
H:={t+xi+yj+zk|t, x, y, z∈R}, wherei, j andksatisfy the relations
i2=j2=k2=ijk=−1.
Endowed with an associative multiplication having 1 as neutral element and satisfying the above relations they form a quaternion algebra over R. Stating the above relations is equivalent to giving the relations
i2=−1, j2=−1, and ij =−ji=k.
In the first section of this thesis we will generalize the above example. Consider an arbitrary fieldF with char(F)6= 2, nonzero elements a, b∈F× as well as anF-vector spaceAwith basis elements 1, i, j, k. We will show that there exists a unique F-bilinear associative multiplication onAhaving 1 as neutral element and satisfying the relations
i2=−a j2=b, and ij=−ji=k.
This multiplication turnsA into anF-algebra, which we denote by a,bF . The algebra a,bF is in fact a quaternion algebra, and conversely any quaternion algebra overF is isomorphic to one of the form a,bF for somea, b∈ F. In this notation the Hamilton quaternions can be written as −1,−1R . One can also show that the matrix algebra Mat2×2(F) is isomorphic to 1,1F .
This characterization allows us to show that every quaternion algebra is either isomorphic to the matrix algebra or it is a division algebra. Furthermore the characterization gives us a very
“hands on” feel for quaternion algebras.
Given a quaternion algebra a,bF
we can define the map
nrd:A→F, t+xi+yj+zk 7→ t2−ax2−by2+abz2.
This map is called reduced norm ofAand it can be seen as a homogeneous polynomial of degree two in four variables. This polynomial yields a quadratic form onAseen asF-vector space. We can also restrict nrd onA0:= span{i, j, k}and get a ternary quadratic form. The reduced norm is a connecting element between the theory of quaternion algebras and the theory of quadratic forms, which constitutes the main topic of section 2.
To each quadratic form Qover anF-vector spaceV we can associate anF-bilinear formT. We will show that a change of the basis ofV affects the determinant of the representation matrix ofT rescaling it by squares. Hence we will introduce the discriminant of a quadratic form as the determinant modulo (F×)2.
We will define two equivalence relations on the set of quadratic forms: similarity and isometry.
Two quadratic forms over two F-vector spacesV and V0 are said to be similar if they can be
“converted” in one another by a linear isomorphism and a scaling. If there is no scaling (or in
These tools will allow us to prove the two main theorems of section 2. The first one states that there are natural bijections:
Quaternion algebras overF up to isomorphism
↔
Ternary quadratic forms overF with discriminant 1∈F×/(F×)2up to isometry
↔
Nondegenerate ternary quadratic forms over
F up to similarity
,
given respectively by [A] 7→[nrd|A0] and [Q] 7→ [Q]. The second theorem gives six equivalent conditions about a quaternion algebra and the corresponding quadratic form. In particular it states that a quaternion algebraAis a division algebra if and only if the corresponding reduced norm nrd is isotropic if and only if the form nrd|A0 is anisotropic.
In section 3 we will introduce the reader to thep-adic numbersQpand we will study the theory of quadratic forms overQp. We will show that for each primepand forp=∞(settingQp:=R) there is a unique ternary anisotropic quadratic form overQp, up to similarity. Combining this with the two main results of section 2 we will get that, up to isomorphy, there is a unique division quaternion algebra over each fieldQp forpprime orp=∞.
Finally in section 4 we well proceed to classify the quaternion algebras over Q. Consider a quaternion algebraA:= a,bQ. For each pprime orp=∞we consider the scalar extension
A⊗QQp∼=a, b Qp
.
From section 3 we know that for eachpthatA⊗QQp is either isomorphic to the matrix algebra Mat2×2(Qp) or to a uniquely determined division quaternion algebra. We will introduce two tools from number theory—the Legendre symbol and the Hilbert symbol—to show that the ramification set ofA, defined as
Ram(A) :={pprime orp=∞ |A⊗QQpis a division algebra}, is finite of even cardinality.
We will formulate (without proof) the Hasse-Minkowski theorem and prove a corollary, which states that two quadratic forms are isometric overQif and only if they are isometric overQpfor every p, includingp=∞. We will also state (without proof) Dirichlet’s theorem on primes in arithmetic progression, wich says that for coprimeaandn∈Zwithn6= 0 there exist infinitely many prime numberspsatisfyingp≡a(modn).
These three results, as well as other small ones will allow us to prove the final theorem of this thesis, stating that there are bijections
Isomorphy classes of quaternion algebras overQ
←→
Finite subsets ofP of even cardinality
←→n
D∈Z>1squarefreeo . given by
[A]7→Ram(A) and Σ7→ Y
p∈Σ p6=∞
p
respectively. This will give us a complete classification of the quaternion algebras over the field of rational numbers.
We will follow closely the bookQuaternion algebrasby John Voight ([Voi17]) for the theory of quaternion algebras, as well as the bachelor thesisDer Satz von Hasse-Minkowski by Charlotte Jergitsch ([Jer17]), which was also supervised by Prof. Dr. Richard Pink, and the bookA Course in Arithmetic by Jean-Pierre Serre ([Ser73]) for the theory of quadratic forms and the results from number theory. Some other auxiliary literature will also be used.
1 General results about algebras over a field
All rings in this thesis are associative and unitary, but not necessarily commutative.
We begin with some very basic general definitions and remarks on algebras. Fix a fieldF. Definition 1.1. AnF-algebra is anF-vector space (A,+,0A) together with anF-bilinear mul- tiplication map
·:A×A→A, and an element 1A such that (A,+,·,0A,1A) is a ring.
