Arithmetic of Quaternion Algebra 2012 1 Quaternion Algebras In this section, F is a field of characteristic 6= 2. Unless stated otherwise, all algebras considered here are finite dimensional algebras over F . If 1A (or simply 1) is the identity of an F -algebra A, then the map α 7! α1A is a monomorphism of F -algebras. This map identifies F as a subalgebra of A. If R is a ring, then R× denotes the group of units in R. 1.1 Basic Definitions Definition 1.1 A quaternion algebra H over F is a 4-dimensional algebra over F with a basis f1; i; j; kg such that i2 = a; j2 = b; ij = k = −ji for some a; b 2 F ×. In this definition, notice that k2 = −ab. The basis f1; i; j; kg is called a standard basis a;b for H and we write H = F . Note that there are infinitely many standard bases for H, and hence there could be another pair of nonzero elements c; b 2 F , different from the c;d a;b a;b ax2;by2 × pair a; b, such that F = F . For instance, F = F for any x; y 2 F , and a;b a;−ab F = F . a;b The notation H = F is functorial in F , that is, if K is a field extension of F , then a; b a; b ⊗ K =∼ as K-algebras: F K Example 1.2 In M2(F ), let 1 0 0 1 0 1 i = ; j = ; k = ij = : 0 −1 1 0 −1 0 2 2 1;1 1;−1 Then i = j = 1 and f1; i; j; kg is a basis for M2(F ). Therefore, M2(F ) = F = F . Example 1.3 Another familiar example of quaternion algebras is Hamilton's quaternions H. It is a quaternion algebra over R with a basis f1; i; j; kg such that i2 = −1; j2 = −1; ij = k = −ji: This shows that = −1;−1 . A simple calculation shows that any two elements from H R fi; j; kg are anti-commutative. Moreover, ij = k; jk = i and ki = j. 1 × a;b Theorem 1.4 Let a; b 2 F . Then F exists. Proof. Fix α; β in an algebraic closure E of F such that α2 = a and β2 = −b. Consider the two matrices α 0 0 β i = ; j = : 0 −α −β 0 Direct computations show that 0 αβ i2 = a; j2 = b; ij = = −ji: αβ 0 Since fI2; i; j; ijg is clearly independent over E, it is also linearly independent over F . a;b Therefore the F -span of fI2; i; j; ijg is a 4-dimensional algebra H over F , and H = F .2 Theorem 1.5 A quaternion algebra over F is central simple, that is, its center is F and it does not have any nonzero proper two-sided ideal. Proof. Let H be a quaternion algebra over F , and f1; i; j; kg be a standard basis of H over F . Consider an element x = α + βi + γj + δk in the center of H, where α; β; γ; δ 2 F . Then 0 = jx − xj = 2k(β + δj): Since k is invertible in H, we must have β = δ = 0. Similarly, γ = 0. Hence x is in F . Next, we need to show that a nonzero two-sided ideal a is H itself. It is sufficient to show that a contains a nonzero element of F . Take a nonzero element y = a + bi + cj + dk in a, where a; b; c; d 2 F . We may assume that one of b; c and d is nonzero. By replacing y by one of iy; jy and ky, we may further assume that a 6= 0. Since yj − jy 2 a and 2k is invertible in H, we see that b + dj, and hence bi + dk as well, are in a. This shows that a + cj is in a. By the same token, a + bi and a + dk are also in a. As a result, −2a = y − (a + bi) − (a + cj) − (a + dk) is a nonzero element of F lying in a.2 So, we may study quaternion algebras using the theory of central simple algebras. Below is a couple of well-known theorems concerning the structure of central simple algebras. Theorem 1.6 (Wedderburn's Structure Theorem) Let A be a finite dimensional simple ∼ algebra over F . Then A is isomorphic to Mn(D), where D = EndA(N) is a division algebra over F with N a nonzero minimal right ideal of A. The integer n and the isomorphism class of the division algebra D is uniquely determined by A. Theorem 1.7 (Skolem-Noether Theorem) Let A be a finite dimensional central simple algebra over F and let B be a finite dimensional simple algebra over F . If φ, are algebra homomorphisms from B to A, then there exists an invertible element c 2 A such that φ(b) = c−1 (b)c for all b 2 B. In particular, all nonzero endomorphisms of A are inner automorphisms. 