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Application of the graded Posner theorem Y. Karasik 1 Department of Mathematics Technion - Israel Institute of Technology Group graded algebras Definition Let G be a group and F be a field. An (associative) F-algebra A is called G-graded if M A= Ag , g ∈G where Ag is an F-subspace of A and Ag Ah ⊆ Agh for every g, h ∈ G. Group graded algebras Definition Let G be a group and F be a field. An (associative) F-algebra A is called G-graded if M A= Ag , g ∈G where Ag is an F-subspace of A and Ag Ah ⊆ Agh for every g, h ∈ G. A is G-simple if it does not have any non-trivial graded ideals. Group graded algebras Definition Let G be a group and F be a field. An (associative) F-algebra A is called G-graded if M A= Ag , g ∈G where Ag is an F-subspace of A and Ag Ah ⊆ Agh for every g, h ∈ G. A is G-simple if it does not have any non-trivial graded ideals. Example (Fine grading) A = Fα H = L g ∈H F · ug , where H is a subgroup of G. More examples Example Take G = Z/2Z = {0, 1} and consider a A0 = c 0 b d 0 0 0 0 ; A1 = 0 e w 0 0 z x 0 In short, 0 y 0 1 0 0 1 1 1 0 More examples Example Take G = Z/2Z = {0, 1} and consider a A0 = c 0 b d 0 0 0 0 ; A1 = 0 e w 0 0 z x 0 In short, 0 y 0 1 0 0 1 1 1 0 Definition → Let G be a group and − g = (g1 , ..., gn ) ∈ G ×n . The G-graded F algebra → A = M− ( F ) is the algebra Mn (F) graded by: g More examples Example Take G = Z/2Z = {0, 1} and consider a A0 = c 0 b d 0 0 0 0 ; A1 = 0 e w 0 0 z x 0 In short, 0 y 0 1 0 0 1 1 1 0 Definition → Let G be a group and − g = (g1 , ..., gn ) ∈ G ×n . The G-graded F algebra → A = M− ( F ) is the algebra Mn (F) graded by: g g1−1 g1 = e g −1 g 2 1 .. . gn−1 g1 g1−1 g2 −1 g2 g2 = e g1−1 g3 g2−1 g3 −1 g3 g3 = e .. . gn−1 g2 ··· ··· ··· g1−1 gn g2−1 gb .. .. . gn−1 gn = e . Posner’s Theorem for G-graded algebras Theorem Suppose A is a G-graded F-algebra such that: Posner’s Theorem for G-graded algebras Theorem Suppose A is a G-graded F-algebra such that: 1 K := Z (A)e is a field. Posner’s Theorem for G-graded algebras Theorem Suppose A is a G-graded F-algebra such that: 1 K := Z (A)e is a field. 2 A satisfies an ordinary PI. Posner’s Theorem for G-graded algebras Theorem Suppose A is a G-graded F-algebra such that: 1 K := Z (A)e is a field. 2 A satisfies an ordinary PI. 3 A is G-semiprime (this holds if G is finite and J (A) = 0). Posner’s Theorem for G-graded algebras Theorem Suppose A is a G-graded F-algebra such that: 1 K := Z (A)e is a field. 2 A satisfies an ordinary PI. 3 A is G-semiprime (this holds if G is finite and J (A) = 0). Then, A is a G-simple finite dimensional K-algebra. Posner’s Theorem for G-graded algebras Theorem Suppose A is a G-graded F-algebra such that: 1 K := Z (A)e is a field. 2 A satisfies an ordinary PI. 3 A is G-semiprime (this holds if G is finite and J (A) = 0). Then, A is a G-simple finite dimensional K-algebra. This holds for F of characteristic zero and G a residually finite group. Posner’s Theorem for G-graded algebras Theorem Suppose A is a G-graded F-algebra such that: 1 K := Z (A)e is a field. 2 A satisfies an ordinary PI. 3 A is G-semiprime (this holds if G is finite and J (A) = 0). Then, A is a G-simple finite dimensional K-algebra. This holds for F of characteristic zero and G a residually finite group. If one replaces “G-semiprime” by “G-semisimple”, the theorem is true for every G. Posner’s Theorem for G-graded algebras Theorem Suppose A is a G-graded F-algebra such that: 1 K := Z (A)e is a field. 