Proposition 1.2. The above definition is equivalent to: Ais a ring with a ring homomorphism ϕ:F →Z(A),
whereZ(A) is the center ofA, i.e. the subring {x∈A| ∀y∈A:xy=yx}.
Given a nonzero F-algebra Awe identifyF with its imageϕ(F)⊂Z(A). Definition 1.3.
i) An isomorphism ϕ: A → B of F-algebras is an F-vector space isomorphism with the additional property ∀x, y∈A:ϕ(xy) =ϕ(x)ϕ(y). In other words it is anF-vector space isomorphism, which is also a ring isomorphism.
ii) Ananti-isomorphismϕ:A→B ofF-algebras is anF-vector space isomorphism with the additional property∀x, y∈A: ϕ(xy) =ϕ(y)ϕ(x).
Definition 1.4. Given anF-algebra Awe define the opposite algebra Aop as the algebra with the same underlying set, 0, 1, and addition, as well as with the multiplication α·opβ :=β·α. An algebra and its opposite algebra are naturally anti-isomorphic.
Definition 1.5. AnF-algebraD is called adivision algebraif:
i) D6= 0,
ii) ∀a∈D ∀b∈Dr{0} ∃x∈D:a=xb, and iii) ∀a∈D ∀b∈Dr{0} ∃y∈D:a=by.
Remark 1.6. An F-algebra D is a division algebra if and only if it is a division ring, i.e. if D6= 0 andD×=Dr{0}.
Proof. First assume thatDis a division algebra. Pick an element a∈Dr{0}. By definition of division algebra there exists an elementa0∈D witha0a= 1. ThusD is a division ring.
Conversely assume thatD is a division ring and pick elementsa∈D and b∈Dr{0}. By settingx:=ab−1 andy:=b−1athe conditions in Def. 1.5 ii) and iii) are satisfied.
Definition 1.7.
i) A nonzeroF-algebraAwhose only two-sided ideals are (0) andA, is calledsimple.
ii) AnF-algebra Ais said to becentral ifZ(A) =F.
Lemma 1.8. Let A and B be F-algebras, with A simple, and ϕ: A → B be an F-algebra homomorphism. Thenϕis injective or it is the zero map.
Proof. The claim follows from the fact that ker(ϕ) is a two-sided ideal ofA. We are now ready to define the main object of this thesis.
Definition 1.9. Aquaternion algebraover a fieldF is a central simpleF-algebra of dimension 4.
We now want to characterize quaternion algebras in terms of generators.
Proposition 1.10.
i) Let char(F) 6= 2, let a, b ∈ F× and let A be a four-dimensional F-vector space with ba- sis elements {1, i, j, k}. Then there exists a unique F-bilinear associative multiplication A×A→A satisfying:
∀α∈A: 1α=α, i2=a, j2=b, andij=−ji=k.
Further this multiplication satisfies
k2=−ab, ik=−ki=aj, andkj=−jk=bi.
This multiplication turns A into anF-algebra.
ii) Let char(F) = 2, let a∈F, b∈F× and let A be a four-dimensional F-vector space with basis elements {1, i, j, k}. Then there exists a unique F-bilinear associative multiplication A×A→A satisfying:
∀α∈A: 1α=α, i2+i=a, j2=b, andij=j(i+ 1) =k.
Further this multiplication satisfies
k2=ab, ki= (i+ 1)k=aj, andkj=jk+b=bi.
This multiplication turns A into anF-algebra.
Proof. Both statements are proven in the same way, therefore we only prove i). To prove the existence we define a multiplication onAvia the relations
∀α∈A: 1α:=α, i2:=a, j2:=b, ij:=kand ji:=−k.
Since we require associativity we get k2 = (ij)2 =i(ji)j =−i2j2 =−ab. In a similar way we can showik=−ki=aj andkj =−jk=bi.Having defined the relations for the basis elements, the multiplication extends uniquely to anF-bilinear multiplication onA. Since∀α∈A: 1α=α,
the multiplication defines anF-algebra structure onA.
Definition 1.11.
i) In the case char(F)6= 2 we denote the algebra from Prop. 1.10 i) by a,bF . ii) In the case char(F) = 2 we denote the algebra from Prop. 1.10 ii) bya,b
F
. In both casesi andj are called thestandard generatorsof the algebra.
Proposition 1.12.
i) Letchar(F)6= 2. Every algebra of the form a,bF
is a quaternion algebra. Conversely for every quaternion algebraA there exista, b∈F× such that A is isomorphic to a,bF
. ii) Letchar(F) = 2. Every algebra of the form a,b
F
is a quaternion algebra. Conversely for every quaternion algebraA there exista∈F, b∈F× such that Ais isomorphic to a,b
F
. Example 1.13. We can use the above notation to describe the classical Hamilton quaternions:
H:={t+xi+yj+zk|t, x, y, z∈R, i2=j2=k2=ijk=−1}=−1,−1 R
.
Example 1.14. The matrix algebra Mat2×2(F) for char(F)6= 2 is isomorphic to (1,1R) via 1,1
F
→Mat2×2(F), i7→
1 0 0 −1
, j7→
0 1 1 0
. Since not all nonzero matrices are invertible, Mat2×2(F) is not a division algebra.
Proof of Prop. 1.12. Since the goal of this thesis is to classify the quaternion algebras overQ, we restrict ourselves to the proof of the case char(F)6= 2. First we consider the algebraA:= a,bF
. It is clear thatF ⊂Z (a,bF ). Let α:=t+xi+yj+zk ∈Z (a,bF ). Since 2j and 2k are both invertible we obtain:
0 =αi−iα = 2j(yi−az) ⇐⇒ yi−az= 0 ⇐⇒ y=z= 0, and 0 =αj−jα= 2xk ⇐⇒ x= 0.
