# Separable algebras over a field are reduced

Let $k$ be a field, and let $A$ be a $k$-algebra which is finite as a $k$-module. Then $A$ is said to be separable over $k$ if the trace map $\operatorname{Tr}$ induces an isomorphism of $A$ with its dual, i.e. if $a \in A$, and $\operatorname{Tr}(aa') = 0$ for all $a' \in A$, then $a = 0$. This may not be a standard definition.

I want to show that if $A$ is separable, then $A$ is a direct sum of finite separable field extensions of $k$.

Here's what I have so far: for each $a \in A$, the ring extension $k[a]$ is finite as a $k$-algebra, so $A$ is an integral extension of $k$. This implies $A$ and $k$ have the same dimension, so $A$ is an artinian $k$-algebra.

Since $A$ is artinian, it has finitely many maximal ideals $\mathfrak m_1, ... , \mathfrak m_t$. As is the case of finitely generated algebras over a field, the nilradical $I$ of $A$ is the intersection of the $\mathfrak m_i$.

If I can show that $A$ is reduced, then I would have $A \cong L_1 \times \cdots \times L_t$ as $k$-algebras for $L_i = A/\mathfrak m_i$. The nondegeneracy of the trace would easily restrict to nondegeneracy on each $L_i$, making each $L_i$ a finite separable field extension of $k$.

So my problem is to show that $A$ is reduced.

Okay, how about this. Let $a \in A$, and $a^m = 0$ for some $m \geq 1$. Let $\phi = \phi_a: A \rightarrow A$ be the linear map $x \mapsto xa$. Our assumption on $a$ implies that $\phi^m = 0$. Since the linear maps $\phi_x : x \in A$ commute with each other, they can be simultaneously upper triangularized. So we can identify $\phi$ with an upper triangular matrix. Since $\phi^m$ is the zero transformation, $\phi$ is a nilpotent matrix, and must have zeroes on the diagonal.
Now let $b \in A$, and let $\psi: A \rightarrow A$ be the linear map $x \mapsto xb$. The trace of $ab$ is by definition the trace of the matrix product $\phi \circ \psi$. But if you multiply a matrix with zeroes on the diagonal with an upper triangular matrix, the product will also have zeroes on the diagonal. So $\operatorname{Tr}(ab) = 0$. Since $b$ is arbitrary, this implies $a = 0$.