Element of with n0n-zero trace Let $F$ be a field of characteristic $p$ and $K$ a finite, separable extension of $F$ such that $p \mid [K : F]$. I want to show that there must exist an element of $K$ with non-zero trace.
One idea I worked on began as follows: Since $K/F$ is separable, it has a primitive element $\alpha$. Suppose that $\mathrm{tr}(\alpha^k) = 0$ for all $k\geq 0$. I then was able to conclude that many coefficients of $m_\alpha(x)$ must be zero, and I can see that with enough crunching I ought to be able to contradict separability.
I wanted to know if there is a more elegant way to show the existence of my desired element? Note, I am doing this in order to prove that the trace form of a separable extension is non-singular, so I don't want to quote this fact.
 A: Write your extension in primitive form, $K=F[x]/(m(x))$. Note then that the nonzeroness of the map $\text{Tr}_{K/F}:K\to F$ is preserved under tensoring by $\overline{F}$. Thus, we see that our map is nonzero, if and only if the map $1\otimes\text{Tr}_{K/F}:\overline{F}\otimes_F K\to \overline{F}\otimes_F F$ is nonzero. But, $\overline{F}\otimes K\cong \overline{F}[x]/(m(x))$. Now, since $m(x)$ is separable, we know that $m(x)=(x-\alpha_1)\cdots(x-\alpha_n)$ where each $\alpha_i$ is distinct. So, in fact, $\overline{F}[x]/(m(x))\cong \overline{F}^n$. But, I leave it to you to check that in general, if $X$ and $Y$ are $F$-algebras, then $\text{Tr}_{X\times Y/F}=\text{Tr}_{X/F}+\text{Tr}_{Y/F}$. So, to prove that $\text{Tr}_{\overline{F}^n/\overline{F}}$ is nonzero, it suffices to check that each $\text{Tr}_{\overline{F}/\overline{F}}$ is nonzero. But, this is trivial since the last map is the identity.
By the way, this seems like a very hard way of showing the non-degeneracy of the trace form.