Hence we haveα∈F, which implies that a,bF is central.
Now let I 6= (0) be a two-sided ideal of a,bF
. In order to show that I is equal to a,bF suffices to show that I contains an element of F×. So let 0 6= α := t+xi+yj+zk ∈ I. Ifit x=y =z = 0 then t∈ F×∩I and we are done, so assume that one of x, y, zis nonzero. By multiplyingαwith i, j or kwe can assume thatt 6= 0. Using the invertibility of−2i,−2j, and
−2kwe obtain:
αi−iα=−2i(yj+zk)∈I ⇒yj+zk∈I⇒t+xi =α−(yj+zk)∈I, αj−jα=−2j(xi+zk)∈I⇒xi+zk∈I ⇒t+yj =α−(xi+zk)∈I, αk−kα=−2k(xi+yj)∈I⇒xi+yj∈I ⇒t+zk=α−(xi+yj)∈I.
So−2t=α−(t+xi)−(t+yj)−(t+zk)∈F×∩I and we are done.
Conversely let A be a quaternion algebra, i.e. a central simple F-algebra of dimension 4.
First we assume that A is a division algebra. We take an i ∈ ArF and consider the ring F[i]. This is a commutative subring of the division ring A. The ringF[i] has finite dimension as F-vector space and therefore it is a field. Since F[i] is commutative whereas A is not we have F ( F[i] ( A, and from the multiplicativity of the degree of an extension of division rings it follows that [F[i]/F] = 2. MoreoverF[i] is its own centralizer inA. Since char(F)6= 2 after replacingi by another element ofF[i]rF we can assume without loss of generality that a:=i2∈F. Consider theF[i]-linear map
(the conjugation withi). This is an endomorphism of the two-dimensional leftF[i]-vector space A. One can check by computation thatϕ2= id, which implies thatϕhas two eigenvalues 1 and
−1 and therefore it is diagonalizable. From
ϕ(α) =iαi−1=α ⇐⇒ iα=αi ⇐⇒ α∈F[i]
it follows thatF[i] is the eigenspace associated to 1. We take now aj∈ArF[i]. In particular j lies in the eigenspace associated to −1 and hence iji−1 =−j ⇒ij = −ji. From ij2 =j2i it follows that j2 ∈F[i], i.e. ∃b, c∈ F:j2 =b+ic. Ifc 6= 0, so i= 1cj2− bc ∈ F[j] and thus F[i](F[j], implying thatF[j] =A, which is a contradiction by the noncommutativity ofA. So c= 0 andj2=b∈F. We claim that{1, i, j, ij} is a basis ofAoverF. Indeed, lett, x, y, z∈F such thatt+xi+yj+zij= 0. By computing and usingi2=a, j2=bandij=−jiwe obtain:
0 =i(αi+iα) = 2a(t+xi)⇒t+xi= 0, 0 =j(αj+jα) = 2b(t+yj)⇒t+yj= 0, 0 =ij(αij+ijα) =−2ab(t+zij)⇒t+zij= 0. Hence
−2t= (t+xi+yj+zij)−(t+xi)−(t+yj)−(t+zij) = 0,
which implies t = 0, hence xi = yj = zij = 0. Since i, j and ij are all nonzero, we have x=y=z= 0, which implies the linear independence of 1, i, j andij.
IfAis not a division algebra, it has a nonzero proper left ideal, because by Rem. 1.6 we can choose a nonunitα∈A, then the left ideal generated byαis a nontrivial left ideal ofA. LetI be a nonzero proper left ideal ofAand letm:= dimF(I). The operation ofAonI corresponds to an F-algebra homomorphism A→ EndF(I), which is injective since A is simple. We know from linear algebra that EndF(I) is isomorphic to Matm×m(F) asF-vector space, and from the injectivity it follows thatm2= dimF(Matm×m(F))>dimF(A) = 4, and som>2. By arguing in the same fashion with the quotient algebraA/I, which has dimension 4−moverF, we obtain m62, hencem= 2. ThusAis isomorphic to Mat2×2(F)∼= 1,1F
.
The proof of Prop. 1.12 yields the following proposition.
Proposition 1.15. For a quaternion algebraA overF the following are equivalent:
i) A∼= 1,1F
∼= Mat2×2(F). ii) A is not a division algebra.
Proof. The implication “i)⇒ii)” follows directly from Rem. 1.6. The converse was shown in the
proof of Prop. 1.12.
LetAbe anF-algebra.
Definition 1.16. For a finite dimensional F-algebra A we define the algebra norm and the algebra traceas
NmA/F:A→F, α7→det(Tα), respectively TrA/F:A→F, α7→tr(Tα),
whereTαis theF-vector space endomorphism ofAgiven byβ 7→αβ.
Definition 1.17. AninvolutiononAis anF-linear map
·:A→A, α7→α with
i) 1 = 1,
ii) ∀α∈A:α=α, iii) ∀α, β∈A:αβ=βα.
The involution·is calledstandard if also iv) ∀α∈A:αα∈F.
Proposition 1.18. AnyF-algebra with an involution is isomorphic to its opposite algebra.
Proof. By linearity and property iii) of Def. 1.17 an involution defines a homomorphism A →
Aop, α→α, and by ii) it is bijective.
Lemma 1.19. Given a standard involution onA we have∀α∈A:α+α∈F.
