A rank-nulity theorem for Frobenius algebras

Posted: May 5, 2011 in Frobenius Algebras, Noncommutative Ring Theory Notes
Tags: , ,

As usual k is a field. In this post I will prove a rank-nulity theorem for Frobenius algebras. First we need a lemma also known as the Riesz representation theorem.

Lemma. Let V be a finite dimensional k-vector space and let B be a non-degenerate bilinear form on V. If f \in V^*, then there exists a unique v_0 \in V such that f(v)=B(v,v_0), for all v \in V.

Proof. The uniqueness is obvious because B is non-degenerate. Now let \{v_1, \ldots , v_n \} be a basis for V and suppose that [B] is the matrix of B with respect to this basis, i.e. the (i,j)-entry of [B] is B(v_i,v_j), for all 1 \leq i,j \leq n.  Let \bold{y} \in k^n be a vector whose i-th coordinate is f(v_i). By the theorem in this post, [B] is invertible because B is non-degenerate. So the system of equations [B] \bold{x} = \bold{y} has a solution \bold{x} \in k^n. Let x_i be the i-th coordinate of \bold{x}. So

\sum_{j=1}^n x_j B(v_i,v_j)=f(v_i),

for all i. Let v_0=\sum_{i=1}^n x_iv_i. Then for any v=\sum_{j=1}^n \alpha_i v_i \in V we have

f(v)=\sum_i \alpha_i f(v_i)=\sum_{i,j} x_j \alpha_i B(v_i,v_j)=\sum_i \alpha_i B(v_i,v_0)=B(v,v_0). \ \Box

Notation. Let V be a k-vector space, W a k-vector subspace of V and B a bilinear form on V. We let W^{\perp}=\{v \in V : \ B(w,v)=0, \ \forall w \in W \}.

Theorem. Let A be a Frobenius k-algebra. Let the bilinear form B and f \in V^* be as stated in Definition 3 and the theorem in this post, respectively. Then

1) L^{\perp} = \{x \in A : \ f(yx)=0, \ \forall y \in L \},

2) \dim_k L + \dim_k L^{\perp} = \dim_k A.

Proof.  Part 1) is clear because B(y,x)=f(yx), for all x,y \in A. So we only need to prove the second part of the theorem. Define the map \varphi : A \longrightarrow L^* by \varphi(x)(y)=f(yx), for all x \in A and y \in L. Obviously \varphi is a k-linear map and \ker \varphi = L^{\perp}, by part 1). We are now going to prove that \varphi is onto. So let g \in L^*. Let B_1 be the restriction of B to L. By the above lemma, there exists y_0 \in L such that

g(y)=B_1(y,y_0)=B(y,y_0)=f(yy_0),

for all y \in L. Hence g = \varphi(y_0) and so \varphi is onto. Thus, by the rank-nulity theorem for vector spaces, we have

\dim_k A = \dim_k \ker \varphi + \dim_k L^*=\dim_k L^{\perp} + \dim_k L. \ \Box

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