2 A satisfies an ordinary PI. 3 A is G-semiprime (this holds if G is finite and J (A) = 0). Then, A is a G-simple finite dimensional K-algebra. This holds for F of characteristic zero and G a residually finite group. If one replaces “G-semiprime” by “G-semisimple”, the theorem is true for every G. The main ingredient in the proof is the existence of a special kind of central polynomials: Strong central polynomials Definition A polynomial f (x1 , ..., xn ) ∈ F hX i is a strong central polynomial of exponent d 2 for the group G, If: Strong central polynomials Definition A polynomial f (x1 , ..., xn ) ∈ F hX i is a strong central polynomial of exponent d 2 for the group G, If: 1 f is central (non-identity) for every G-simple algebra A of exponent d 2 . Strong central polynomials Definition A polynomial f (x1 , ..., xn ) ∈ F hX i is a strong central polynomial of exponent d 2 for the group G, If: 1 f is central (non-identity) for every G-simple algebra A of exponent d 2 . Moreover, For every homogeneous elements a1 , ..., an (i.e. they are inside ∪g ∈G Ag ), f (a1 , ..., an ) 6= 0 ⇐⇒ f (a1 , ..., an )e 6= 0. Strong central polynomials Definition A polynomial f (x1 , ..., xn ) ∈ F hX i is a strong central polynomial of exponent d 2 for the group G, If: 1 f is central (non-identity) for every G-simple algebra A of exponent d 2 . Moreover, For every homogeneous elements a1 , ..., an (i.e. they are inside ∪g ∈G Ag ), f (a1 , ..., an ) 6= 0 ⇐⇒ f (a1 , ..., an )e 6= 0. 2 f is an identity for every G-simple algebra of smaller exponent. Strong central polynomials Definition A polynomial f (x1 , ..., xn ) ∈ F hX i is a strong central polynomial of exponent d 2 for the group G, If: 1 f is central (non-identity) for every G-simple algebra A of exponent d 2 . Moreover, For every homogeneous elements a1 , ..., an (i.e. they are inside ∪g ∈G Ag ), f (a1 , ..., an ) 6= 0 ⇐⇒ f (a1 , ..., an )e 6= 0. 2 f is an identity for every G-simple algebra of smaller exponent. It turns out that when F is of characteristic zero, every central polynomial for A is strong for every group G. A theorem of Aljadeff and Haile Theorem Let G be a group and F an algebraically closed of 0 characteristic and A, B two f.d. G-simple F-algebras. A theorem of Aljadeff and Haile Theorem Let G be a group and F an algebraically closed of 0 characteristic and A, B two f.d. G-simple F-algebras. Then, A is G-isomorphic to B iff idG (A) = idG (B ). A theorem of Aljadeff and Haile Theorem Let G be a group and F an algebraically closed of 0 characteristic and A, B two f.d. G-simple F-algebras. Then, A is G-isomorphic to B iff idG (A) = idG (B ). Notice that the theorem is easy when G = {e }. A theorem of Aljadeff and Haile Theorem Let G be a group and F an algebraically closed of 0 characteristic and A, B two f.d. G-simple F-algebras. Then, A is G-isomorphic to B iff idG (A) = idG (B ). Notice that the theorem is easy when G = {e }. Indeed, A = Mn (F), B = Mm (F) and n2 = exp(A) is determined by id (A). A theorem of Aljadeff and Haile Theorem Let G be a group and F an algebraically closed of 0 characteristic and A, B two f.d. G-simple F-algebras. Then, A is G-isomorphic to B iff idG (A) = idG (B ). Notice that the theorem is easy when G = {e }. Indeed, A = Mn (F), B = Mm (F) and n2 = exp(A) is determined by id (A). We now show how one can use Posner’s theorem in order to get a quick proof of this theorem. Generic algebras Let A be a G-graded F-algebra. The (G-graded) relatively free algebra of A is defined to be UA := F hXG i /idG (A). Generic algebras Let A be a G-graded F-algebra. The (G-graded) relatively free algebra of A is defined to be UA := F hXG i /idG (A). Every graded algebra B with the same graded identities as A, is a homomorphic image of UA . Generic algebras Let A be a G-graded F-algebra. The (G-graded) relatively free algebra of A is defined to be UA := F hXG i /idG (A). Every graded algebra B with the same graded identities as A, is a homomorphic image of UA . If A is also f.d., then it is possible to embed UA in a form of A: Generic algebras Let A be a G-graded F-algebra. The (G-graded) relatively free algebra of A is defined to be UA := F hXG i /idG (A). Every graded algebra B with the