Proof. For anyα∈Awe haveα+α= (α+ 1)(α+ 1)−αα−1∈F. Definition 1.20. Given a standard involution onAthereduced norm ofαis the map
nrd:A→F, α7→αα, and thereduced trace ofαis the map
trd:A→F, α7→α+α.
Forα∈Awe call the polynomial
X2−trd(α)X+ nrd(α)∈F[X] thereduced characteristic polynomial ofα.
Lemma 1.21. Given any standard involution onA(which induces a reduced norm and a reduced trace), any α∈Ais a root of its reduced characteristic polynomial.
Proof. For anyα∈Awe have:
α2−trd(α)α+ nrd(α) =α2−αα−αα+αα=−αα+αα=−αα+αα= 0.
Proposition 1.22. Let K be a quadratic F-algebra (i.e. an F-algebra with dimF(K) = 2).
ThenK is commutative and has a unique standard involution.
Proof. Letα∈KrF. Since dimF(K) = 2 we can writeK=F[α] and henceKis commutative.
Since 1 andαbuild a basis ofK overF we can writeα2=tα−nfor uniquet, n∈F.
Assume that there is a standard involution onKand let ˜·be any such. Then by Lemma 1.21 we haveα2= trd(α)α−nrd(α) and by uniquenesst= trd(α) =α+ ˜αandn= nrd(α) and hence every standard involution must satisfy ˜α=t−α.
We can extendα7→t−αonK usingF-linearity and get the map
·: K→K, x+yα7→x+y(t−α),
for any x, y ∈F. By computing one can check that· satisfies the conditions in Def. 1.17 and
hence it defines the unique standard involution onK.
Corollary 1.23. If Apossesses a standard involution, then it is unique.
Proof. Let·be a standard involution onAand pick an arbitraryα∈ArF. By Lemma 1.21 we haveα2−trd(α)α+nrd(α) and hence dimFF[α] = 2. By the previous proposition the restriction of·toF[α] is unique and sinceαis arbitrary the standard involution is unique onAas well.
Corollary 1.24. Let char(F) 6= 2. Any quaternion algebra A = a,bF
possesses a unique standard involution. It is given by
α=t+xi+yj+zk7→α=t−xi−yj−zk, and hencetrd(α) = 2t andnrd(α) =t2−ax2−by2+abz2.
Proof. By checking that the above defined map satisfies the axioms of a standard involution one
proves the existence; the uniqueness is given by Cor. 1.23.
Proposition 1.25. Let Abe a nontrivialF-algebra with a standard involution andα∈Ar{0}.
Then the following are equivalent:
i) αis not a left zero divisor.
ii) αis not a right zero divisor.
iii) nrd(α)6= 0. iv) α∈A×.
Proof. Suppose nrd(α) =αα=αα6= 0 and pickβ ∈Awithαβ= 0. Then nrd(α)β=ααβ= 0 and since nrd(α) ∈ F× we have β = 0. Conversely if α is not a left zero divisor we have nrd(α) = αα 6= 0. This proves “i) ⇐⇒ iii)”; the equivalence “ii) ⇐⇒ iii)” can be proven analogously.
Once again suppose that nrd(α) 6= 0. We have αnrd(α)α = 1 and hence α ∈ A×. Finally suppose thatα∈A× and pick β ∈A withαβ= 0. Then β = 1β =α−1αβ = 0 and henceαis
not a left zero divisor.
2 Quadratic forms and quadratic spaces
We introduce the reader to some basic theory of quadratic forms, which shall serve as foundation for our further discussion of quaternion algebras.
Definition 2.1. A quadratic formon anF-vector spaceV is a mapQ: V →F with i) ∀λ∈F ∀x∈V:Q(λx) =λ2Q(x), and
ii) the mapT:V ×V →F, (x, y)7→Q(x+y)−Q(x)−Q(y) is bilinear.
The pair (V, Q) is calledquadratic space andT the(symmetric) bilinear form associated to Q. Remark 2.2. Since for a quadratic form Q on V we have ∀x ∈ V: T(x, x) = 2Q(x), if char(F)6= 2 we can uniquely determineQby knowingT.
Let us assume that char(F)6= 2 andn:= dimF(V)<∞and letB:= (bi)16i6n be a basis of V overF.
Definition 2.3. The symmetricn×n-matrix
MB(T) = (T(bi, bj))i,j=: (mij)i,j
is called theGram matrix ofQin the basisB.
By identifyingV withFn (endowed with the standard basis (ei)16i6n) via the isomorphism ϕ: V →Fn, bi7→ei we get:
∀x, y∈V:T(x, y) =ϕ(x)TMB(T)ϕ(y) as well as
∀x=
n
X
i=1
xibi∈V:Q(x) = 1
2ϕ(x)TMB(T)ϕ(x) =1 2
n
X
i,j=1
mijxixj. This yields a homogeneous polynomial of degree 2:
fQ,B(X1, . . . , Xn) =1 2
n
X
i,j=1
mijXiXj ∈F[X1. . . Xn].
Definition 2.4. The polynomialfQ,B is calledpolynomial representation ofQ in the basisB.
We consider another basisB0:= (b0i)16i6n with basis change matrix MBB0. Then the Gram matrix ofQwith respect toB0 is given byMB0(T) =MBBT 0MB(T)MBB0. Since det(MB0(T)) = det(MBB0)2det(MB(T)) we notice that the determinant of the Gram matrix ofQ is uniquely determined up to squares.
Definition 2.5. Thediscriminant ofQis defined as
disc(Q) := det(M) (mod(F×)2)∈F/(F×)2, and is independent of the choice of the basis.
