Quaternion algebras over ¯elds De¯nitions and ¯rst results Let K be a ¯eld and L j K a quadratic extension. Let θ 2 K. De¯nition: H := L © Lu with u 2 H such that u2 = θ and u¸ = ¸u for all ¸ 2 L, with ¹ : L ¡! L the non-trivial automorphism from L to L. Notation: H := (L; ¹ ; θ) Remark: H is a central simple algebra over K. A quaternion algebra is a central simple algebra of dimension 4 over K. We also introduce a second ¤ a;b De¯nition: Let a; b 2 K and H := ( K ) be the algebra generated by i; j with the relations i2 = a; j2 = b; ij = ¡ji = k. We have k2 = ¡ab. As K-vectorspace H is generated by the elements f1; i; j; kg. 2 » a;b Remark: For L = K(i); θ = b; u = j; i = a we get: L © Lu = ( K ). The ¤ a;b » b;a » elements a; b 2 K as well as L; θ are not unique, because of: ( K ) = ( K ) = a;¡ab » ac2;b ( K ) = ( K ). a;b » a;b For a ¯eld extension M j K we have: M ­K ( K ) = ( M ), respectivly: » M ­K (L; ¹ ; θ) = (M ­K L; ¹ ; θ) De¯nition: Let h = ® + i¯ + jγ + k± 2 H with ®; ¯; γ; ± 2 K. h¹ := ¹ ® ¡ i¯ ¡ jγ ¡ k±, respectivly: for h = ¸1 + ¸2u de¯ne: h = ¸1 ¡ ¸2u. ¹ : H ¡! H is an involution, extending the non-trivial K-automorphism ¹ : L ¡! L, called conjugation. This conjugation is additive and anti- communtative. For h 2 H we de¯ne n(h) := h ¢ h;¹ t(h) := h + h¹. This is the reduced norm and the reduced trace. The reduced norm de¯nes a quadratic form on H Also de¯ne: H £ H ¡! K given by: (h1; h2) 7!< h1; h2 >:= t(h1 ¢ h2). Lemma ² h 2 H is invertible () n(h) 6= 0 ² <; > de¯nes a K-linear, non-degenerate bilinearform. 1 Remark: Let H ¡! EndK (H); h 7! lh : a 7! h ¢ a be the left regular 2 representation. Then it holds: 2t(h) = T r(lh) and n(h) = det(lh). De¯nition: Let F j K be an extension. F is called splitting ¯eld for H () » HF := F ­K H = M2(F ). A F -representation is an inclusion H ¡! M2(F ). Examples: 1. M (K) » ( 1;¡1 ). Let L = K(i) with i2 = ¡1 and j2 = 1. Consider 2 = K à ! à ! à ! 1 0 0 ¡1 0 1 the matrices e1 = ; ei = ; ej = ; eij = 0 1 1 0 1 0 à ! ¡1 0 . This gives an K-algebra-isomorphism: 0 1 e1 7! 1; ei 7! i; ej 7! j; eij 7! ij ¡1;¡1 2. H := ( R ) the real quaternions have a C-representation: à ! z1 z2 H = f 2 M2(C)g ¡z¹2 z¹1 2 2 » H = R(i) © R(i)j with i = ¡1; j = ¡1. We have: H ­R C = M2(C) with H M2(R). 3. Let a; b 2 K¤. Consider: à ! à p ! à ! 0 1 b 0 1 0 I = ;J = p ;E2 = a 0 0 ¡ b 0 1 à p ! 2 2 0 ¡ b Then it holds: I = aE2;J = bE2;IJ = ¡JI = p a b 0 For ®; ¯; γ; ± 2 K¤ one gets: à p p ! ® + γ (b) ¯ ¡ ± b ®E2 + ¯I + γJ + ±IJ = p p a(¯ + ± b) ® ¡ γ (b) p With ¸s 2 L = K( b) for s = 1; 2, we obtain: à ! a; b ¸ ¸ p ( ) » f 1 2 2 M (K( b))g = ¹ ¹ 2 K a¸2 ¸1 p and recover the splitting ¯eld as K( b). 2 Remark: Let H be a quaternion algebra over K. 1 ¹ Then B(h1; h2) := 2 t(h1 ¢ h2) 2 K gives a symmetric, non-degenerate, bilin- earform. We get B(h; h) = n(h). For pure quaternions H0 = fh 2 H j h = ¯i + γj + ±ijg the set fi; j; ijg forms an orthogonal basis. K is orthogonal to the span of this set, so that (H; B) as quadratic space, has f1; i; j; ijg as orthogonal basis. We have the following a;b 0 c;d Proposition: For H = ( K ) and H = ( K ) with a; b; c; d 2 K the following statements are equivalent: 1. H and H0 are isomorphic K-algebras 2. H and H0 are isometric as quadratic spaces 0 3. H0 and H0 are isometric as quadratic spaces Proof: done in the seminar a;a » a;¡1 Corollary:( K ) = ( K ) because of their isometric normforms. a;b Theorem: For H = ( K ) the following statements are equivalent: 1;¡1 » 1. H = ( K ) = M2(K) 2. H is not a division algebra 3. H is isotrop as quadratic space p 4. a 2 NormF jK (F ) with F = K( b) Proof: done in the seminar ¤ a;¡a » 1;a » Corollary: Let a 2 K , then ( K ) = ( K ) = M2(K). If a 6= 0; 1 then a;1¡a » ( K ) = M2(K). a;b » Remark: For a ¯nite ¯eld K, not of characteristic 2, we get: ( K ) = M2(K) for all a; b 2 K 3 The Skolem-Noether-Theorem Lemma: ' : H ­K H ¡! EndK (H); h1 ­K h2 7! (h 7! h1 ¢ h ¢ h2) is a K-algebra isomorphism. Proof: done in the seminar Theorem: Let L j K be a quadratic extension contained in H. ¿ : L ¡! L the non-trivial K-automorphism of L. Then there exists a K-algebra- automorphism σ : H ¡! H which extends ¿, and is of the form: σ(h) = a ¢ h ¢ a¡1 for an a 2 H¤. Every K-automorphism σ : H ¡! H is an inner automorphism. Proof: done in the seminar Corollary: Let L j K be a quadratic extension contained in H. Then there exists θ 2 K¤ such that H »= (L; ¹ ; θ). θ depends on the non-trivial automorphism ¿ : L ¡! L. Proof: done in the seminar Remark: Let Aut(H j K) be the group of K-automorphisms of H. Then Aut(H j K) »= H¤=K¤. » Proposition: Let H = (L; ¹ ; θ). Then it holds: H = M2(K) () θ 2 n(L). Proof: =): If H is not a division algebra then there exists an element h 6= 0 2 H with n(h) = 0. Let h be written in the form h = ¸ + ¹u with u2 = θ 2 K¤. Then it follows: 0 = n(h) = ¸ ¢ ¸¹ ¡ ¹ ¢ ¹ ¢ θ with ¸; ¹ 6= 0. This ¸ gives θ = n( ¹ ) 2 n(L). (=: θ 2 n(L) () θ = ¸ ¢ ¸¹ for a ¸ 2 L. Consider h := ¸ + u 6= 0 2 H. For this element we get: n(h) = 0. So H is not a division algebra. 4 Further results Proposition (Frobenius): Let D be a ¯nite dimensional division algebra with center R. Then D »= H (the real quaternions) Proof: Let d 2 D ¡ R. Then R(d) »= R(i) »= C with i2 = ¡1. So, R(d) is proper contained in D. Let d~2 D be an element such that R(d~) »= R(u) with u2 = ¡1 and u2 = R(i). The elements i; u do not commute, because otherwise R(i; u) would be a ¯eld extension. But then u 2 R(i). Consider now the element j := iui + u. For this we get: ij = ¡ji. Let H := R(i) + R(i)j ½ D, and ¿ : i 7! ¡i be conjugation in R(i). Then 'j : H ¡! H, de¯ned as ¡1 2 x 7! j ¢ x ¢ j extends ¿, because of 'j(i) = ¡i. We now claim that j 2 R: 2 2 2 2 2 2 2 2 j ¢i = i¢j So, j 2 R(i). ¿(j ) = 'j(j ) = j . So, j 2 R and j < 0 because otherwise j 2 R = center(D). But then we would get jij¡1 = ¡i = i, which is absurd. So we conclude that H »= H ½ D. If H is proper contained in D, we can choose an element d^ 2 D ¡ H such that di^ = id^ and d^2 2 R. We proceed as done before. But then we get: dji^ = d^(¡i)j = idj^ , which means that dj 2 R(i). But then d^2 H which contradicts our choise. =) H »= D. We conclude the section with a remarkable property of quaternion algebras: Proposition: Let a; b; c 2 K¤. Then it holds: a; b a; c a; bc ( ) ­ ( ) »= ( ) ­ M (K) K K K K K 2 a;b » 2 2 2 Proof: Let H1 := ( K ) =< 1; i1; j1; k1 >K with i1 = a; j1 = b; k1 = ¡ab a;c » and similar H2 := ( K ) =< 1; i2; j2; k2 >K . Consider H3 :=< 1 ­ 1; i1 ­ 1; j1 ­ j2; k1 ­ j2 >K =:< 1; I; J; IJ >K . This is a 4-dimensional subalgebra 2 2 » a;bc of H1 ­H2. We have: I = a; J = bc; IJ = ¡JI. This gives H3 = ( K ). ~ ~ ~ De¯ne H4 :=< 1 ­ 1; 1 ­ j2; i1 ­ k2; ¡ci1 ­ i2 >K =< 1; I; J; K >K . We then ~2 ~2 2 ~ ~ ~~ c;¡a2c get: I = c; J = ¡a c; IJ = JI. This gives: H4 = ( K ). Using » the classi¯cation by quadratic forms we obtain that H4 = M2(K). Further » » more we get that H1 ­ H2 = H3 ­ H4 = H3 ­ M2(K). This follows from the fact that f1; I; J; Kg commutes with f1; I;~ J;~ K~ g elementwise. Evaluating the two tensorproducts gives the proposition. » Corollary: H ­ H = M4(R). ¡1;¡1 ¡1;¡1 » ¡1;1 » » Proof:( R ) ­R ( R ) = ( R ) ­R M2(R) = M2(R) ­R M2(R) = M4(R). 5.