# Division algebras are frobenius algebras

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 $$A \cong A^b$$.

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

## 2 Answers

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.

• Thanks for the answer, but this is not what I asked in my post. Apr 12, 2023 at 17:55
• 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.) Apr 12, 2023 at 18:10
• 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 $a,b\in A$. Apr 13, 2023 at 10:10
• @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 $\rho: A \rightarrow A/\mathcal{C}(A) \rightarrow K$. Apr 21, 2023 at 19:15
• 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 $K\to k$. 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 $$D$$.