Remark 2.6. For a given homogeneous polynomial f of degree two in n variables and any given basisB ofV there exists precisely one quadratic formQonV such thatf =fQ,B. More explicitly, given f = Pn
i,j=1aijXiXj (without loss of generality ∀i, j: aij = aji) and a basis B:= (bi)16i6n ofV the form
QXn
i=1
xibi
:= 1
2f(x1, . . . , xn) satisfiesfQ,B =f.
Definition 2.7.
i) Two quadratic spaces (V, Q) and (V0, Q0) are said to besimilar if there exists a pair (f, u), where f:V →V0 is anF-vector space isomorphism andu∈F× satisfying
∀v∈V:Q0(f(x)) =uQ(x). We denote the similarity byQ∼Q0.
ii) Two quadratic forms Q : V → F and Q0:V0 → F are said to be isometric if they are similar with similarity factor u = 1. We denote the isometry by Q ∼= Q0. We call the isomorphism f anisometry.
iii) In the special case (V, Q) = (V0, Q0) the isometry is called anautometry. The setO(V) of all autometries of V is a subgroup of AutF(V).
If the underlying spaces are understood we speak simply of similar, respectively isometric quadratic forms. From the definition it follows directly that both similarity and isometry of quadratic forms are equivalence relations.
Proposition 2.8. Let (V, Q)and(V0, Q0)be similar quadratic spaces,(f, u)as in Def. 2.7, and B, B0 two bases of V andV0 respectively. Thendisc(Q0) =undisc(Q)∈F/(F×)2.
Proof. The proof can be done with similar matrix manipulations as in Def. 2.3 – 2.5.
Definition 2.9.
i) We say thatx, y∈V areorthogonalwith respect to the quadratic form QifT(x, y) = 0.
ii) For a subsetS⊂V the setS⊥ :={x∈V | ∀y∈S: T(x, y) = 0}is a subspace of V called theorthogonal complement of S.
iii) A basis of the quadratic space (V, Q) is said to be orthogonal, if the basis elements are pairwise orthogonal with respect toQ.
Definition 2.10.
i) Theorthogonal sum of two quadratic spaces (V0, Q0) and (V00, Q00) is the quadratic space (V, Q), where
Q:V :=V0⊕V00→F, x0+x007→Q(x) +Q(x0).
We denote the quadratic form Q by Q0 ⊥Q00 and the orthogonal direct sum of the two subspaces V0 andV0 byV0kV00.
ii) Fora∈F we writehaifor the quadratic formQ(x) =ax2 onF.
iii) More generally fora1, . . . , an∈F we defineha1, . . . , ani:=ha1i ⊥. . .⊥ hani. Remark 2.11. In the notation from Def. 2.10 iii) we have:
i) ∀c∈F:ha1, . . . , ani ∼cha1, . . . , ani=hca1, . . . , cani. ii) ∀c∈F, ∀16i6n: ha1, . . . , c2ai, . . . , ani ∼=ha1, . . . , ani. iii) ∀σ∈Sn:ha1, . . . , ani ∼=haσ1, . . . , aσni.
Definition 2.12.
i) The subspace rad(V) :=V⊥ is called theradical of V. By choosing a subspaceW comple- mentary to rad(V) we obtain a (non unique) orthogonal decomposition
V = rad(V)kW, Q= 0⊥Q|W.
ii) The rank of the quadratic formQ, denoted by rank(Q), is defined as the codimension of rad(V).
Proposition 2.13. Let Abe an F-algebra with a standard involution. Then:
i) The reduced norm nrd defines a quadratic form on A and the associated bilinear form is given by T(α, β) = trd(αβ).
ii) Two elementsαandβ of Aare orthogonal if and only iftrd(αβ) =αβ+αβ= 0.
iii) If 1, α, β ∈ A are linearly independent, then they are pairwise orthogonal if and only if αβ=−βα.
Proof.
i) We check the axioms of a quadratic form. In the first place for any α∈ A and for any λ ∈ F we have nrd(λα) = λαλα = λ2αα = λ2nrd(α). For any α, β ∈ A we define T(α, β) := nrd(α+β)−nrd(α)−nrd(β). This is clearly symmetric. By computing we obtain:
T(α, β) = (α+β)(α+β)−αα−ββ
=αα+αβ+βα+ββ−αα−ββ
=αβ+αβ= trd(αβ)
By symmetry it suffices to prove the linearity of T only in the first component. For any α, β, γ∈Aand for any λ, µ∈F we have:
T(λα+µβ, γ) = (λα+µβ)γ+γ(λα+µβ)
=λαγ+λγα+µβγ+µγβ
=λT(α, γ) +µT(β, γ), which proves the bilinearity.
ii) follows directly from i).
iii) Using the computations from i) we obtain:
αβ+βα=α(β+β) +β(α+α)−αβ−βα
=αtrd(β) +β trd(α)−trd(αβ)
=αtrd(β1) +β trd(α1)−trd(αβ).
Assume that 1, α, β are linearly independent. If they are pairwise orthogonal then αβ+ βα= 0. Conversely, ifαβ+βα= 0 by linear independence we obtain trd(α) = trd(β) = trd(αβ) = 0 and hence 1, α, β are pairwise orthogonal.
Definition 2.14. A quadratic formQ:V →F is called nondegenerate if for allx ∈V r{0} the homomorphism Tx:V → F, y7→T(x, y) is nonzero, or equivalently if the homomorphism V →V∗, x7→Tx is injective. OtherwiseQis called degenerate.
Proposition 2.15. Let Qbe a quadratic form on ann-dimensional spaceV. The following are equivalent:
i) Qis nondegenerate.
ii) rad(V) = 0. iii) rank(Q) =n.
iv) disc(V)6= 0.
