I am following the book Frobenius Algebras I by Andrzej Skowronski and Kunio Yamagata to learn about Frobenius algebras. The goal of Chapter IV, section 5 is to show that finite dimensional semisimple $K$ algebras are symmetric $K$-algebras using the Noether-Skolem theorem.
A finte dimensional $K$-algebra $A$ is a Frobenius $K$-algebra if one of the following conditions is satisfied (Theorem 2.1 in the book):
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There exists a nondegenerate associative $K$-bilinear form $A \times A \rightarrow K$.
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There exists a $K$-linear form $\varphi : A \rightarrow K$ such that Ker $\varphi$ does not contain a nonzero right (resp. left) ideal of A.
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There exists an isomorphism of right (resp. left) A-modules $A_A \cong$ Hom$(A,K)_A$ (resp. s $_A A \cong$ $_A$Hom$(A,K)$ ).
In Proposition IV.5.16 it is claimed that if $F$ is a finite dimensional division $K$-algebra, then $F$ is a Frobenius $L=C(F)$ algebra, where $C(F)$ is the center of $F$. To justify this claim the authors use that $F$ is a finite dimensional basic selfinjective $L$ algebra, and then they use Proposition IV.3.9, which claims that finite dimensional basic self-injective algebras over a field are Frobenius.
My question is, is there a way to show this claim more directly? Is there a way to show that a finite dimensional division $K$-algebra is a Frobenius algebra?
My idea was to check that it satisfies condition 2., since $F$ is simple, but I do not know… Please tell me if someone knows an alternative way to show this fact.
EDIT: In the book the following definition of basic $K$-algebra is given. Let $A$ be a finite dimensional $K$-algebra and $eA$ a minimal progenerator of mod $A$ with $e^2=e$. The algebra $A^b=eAe$ is called basic algebra of $A$. The $K$-algebra $A$ is basic if $A \cong A^b$.
Best Answer
Every finite dimensional semisimple algebra $A$ is actually a symmetric algebra, meaning that there is an isomorphism of $A$-bimodules $A\cong D(A)$. Here $D$ is the usual vector space duality $D=\mathrm{Hom}(-,K)$.
To see this one first notes that products and matrix algebras of symmetric algebras are again symmetric, so one just needs to show it for division algebras.
Let $A$ be a finite dimensional division algebra with centre $K$. Let $F$ be a splitting field, so that $A\otimes_KF\cong M_n(F)$. Let $\mathcal C(A)$ be the subspace spanned by all commutators $ab-ba$ in $A$. Then $\mathcal C(A)\otimes_KF$ is contained in the space of commutators for $A\otimes_KF=M_n(F)$, which is a proper subspace. Thus $A/\mathcal C(A)$ is not zero, and any nonzero linear form $\rho\in D(A)$ vanishing on $\mathcal C(A)$ yields a bimodule isomorphism $$ A\cong D(A), \quad a\mapsto (\rho a\colon b\mapsto \rho(ab)). $$ This shows that $A$ is symmetric.