The Stacks project

Proof. It is clear that (3) implies (2). If (2) holds, then pick $\gamma \in L$ with $\text{Trace}_{L/K}(\gamma ) \not= 0$. Then if $\alpha \in L$ is nonzero, we see that $Q_{L/K}(\alpha , \gamma /\alpha ) \not= 0$. Hence $Q_{L/K}$ is nondegenerate. This proves the equivalence of (2) and (3).

Suppose that $K$ has characteristic $p$ and $L = K(\alpha )$ with $\alpha \not\in K$ and $\alpha ^ p \in K$. Then $\text{Trace}_{L/K}(1) = p = 0$. For $i = 1, \ldots , p - 1$ we see that $x^ p - \alpha ^{pi}$ is the minimal polynomial for $\alpha ^ i$ over $K$ and we find $\text{Trace}_{L/K}(\alpha ^ i) = 0$ by Lemma 9.20.3. Hence for this kind of purely inseparable degree $p$ extension we see that $\text{Trace}_{L/K}$ is identically zero.

Assume that $L/K$ is not separable. Then there exists a subfield $L/K'/K$ such that $L/K'$ is a purely inseparable degree $p$ extension as in the previous paragraph, see Lemmas 9.14.6 and 9.14.5. Hence by Lemma 9.20.5 we see that $\text{Trace}_{L/K}$ is identically zero.

Assume on the other hand that $L/K$ is separable. By induction on the degree we will show that $\text{Trace}_{L/K}$ is not identically zero. Thus by Lemma 9.20.5 we may assume that $L/K$ is generated by a single element $\alpha $ (use that if the trace is nonzero then it is surjective). We have to show that $\text{Trace}_{L/K}(\alpha ^ e)$ is nonzero for some $e \geq 0$. Let $P = x^ d + a_1 x^{d - 1} + \ldots + a_ d$ be the minimal polynomial of $\alpha $ over $K$. Then $P$ is also the characteristic polynomial of the linear maps $\alpha : L \to L$, see Lemma 9.20.2. Since $L/k$ is separable we see from Lemma 9.12.4 that $P$ has $d$ pairwise distinct roots $\alpha _1, \ldots , \alpha _ d$ in an algebraic closure $\overline{K}$ of $K$. Thus these are the eigenvalues of $\alpha : L \to L$. By linear algebra, the trace of $\alpha ^ e$ is equal to $\alpha _1^ e + \ldots + \alpha _ d^ e$. Thus we conclude by Lemma 9.13.2. $\square$