Proof. The equivalence of the first three statements follows directly from the definitions, hence we prove only “i) ⇐⇒iv)”. Choose a basis B ofV and identify V with Fn endowed with the standard basis. LetMB(T) be the Gram matrix ofQin the basisB. Then:
Qis nondegenerate ⇐⇒ ∀x∈V: (∀y∈V:T(x, y) = 0⇒x= 0)
⇐⇒ ∀x∈Fn: (∀y∈Fn:xTMB(T)y= 0⇒x= 0)
⇐⇒ det(MB(T))6= 0
⇐⇒ disc(Q)6= 0.
Example 2.16. The reduced norm of a quaternion algebra a,bF
with char(F) 6= 2, as seen in Cor. 1.24, defines a nondegenerate quadratic form of rank 4, which is isometric to the form h1,−a,−b, abionF4.
Proposition 2.17. Let (V, Q) be a quadratic space and letv∈V withQ(v)6= 0. Then we get an orthogonal decompositionV = span{v}kspan{v}⊥, Q=Q|span{v}⊥Q|span{v}⊥.
Proof. For everyx∈V we consider the decomposition x=T(v, x)
T(v, v)v+
x−T(v, x) T(v, v)v
.
Then we have T v, x− TT(v,v)(v,x)v = 0 and hence x− T(v,x)T(v,v)v ∈ span{v}⊥. Geometrically we can interpret the second summand as the projection of xto span{v}. Since Q(v)6= 0 we have span{v} ∩span{v}⊥ = {0}, which impliesV = span{v} ⊕span{v}⊥ and by definition of the
orthogonal complement we haveV = span{v}kspan{v}⊥.
Proposition 2.18. Every quadratic space(V, Q)possesses an orthogonal basis. This implies that every quadratic formQhas a representation of the typefQ,B(X1, . . . , Xn) =a1X12+· · ·anXn2with ai∈F for some basisB, or in other words for every quadratic formQthere exista1, . . . , an∈F such that Q is isometric to the form ha1, . . . , ani.
Proof. The claim follows from Prop. 2.17 by induction over dimF(V) and by taking into account that every basis of a degenerate quadratic space is orthogonal.
Definition 2.19. LetQ:V →F be a quadratic form.
i) We say thatQrepresents an elementa∈F if there exists ax∈V such thatQ(x) =a. ii) The quadratic formQ(or the quadratic space (V, Q)) is said to beisotropicif it represents
0∈F nontrivially, i.e. if there exists ax∈V r{0} such thatQ(x) = 0.
iii) The quadratic formQ(or the quadratic space (V, Q)) is said to beuniversalif it represents every element ofF.
Remark 2.20.
i) A degenerate quadratic form is always isotropic.
ii) IfQandQ0 are two similar quadratic forms, thenQis isotropic if and only ifQ0 is.
iii) Two isometric quadratic forms represent the same elements ofF, so ifQand Q0 are two isometric quadratic forms, then Qis universal if and only ifQ0 is.
Example 2.21. Given a two-dimensional F-vector space V and a basis (e1, e2) the quadratic form given by f(X, Y) =XY (as explained in Rem. 2.6) is calledhyperbolic plane. It is clearly universal, and the Gram matrix of the quadratic form with respect to the basis (e1, e2) is given by
M :=
0 11 0
.
Equivalently a quadratic form QonV is a hyperbolic plane if there exists a basis (e1, e2) of V withT(e1, e1) =T(e2, e2) = 0 andT(e1, e2) = 1.
Proposition 2.22. A nondegenerate quadratic form Q is isotropic if and only if it is the or- thogonal sum of a hyperbolic planeH and a nondegenerate quadratic form Q0.
Proof. We follow the proof from Ch. 2, Lemma 2.1 of [Cas08].
“⇐” Clear since the hyperbolic plane is isotropic.
“⇒” SinceQis isotropic there exists an elemente1∈V withQ(e1) = 0 (and henceT(e1, e1) = 0). SinceQis nondegenerate, there existsx∈V withT(e1, x)6= 0. By rescalingxwe can assume thatT(e1, x) = 1. By puttinge2:=x−Q(x)(e1) we get:
T(e2, e2) =T(x, x)−2Q(x)T(x, e1) +Q(x)2T(e1, e1) = 0 and T(e1, e2) =T(e1, x)−Q(x)T(e1, e1) = 1.
H :=Q|span{e1,e2}andQ0 :=Q|{e1,e2}⊥ do the job.
Proof. A nondegenerate isotropic quadratic form contains a hyperbolic plane, which is universal.
Corollary 2.24. LetQ:V →F be ann-dimensional nondegenerate quadratic form, leta∈F×. Then the following are equivalent:
i) Qrepresents a.
ii) Qis isometric to Q0⊥ haifor somen−1-dimensional nondegenerate quadratic formQ0. iii) The quadratic formQ⊥ h−airepresents 0nontrivially.
Proof. It is clear that ii) implies both i) and iii). IfQrepresents a, so there exists x∈V with Q(x) = a. Define Q0 := Q|span{x}⊥. Since Q|span{x} is isometric to hai we get the isometry Q ∼= Q0 ⊥ hai. Thus i) implies ii). Finally we prove “iii) ⇒ ii)”. Let Q ⊥ h−ai represent 0 nontrivially. Then there existx∈V andy∈F withQ(x)−ay2 = 0. Ify = 0, thenx6= 0 and henceQis isotropic, thus by Cor. 2.23 Qrepresents a. Ify 6= 0 we haveQ(xy) = aand we are
done.
