Separable field extensions are Frobenius algebras Wikipedia says that if $L/K$ is finite extension then $L/K$ is separable if and only if $L$ is a separable $K$-algebra. I am interested in the "only if" direction, which is outlined in the article.
$L/K$ has a primitive element $a$, and let $p(x) \in K[x]$ be it's minimal polynomial. Writing
$p(x) = (x-a) \sum_{i=0}^{n-1} b_i x^i$
we obtain a separability idempotent for $L/K$:
$\sum_{i=0}^{n-1} a^i \otimes_K \frac{b_i}{p'(a)}$
I would like to prove that this is indeed a separability idempotent by showing that $\{a^i\}$ and $\{\frac{b_i}{p'(a)}\}$ are dual bases with respect to the inner product $(y,z) \mapsto \mathrm{Tr}_{L/K}(yz)$. In particular: 
$z = \sum_{i=0}^{n-1}a^i\mathrm{Tr}_{L/K}(\frac{b_i z}{p'(a)})$
or
$z = \sum_{i=0}^{n-1} \mathrm{Tr}_{L/K}(za^i)\frac{b_i}{p'(a)}$
for all $z \in L$, but my Galois theory is a little rusty.
This is more than separability: such algebras, (with non-degenerate trace) are called strongly separable (see A note on strongly separable algebras)
 A: This only answers the question of the element being a seperability idempotent, not the dual basis claim.
In the first place, in the Wikipedia article the definition of a separability idempotent $$p=\sum x_i\otimes y_i$$ is one that commutes with all elements in $A$ (automatically verified here) and such that $\sum x_i y_i=1$.  This is achieved by the computation:
$p'(a) = \sum_{i=0}^{n-1}b_i a^i$ so $$\sum_{i=0}^{n-1} a^i \frac{b_i}{\sum_{i=0}^{n-1}b_i a^i}=1.$$
To show that the sets $\langle a^i\rangle$ and $\langle \frac{b_i}{p'(a)}\rangle$ are dual bases under the trace form, we need to show $$\langle a^i,\frac{b_j}{p'(a)}\rangle = \delta_{ij}$$ because we already know that $a^i$ span $L$ over $K$.  From these Kroneckers you can immediately get the representations for $z\in L$ you have above.
I started doing a computation to show this but seems to require more time than I can spare now.  I started with $$ \langle a^i,\frac{b_j}{p'(a)}\rangle = \sum_{\sigma} \sigma( a^i\frac{b_j}{p'(a)}) = \sum_{\sigma} \frac{\sigma(a)^i \sigma(b_j)}{p'(\sigma(a))}$$  and noted that $$p'(\sigma(a)) = \prod_{\sigma'\neq \sigma}(\sigma'(a)-\sigma(a))$$ (the product ranges over $\sigma'$ only) but I have not been able to simplify this somehow to show the required identity.
