Prove two bases are dual in a finite field. Let K be a finite field, $F=K(\alpha)$ a finite simple extension of degree $n$, and $ f \in K[x]$ the minimal polynomial of $\alpha$ over $K$. Let $\frac{f\left( x \right)}{x-\alpha }={{\beta }_{0}}+{{\beta }_{1}}x+\cdots +{{\beta }_{n-1}}{{x}^{n-1}}\in F[x]$ and $\gamma={f}'\left( \alpha  \right)$.
Prove that the dual basis of $\left\{ 1,\alpha ,\cdots ,{{\alpha }^{n-1}} \right\}$ is $\left\{ {{\beta }_{0}}{{\gamma }^{-1}},{{\beta }_{1}}{{\gamma }^{-1}},\cdots ,{{\beta }_{n-1}}{{\gamma }^{-1}} \right\}$.
I met this exercise in "Finite Fields" Lidl & Niederreiter Exercises 2.40, and I do not how to calculate by Definition 2.30. It is
Definition 2.30 Let $K$ be a finite field and $F$ a finite extension of $K$. Then two bases $\left\{ {{\alpha }_{1}},{{\alpha }_{2}},\cdots ,{{\alpha }_{m}} \right\}$ and $\left\{ {{\beta }_{1}},{{\beta }_{2}},\cdots ,{{\beta }_{m}} \right\}$ of $  F$ over $K$ are said to be dual bases if for $1\le i,j\le m$ we have $T{{r}_{{F}/{K}\;}}\left( {{\alpha }_{i}}{{\beta }_{j}} \right)=\left\{ \begin{align}
  & 0\;\;\text{for}\;\;i\neq j, \\ 
 & 1\;\;\text{for}\;\;i=j. \\ 
\end{align} \right.$
I think $\gamma =\underset{x\to \alpha }{\mathop{\lim }}\,\frac{f(x)-f{{(\alpha )}_{=0}}}{x-\alpha }={{\beta }_{0}}+{{\beta }_{1}}\alpha +\cdots {{\beta }_{n-1}}{{\alpha }^{n-1}}$.
How can I continue? The lecturer did not teach the "dual bases" section.
 A: If $\;G=Gal(F/K)=\{\sigma_1:=Id,\sigma_2,...,\sigma_n\}\;$, then using your nice characterization 
$$\;\gamma:=f'(\alpha)=\sum_{k=0}^{n-1}\beta_k\alpha^k\;$$
we get:
$$tr.(\alpha^i\beta_j\gamma^{-1})=\sum_{k=1}^m\sigma_k(\alpha^i\beta_j\gamma^{-1})=\sum_{k=1}^n\sigma_k(\alpha)^i\sigma_k(\beta_j)\sigma_k(\gamma^{-1})=$$
$$=\sum_{k=1}^n\sigma_k(\alpha)^i\sigma_k(\beta_j)\sigma_k\left(\left(\sum_{t=0}^{n-1}\beta_t\alpha^t\right)^{-1}\right)=\sum_{k=1}^n\sigma_k(\alpha)^i\sigma_k(\beta_j)\left(\sum_{t=0}^{n-1}\sigma_k(\beta_t)\sigma_k(\alpha)^k\right)^{-1}=$$
$$=\frac{\alpha^i\beta_j}{\beta_0+\beta_1\alpha+\ldots+\beta_{n-1}\alpha^{n-1}}+$$
$$\frac{\sigma_2(\alpha)^i\sigma_2(\beta_j)}{\sigma_2(\beta_0)+\sigma_2(\beta_1)\sigma_2(\alpha)+\ldots+\sigma_2(\beta_{n-1})\sigma_2(\alpha^{n-1})}+\ldots+$$
$$+\frac{\sigma_n(\alpha)^i\sigma_n(\beta_j)}{\sigma_n(\beta_0)+\sigma_n(\beta_1)\sigma_n(\alpha)+\ldots+\sigma_n(\beta_{n-1})\sigma_n(\alpha^{n-1})}$$
But $\;\sigma_k(\beta_j)=\beta_j\;$ since $\;\beta_j\in K\;$ , so the above is
$$=\frac{\alpha^i\beta_j}{\beta_0+\beta_1\alpha+\ldots+\beta_{n-1}\alpha^{n-1}}+$$