Now, similarly as in Prop. 2.17, for anyv∈V withQ(v)6= 0 we consider the endomorphism
τv:V →V, x7→x−2T(v, x) T(v, v)v.
Lemma 2.25. The endomorphism τv satisfies the following properties:
i) ∀x∈V:Q(τv(x)) =Q(x). ii) ∀x∈V:τv(τv(x)) =x.
iii) ∀x∈span{v}:τv(x) =−x.
iv) ∀x∈span{v}⊥:τv(x) =x.
In particular the first two statements imply that τv is an autometry ofV.
Proof. The proof can be done by simple calculation.
Lemma 2.26. For anyx, y∈V with Q(x) =Q(y)6= 0there exists an autometry τ of V with τ(x) =y.
Proof. SinceQ(x) =Q(y) impliesT(x, x) =T(y, y) we have
Q(x−y) +Q(x+y) =Q(x−y+x+y)−Q(x−y+x+y) +Q(x−y) +Q(x+y)
=Q(2x)−T(x−y, x+y)
= 4Q(x)−T(x, x) +T(y, y)
= 4Q(x)6= 0,
which implies thatQ(x−y) andQ(x+y) cannot be both equal zero. IfQ(x−y)6= 0 then τx−y(x) =x− 2T(x, x−y)
T(x−y, x−y)(x−y) =x−2T(x, x−y)
2T(x, x−y)(x−y) =y
and by settingτ:=τx−y we are done. Otherwise we haveQ(x+y)6= 0 and τx+y(x) =x− 2T(x, x+y)
T(x+y, x+y)(x+y) =x−2T(x, x+y)
2T(x, x+y)(x+y) =−y.
By the previous lemmaτyτx+y(x) =τy(−y) =y and we conclude by settingτ :=τyτx+y. Remark 2.27. We can interpretτvgeometrically as the reflection with respect to the hyperplane orthogonal tov. In the special caseV =RnandQ(x) = 12||x||2it is calledHouseholder reflection; the reader may have encountered it in linear algebra or numerical analysis.
With these both lemmas we can prove the following important theorem of the theory of quadratic forms.
Theorem 2.28(Witt). Let(V, Q)and(V0, Q0)be two isometric quadratic spaces with orthogonal decompositions W1kW2 and W10 kW20, with nondegenerate subspaces W1 and W10. Then the following two equivalent statements hold:
i) (Witt cancellation) IfW1 andW10 are isometric, then W2 andW20 are isometric as well.
ii) (Witt extension) Ifg:W1→W10 is an isometry, then there exists an isometryf:V →V0 with f|W1 =g andf(W2) =W20.
Proof. We follow the proof of Satz 3.1 of [Kne02]. In the first place let us prove the equivalence of the two statements.
“⇒” Let g: W1 → W10 be an isometry. By i) there is an isometry h:W2 → W20. Then the isometryf :=g⊕h:W1kW2→W10kW20 is an extension ofg.
“⇐” Let g: W1 →W10 be an isometry. By ii) there is an extension f of g with f(W2) =W20. Thenh:=f|W2 is an isometryW2→W20.
We prove ii) using induction on m:= dimF(W1) = dimF(W10). For m= 0 there is nothing to show. Ifm >0 chose an orthogonal basis (bi)16i6mforW1. Thus we can write
V =W1kW2=m−1ë
i=1
F bikF bmkW2.
By induction hypothesis there exists an isometry ˜f:V →V0 satisfying∀16i6m−1: ˜f(bi) = g(bi). We define ˜bm := ˜f−1 g(bm)
. Since f−1 ◦ g is an autometry of V it preserves or- thogonality, which implies that ˜bm is orthogonal to bi for all i < m. Moreover we have Q(˜bm) =Q f−1 g(bm)
=Q(bm)6= 0 since span{bm} is a regular subspace ofV. By Lemma 2.26 there is aτ ∈ {τb
m−˜bm, τ˜bmτb
m+˜bm} sendingbm to ˜bm. By orthogonality τ fixes bi for any i < m. We conclude by settingf := ˜f◦τ. This is in fact an isometryV →V0 and its restriction
onW1is by construction equal to g.
Now we can apply the theory of quadratic forms which we developed in this section to prove three theorems which build the first step towards the classification of the rational quaternion algebras.
Let A be a quaternion algebra over F. Then we define the subspace of pure quaternions A0:={α∈A|trd(A) = 0}={1}⊥. By Ex. 2.16 and Thm. 2.28 i) the restriction of the reduced norm onA defines a quadratic form in three variables, which is isometric to h−a,−b, abifora
Theorem 2.29. Let A and A0 be two quaternion algebras overF. According to Cor. 1.24 and Prop. 2.13 both algebras with their respective reduced norms are quadratic spaces. Then the following are equivalent:
i) A andA0 are isomorphic asF-algebras.
ii) A andA0opare isomorphic asF-algebras.
iii) A andA0 are isometric as quadratic spaces.
iv) A0 andA00 are isometric as quadratic spaces.
If f: A0→A00 is an isometry, thenf extends uniquely to either an isomorphismf:A→A0 or to an isomorphismf:A→A0op ofF-algebras.
Proof. The equivalence between i) and ii) follows from the fact that any any algebra with a standard involution is isomorphic to its opposite algebra. The equivalence between iii) and iv) follows directly from Thm. 2.28.
