1. ## 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!!!

2. Originally Posted by chipai
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 $\dim_K A = \dim_F B = \dim_F \mathbb{M}_n(F)=n^2.$