Trace as Bilinear form on a field extension

Solution 1:

For what it's worth, a complete proof of the following fact can be found in $\S 6$ of these notes.

Theorem: For a finite degree field extension $K/F$, the following are equivalent:
(i) The trace form $(x,y) \in K^2 \mapsto \operatorname{Tr}(xy) \in F$ is a nondegenerate $F$-bilinear form.
(ii) The trace form is not identically zero.
(iii) The extension $K/F$ is separable.

The specific answer to the OP's question appears in there, but here it is: always be careful to read the fine print when dealing with inseparable extensions. In this case, it turns out that when $K/F$ is inseparable, the trace from $K$ down to $F$ of $x \in K$ comes out as a power of $p$ (the characteristic of $F$) times the more familiar sum of Galois conjugates. But in characteristic $p$ multiplying something by a power of $p$ is the same as multiplying it by zero...so the trace is identically zero in inseparable extensions.

[Note also that the proof of another part of the theorem makes use of the Primitive Element Theorem, which is currently to be found in $\S 7$ of the notes, i.e., the following section. This is an obvious mistake which will have to be remedied at some point. There are plenty of other issues with these notes, which are still quite rough and incomplete.]

Solution 2:

I don't know about the separability of the extension but I can see how to prove the proposition. Take your $\alpha$ and hit it with $\alpha^{-1}\beta$ and the conclusion follows quickly.