We prove “i)⇒iii)”. Letf:A→A0 be anF-algebra isomorphism. From Cor. 1.23 we know that the standard involutions (and hence the reduced norms) onAand onA0 are unique. Let us denote both standard involutions by·and both reduced norms by nrd. Since α07→f f−1(α0) defines a standard involution onA0 as well, we have∀α0 ∈A0:α0=f f−1(α0)
. So we have
∀α∈A: nrd(f(α)) =f(α)f(α) =f(α)f(α) =f(αα) =f(nrd(α)) = nrd(α),
where the last equality follows from the fact that nrd(α)∈F. Sof is an isometry of quadratic spaces.
Finally we prove “iv)⇒i)”. Letf:A0→A00be an isometry. WriteA= a,bF witha, b∈F× and standard generatorsi, j,as in Def. 1.11. Sincef(i)∈A00, we havef(i) =−f(i) and hence
f(i)2=−f(i)f(i) =−nrd(f(i)) =−nrd(i) =a.
Analogously we have f(j)2 = b. Since i, j and ij are pairwise orthogonal and isometries preserve orthogonality, f(i), f(j) and f(ij) are pairwise orthogonal as well. In particular by Prop. 2.13 iii) we getf(i)f(j) =−f(j)f(i). Moreover from trd f(i)f(j)f(i)
=atrd(f(j)) = 0 and trd f(i)f(j)f(j)
=−btrd(f(i)) = 0 it follows that f(i)f(j) is orthogonal to bothf(i) and f(j). Since dimF(A00) = 3 there exists an u∈F× with f(ij) =uf(i)f(j). By taking reduced norms we get
nrd(i)nrd(j) = nrd(ij) = nrd(f(ij)) = nrd(uf(i)f(j))
=u2nrd(f(i))nrd(f(j)) =u2nrd(i)nrd(j),
which implies that u = ±1. If u= 1 we have f(ij) = f(i)f(j) and f extends to an algebra isomorphism A → A0 via 1 7→ 1. Otherwise f(ij) = −f(i)f(j) and f extends to an algebra isomorphismA→A0opvia 17→1. By postcomposing it with the standard involutionA0op→A0
we get anF-algebra isomorphismA→A0.
Theorem 2.30. There is a bijection:
Quaternion algebras overF up to isomorphism
↔
Ternary quadratic forms overF with discriminant 1∈F×/(F×)2 up to isometry
↔
Nondegenerate ternary quadratic forms over
F up to similarity
Proof. We begin with the proof of the first bijection. From the equivalence “i) ⇐⇒ iv)” of Thm. 2.29 the map A 7→ nrd|A0 yields a well-defined injective map [A] 7→ [nrd|A0] between isomorphy and isometry classes. In order to prove the surjectivity of this map we consider a quadratic space (Q, V) with disc(Q) = 1 ∈ F×/(F×)2. By choosing an orthogonal basis according to Prop. 2.18 we obtain an isometryQ∼=h−a,−b, cifor some a, b, c∈F with abc∈ (F×)2 (and following abc ∈(F×)2). Rescaling the third basis vector with Rem. 2.11 ii) yields
Q∼=h−a,−b, ci=D
−a,−b, ab c ab
E∼=h−a,−b, abi ∼= nrd|A0, withA= a,bF , so the map is surjective.
Next we prove the second bijection. The natural map sending the isometry class of a quadratic form to the similarity class of the quadratic form is clearly well-defined. To prove the surjectivity we need to prove that every nondegenerate ternary quadratic form is similar to some form with discriminant 1. Let Qbe a nondegenerate ternary quadratic form. By choosing an orthogonal basis and using Rem. 2.11 we obtain
Q∼=ha, b, ci ∼abcha, b, ci=ha2bc, ab2c, abc2i ∼=hbc, ac, abi,
with the last quadratic form having discriminant 1∈F×/(F×)2. This proves the surjectivity.
In order to prove the injectivity we consider two quadratic spaces (Q, V) and (Q0, V0) with discriminant 1∈F×/(F×)2whose isometry classes map to the same similarity class (henceQand Q0 are similar). We want to prove that they are in fact isometric. By the definition of similarity there exists a u∈ F× and an isomorphism f:V → V0 with ∀x ∈ V: Q0(f(x)) = uQ(x). By Prop. 2.8 we have 1 = disc(Q0) =u3disc(Q) =udisc(Q)∈F×/(F×)2 and following u∈(F×)2. We writeu=c2 for ac∈F× and obtain
Q0(c−1f(x)) =c−2Q0(f(x)) =u−1uQ(x) =Q(x).
SoQand Q0 are isometric viac−1f and hence the map is injective (and bijective).
Lemma 2.31. Let a∈ F× and let i 6∈F with i2 =a and consider the F-algebra K := F[i], which is isomorphic to F[X]/(X2−a)as a ring.
i) Ifais a square in F thenK is isomorphic to F×F as a ring, thus it is not a field.
ii) Ifa is not a square inF thenK is a quadratic field extension of F and is therefore equal to its own fraction fieldF(i).
Proof. Ifais a square inF we choose a square root ofainF and denote it by√
a. We consider the ring homomorphismϕ:F[X] →F ×F, f 7→ f(√
a), f(−√
a). It is surjective as for any (b1, b2)∈ F ×F the polynomial f(X) = 2√1a(b1−b2)X +12(b1+b2) is a preimage of (b1, b2) underϕ. The ideal (X2−a) is contained in the kernel ofϕand by the universal property of the factor rings there exists a well-defined surjective homomorphism ϕ:F[X]/(X2−a)→F ×F, which is also anF-linear map between two-dimensionalF-vector spaces. Dimensions being the reason for it,ϕis bijective and hence a ring isomorphism, which proves i).
If a is not a square in F, the polynomial X2−a is irreducible in F[X] and hence K ∼=
F[X]/(X2−1) is a field, which proves ii).
