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Math Help - Central simple algebra

  1. #1
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    Central simple algebra

    These two questions are the exercises in Hungerford's Algebra(p.462 ex.1 and ex.5).

    1. If A is afinite dimensional central simple algebra over the field K, then
    A tensor the opposit ring of A over K is isomorphic to a space of n by n
    matrix over K. Where n is the dimension of A over K.

    2. If A is a finite dimensional central simple algebra over a field K, then
    the dimension of A over K is a perfect square.

    Thank you!!!
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  2. #2
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    Quote Originally Posted by chipai View Post
    These two questions are the exercises in Hungerford's Algebra(p.462 ex.1 and ex.5).

    1. If A is afinite dimensional central simple algebra over the field K, then
    A tensor the opposit ring of A over K is isomorphic to a space of n by n
    matrix over K. Where n is the dimension of A over K.
    define the map f: A \times A^{op} \longrightarrow \text{End}_K(A) by f(a , b)(x)=axb, for all a,b,x \in A. this map is K-bilinear and so it induces a K-module homomorphism f: A \otimes_K A^{op} \longrightarrow \text{End}_K(A) defined by

    f(a \otimes b)(x)=axb, for all a,b,x \in A. this map now is also an algebra homomorphism because f((a \otimes b)(c \otimes d))(x)=f(ac \otimes db)(x)=acxdb=f(a \otimes b)(cxd)=f(a \otimes b)f(c \otimes d)(x).

    also f is injective because A \otimes_K A^{op} is a (central) simple algebra and so \ker f, which is an ideal of A \otimes_K A^{op}, must be trivial. finally \dim_K A \otimes_K A^{op} = n^2 = \dim_K \mathbb{M}_n(K)=\dim_K \text{End}_K (A)

    and therefore A \otimes_K A^{op} \cong \text{End}_K(A) \cong \mathbb{M}_n(K).


    Note: some facts that are used here:

    1) \text{End}_K (V) \cong \mathbb{M}_n(K), for any n dimensional K vector space V. we know this from elementary linear algebra.

    2) the tensor product of a central simple algebra and a simple algebra is a simple algebra. also the tensor product of two central simple algebras is central simple.

    3) for any finite dimensional K vector spaces V,W we have \dim_K V \otimes_K W = (\dim_K V)(\dim_K W).

    finally note that A^{op} cannot be replaced by A because then f would not be a ring homomorphism.


    2. If A is a finite dimensional central simple algebra over a field K, then
    the dimension of A over K is a perfect square.

    Thank you!!!
    first note that if F/K is any field extension, then A \otimes_K F is an F-algebra and \dim_F A \otimes_K F=\dim_K A. now if we assume that F is the algebraic closure of K and put B=A \otimes_K F, then

    B is a finite dimensional simple algebra. thus B is semisimple and so by Artin-Wedderburn we have B \cong \mathbb{M}_n(D), for some finite dimensional F division algebra D and some positive integer n.

    since F is algebraically closed and D is algebraic over F, we must have D=F. thus B \cong \mathbb{M}_n(F) and therefore \dim_K A = \dim_F B = \dim_F \mathbb{M}_n(F)=n^2.
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