If $K = F(\alpha)$ is a non separable and finite extension, then $T^K_F = 0$

Question

I am trying to prove that if $K = F(\alpha)$ is a non separable and finite extension then $T^K_F = 0$. Now, I looked at a solution and it goes as follows: Suppose the minimal polynomial of $\alpha$ is $f=x^n+a_{n-1}x^{n-1}+...+a_0$, then $\{1, \alpha, ... \alpha^{n-1}\}$ is a basis for $K$. Now, consider the linear transformation $x \to \alpha \cdot x$. It is easy to check that the trace of the matrix which represents this linear transformation with respect to the above basis is exactly $-a_{n-1}$. But since the extension is not seperable, $f$ must be a polynomial in $x^p$ (where $p=char(F)$) and therefore $a_{n-1}=0$. Now, I understand that it proves that $T^{K}_F (\alpha) = 0$, but why does it prove that $T^K_F =0$? Say, I don't understand why for example $T^K_F(\alpha^2)=0$. Why is that?

Answer 1

Since you have $T^K_F=T^M_F\circ T^K_M$ for any intermediate field $M$, it follows that $$ T^K_F(\alpha^2)=T^{F(\alpha^2)}_F\circ T^K_{F(\alpha^2)}(\alpha^2)=[K:F(\alpha^2)]T^{F(\alpha^2)}_F(\alpha^2)=0. $$ In the same way, $T_F^K(\alpha^i)=0$ for all $i$. Then by the $F$-linearity of the trace you get $T_F^K=0$.