## A rank-nulity theorem for Frobenius algebras

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

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