same graded identities as A, is a homomorphic image of UA . If A is also f.d., then it is possible to embed UA in a form of A: That is, there is a field LA for which UA is embedded inside ALA := A ⊗F LA . Proof of the theorem We know that idG (A) = idG (B ), so F hX i /idG (B ) ' F hX i /idG (B ). Proof of the theorem We know that idG (A) = idG (B ), so F hX i /idG (B ) ' F hX i /idG (B ). In other words, / U B UA _ ' _ ALA BLB Proof of the theorem We know that idG (A) = idG (B ), so F hX i /idG (B ) ' F hX i /idG (B ). In other words, / U B UA _ ' _ ALA BLB We want to use Posner’s theorem, but the center is not be a field... Proof of the theorem We know that idG (A) = idG (B ), so F hX i /idG (B ) ' F hX i /idG (B ). In other words, / U B UA _ ' _ ALA BLB We want to use Posner’s theorem, but the center is not be a field... The evaluation of Z (UA )e to A, is inside Z (A)e = F · 1A . So, for 0 6= f ∈ Z (UA )e and 0 6= g ∈ UA , we have an evaluation → → → f (− a ) · g (− a ) = αg (− a ) 6= 0, where 0 6= α ∈ F. Proof of the theorem We know that idG (A) = idG (B ), so F hX i /idG (B ) ' F hX i /idG (B ). In other words, / U B UA _ ' _ ALA BLB We want to use Posner’s theorem, but the center is not be a field... The evaluation of Z (UA )e to A, is inside Z (A)e = F · 1A . So, for 0 6= f ∈ Z (UA )e and 0 6= g ∈ UA , we have an evaluation → → → f (− a ) · g (− a ) = αg (− a ) 6= 0, where 0 6= α ∈ F. We may consider the algebras SA := Z (UA )e−1 UA ' Z (UB )e−1 UB =: SB . Both are over K =the quotient field of Z (UA )e = Z (UB )e and have the same G-graded identities as A (and B). Using Posner’s theorem Now we can use Posner’s theorem: Using Posner’s theorem Now we can use Posner’s theorem: We made Z (SA )e = K which is a field. Using Posner’s theorem Now we can use Posner’s theorem: We made Z (SA )e = K which is a field. Clearly A is PI (since f.d.), so also SA . Using Posner’s theorem Now we can use Posner’s theorem: We made Z (SA )e = K which is a field. Clearly A is PI (since f.d.), so also SA . J (UA ) = 0, since an evaluation of f ∈ J (UA ) into A is inside J (A) = 0. Hence, also J (SA ) = 0. Using Posner’s theorem Now we can use Posner’s theorem: We made Z (SA )e = K which is a field. Clearly A is PI (since f.d.), so also SA . J (UA ) = 0, since an evaluation of f ∈ J (UA ) into A is inside J (A) = 0. Hence, also J (SA ) = 0. So, SA and SB are G-simple and f.d. over K. Using Posner’s theorem Now we can use Posner’s theorem: We made Z (SA )e = K which is a field. Clearly A is PI (since f.d.), so also SA . J (UA ) = 0, since an evaluation of f ∈ J (UA ) into A is inside J (A) = 0. Hence, also J (SA ) = 0. So, SA and SB are G-simple and f.d. over K. We have the picture: / U B UA ' _ _ _ / S B _ SA ALA ' BLB Finishing the proof What is left to show is that SA is a form of A. Since then it follows that A and B are graded forms of each other. Finishing the proof What is left to show is that SA is a form of A. Since then it follows that A and B are graded forms of each other. For this it is enough to show that the natural map SA ⊗K LA → SA · LA is a graded isomorphism and SA · LA = ALA . Finishing the proof What is left to show is that SA is a form of A. Since then it follows that A and B are graded forms of each other. For this it is enough to show that the natural map SA ⊗K LA → SA · LA is a graded isomorphism and SA · LA = ALA . The first part follows from the fact that SA ⊗K LA is G-simple. Finishing the proof What is left to show is that SA is a form of A. Since then it follows that A and B are graded forms of each other. For this it is enough to show that the natural map SA ⊗K LA → SA · LA is a graded isomorphism and SA · LA = ALA . The first part follows from the fact that SA ⊗K LA is G-simple. The second part follows from dimLA ALA = expG (ALA ) = expG (SA · LA ) = dimLA SA LA . The End! Thank you!