2 Theorem 1.8 Let H be a quaternion algebra over F . ∼ (a) Either H is a division algebra or H = M2(F ). (b) Let E be a subfield of H which is a quadratic extension of F , and τ be the nontrivial automorphism of E=F . Then there exists j 2 H× such that j2 2 F ×, H = E + Ej, and jx = τ(x)j for all x 2 E. Proof. Part (a) is a direct consequence of Wedderburn's structure theorem. For part (b), since the characteristic of F is not 2, we can write E = F (i) so that i2 2 F ×. Let τ be the nontrivial automorphism of E=F . By the Skolem-Noether Theorem, −i = τ(i) = jij−1 for some invertible element j 2 H. Clearly j 62 E and 1; i; j are linearly independent over F . If ij = α + βi + γj with α; β; γ 2 F , then (i − γ)j = α + βi. But i − γ 6= 0; thus j 2 F (i) = E which is impossible. Therefore, f1; i; j; ijg is a basis for H. Note that ij = −ji and so j2ij−2 = i. Therefore, j2 is in the center of H which is F , and this means that j2 = b 2 F . Clearly, H = E + Ej.2 Definition 1.9 Let f1; i; j; kg be a standard basis for a quaternion algebra H. The elements in the subspace H0 spanned by i; j and k are called the pure quaternions of H. The next proposition shows that H0 does not depend on the choice of the standard basis for H. Proposition 1.10 A nonzero element x 2 H is a pure quaternion if and only if x 62 F and x2 2 F . a;b Proof. Let f1; i; j; kg be a standard basis for H = F . Let x be a nonzero element in H. We can write x = a0 + a1i + a2j + a3k with a` 2 F for all `. Then 2 2 2 2 2 x = (a0 + aa1 + ba2 − aba3) + 2a0(a1i + a2j + a3k): 2 If x is in the F -space spanned by i; j and k, then a0 = 0 and hence x 62 F but x 2 F . 2 Conversely, suppose that x 62 F and x 2 F . Then one of a1; a2 and a3 is nonzero, and hence a0 = 0. Thus x is a pure quaternion.2 Thus each x 2 H has a unique decomposition x = a + α, where a 2 F and α 2 H0. The conjugate of x, denoted x, is defined by x = a − α. For any x; y 2 H, (i) x + y = x + y; (ii) xy = y x; (ii) x = x; (iv) rx = rx for all r 2 F . (v) x = x if and only if x 2 F . 3 In particular, the conjugation is an involution on H (or, equivalently, an algebra isomor- ◦ phism from H to its opposite algebra H ). In M2(F ), a b d −b = : c d −c a Equivalently, if M 2 M2(F ), M = adj(M), the adjoint of M. Definition 1.11 For x 2 H, the (reduced) norm and (reduced) trace of x are the elements nr(x) = xx and tr(x) = x + x, respectively. A direct computation shows that both nr(x) and tr(x) are elements of F . The norm is multiplicative, that is, nr(xy) = nr(x)nr(y) for all x; y 2 H. The invertible elements of H are precisely those with nonzero norm. The trace, however, is linear as tr(ax + by) = atr(x) + btr(y) for all a; b 2 F . For M2(F ), the norm of an element is just its determinant. 1.2 The Matrix Algebras In this subsection, we discuss when a quaternion algebra H over F is isomorphic to M2(F ). Definition 1.12 A nonzero element x in H is said to be isotropic if nr(x) = 0. a;b Theorem 1.13 For H = F , the following are equivalent: ∼ 1;1 ∼ (a) H = F = M2(F ). (b) H is not a division algebra. (c) H has an isotropic element. (d) H0 has an isotropic element. (e) The equation ax2 + by2 = 1 has a solution (x; y) 2 F × F . p (f) If E = F ( b), then a 2 NE=F (E). Proof. We have seen in Theorem 1.8 that (a) is equivalent to (b). Suppose that H is not a division algebra.