$$\frac{\sigma_2(\alpha)^i\beta_j}{\beta_0+\beta_1\sigma_2(\alpha)+\ldots+\beta_{n-1}\sigma_2(\alpha^{n-1})}+\ldots+$$
$$+\frac{\sigma_n(\alpha)^i\beta_j}{\beta_0+\beta_1\sigma_n(\alpha)+\ldots+\beta_{n-1}\sigma_n(\alpha^{n-1})}$$
Now, if $\;i=j\;$ we simply have
$$tr.\left(\alpha^i\beta_j\gamma^{-1}\right)=\sum_{k=1}^n\frac{\sigma_k(\alpha)^i\beta_i}{\sum_{k=1}^n\sigma_k(\alpha)^i\beta_i}=1$$
and if $\;i\neq j\;$ 
$$tr.\left(\alpha^i\beta_j\gamma^{-1}\right)=\sum_{k=1}^n\frac{\sigma_k(\alpha)^i}{\sum_{t=0}^{n-1}\sigma_k(\alpha)^t\beta_t}\beta_j$$
I shall come back to this later...perhaps.
A: This is a standard fact of algebraic field extensions, and doesn't need the assumption that $K$ is finite.
Let $f(x)=\prod_{i=1}^n(x-\alpha_i),$ where $\alpha_1=\alpha$, so the conjugates of $\alpha$ are $\alpha_i, i=1,2,\ldots,n$. Let $0\le r\le n-1$. Consider the polynomial
$$
g_r(x)=\sum_{i=1}^n\frac{f(x)}{x-\alpha_i}\cdot\frac{\alpha_i^r}{f'(\alpha_i)}.
$$
If $i\neq j$, then $f(x)/(x-\alpha_i)$ evaluated at $\alpha_j$ is obviously zero. OTOH, if $i=j$, then $f(x)/(x-\alpha_i)$ evaluated at $\alpha_j=\alpha_i$ is equal to $f'(\alpha_i)$.
Thus we conclude that $g_r(\alpha_i)=\alpha_i^r$ for all $i=1,2,\ldots,n$.
But $g_r(x)$ is clearly of degree $\le n-1$, and the polynomial $g_r(x)-x^r$ has $n$ distinct zeros, namely $x=\alpha_i, 1\le i\le n$. This is possible only, if $g_r(x)=x^r$.
Let us extend the definition of the Galois group actioin and the trace from $F$ to $F[x]$ by declaring that
$$
\sigma(\sum_ic_ix^i)=\sum_i\sigma(c_i)x^i,
$$
for all the automorphisms $\sigma$ (if $F/K$ were not Galois, we would use the various embeddings of $F$ into an algebraic closure), and then declare that
$$
tr(\sum_i c_ix^i)=\sum_{\sigma}\sigma(\sum_i c_ix^i)=\sum_i tr(c_i)x^i,
$$
i.e. we let the Galois group and the trace act on the coefficients. 
We are nearly done, because we can now conclude that
$$
tr\left(\frac{f(x)\alpha^r}{(x-\alpha)f'(\alpha)}\right)=\sum_{i=1}^n\frac{f(x)\alpha_i^r}{(x-\alpha_i)f'(\alpha_i)}=g_r(x)=x^r.\qquad(1)
$$
To verify the first equality above let the Galois group act on $f(x)/(x-\alpha)$.
On the other hand
$$
tr\left(\frac{f(x)\alpha^r}{(x-\alpha)f'(\alpha)}\right)=
\sum_{i=0}^{n-1}tr\left(\frac{\beta_i\alpha^r}{f'(\alpha)}\right)x^i.\qquad(2)
$$
Equating the coefficients of like powers of $x$ in both $(1)$ and $(2)$ for all $i$ and all $r$ gives you the claim:
$$
tr\left(\frac{\beta_i\alpha^r}{f'(\alpha)}\right)=\delta_{ir}.
$$
