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):

  1. There exists a nondegenerate associative $K$-bilinear form $A \times A \rightarrow K$.

  2. 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.

  3. 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 $A \cong A^b$.

  • $\begingroup$ Can you remind me what notion of basic is at play here? $\endgroup$
    – rschwieb
    Apr 11, 2023 at 17:23
  • $\begingroup$ I eddited the post with the definition of basic that is given in the book. $\endgroup$
    – Andarrkor
    Apr 12, 2023 at 18:02

2 Answers 2


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.

  • $\begingroup$ Thanks for the answer, but this is not what I asked in my post. $\endgroup$
    – Andarrkor
    Apr 12, 2023 at 17:55
  • $\begingroup$ It is what you asked. You asked how to prove that finite-dimensional division algebras are Frobenius algebras, and this shows that they are symmetric Frobenius, so it's even a bit better. This show in particular property 3 of your post. (Just take $\rho$ to be the reduced trace if you need that extra bit of detail.) $\endgroup$ Apr 12, 2023 at 18:10
  • $\begingroup$ You asked for a direct proof that division algebras are Frobenius, so $A\cong DA$ as left (or right) $A$-modules. I gave a direct proof that division algebras are symmetric, so $A\cong DA$ as $A$-bimodules, by showing that there is always some nonzero $\rho\in DA$ satisfying $\rho(ab)=\rho(ba)$ for all $a,b\in A$. $\endgroup$ Apr 13, 2023 at 10:10
  • $\begingroup$ @AndewHubery Ok, I noticed that the proof of the book is overcomplicated. Now, I am trying to understand yours. Some questions: 1. If I start with a $K$-algebra $A$, why can you assume that the center of $A$ is $K$? 2. If I understand the proof correctly, since $A/\mathcal{C}(A)$ is nonzero, the dual is nonzero so there is a K-linear functional $A/\mathcal{C}(A) \rightarrow K$, which gives the desired $\rho: A \rightarrow A/\mathcal{C}(A) \rightarrow K$. $\endgroup$
    – Andarrkor
    Apr 21, 2023 at 19:15
  • $\begingroup$ For 2 that is correct. For 1 if the centre is $K$ and if we have a subfield $k$ with $\dim_kK$ finite, then we can compose $\rho$ with any nonzero linear $K\to k$. $\endgroup$ Apr 22, 2023 at 5:33

The OP has asked for a more direct proof that a finite-dimensional division algebra $D$ over a field $K$ is Frobenius: Choose any $K$-linear map $\lambda:D\to K$ with $\lambda(1)=1$. It is easy to check that $\beta(x,y)=\lambda(xy)$ defines a non-degenerate, associative bilinear form on $D$.


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