Lemma 9.20.7. Let L/K be a finite extension of fields. The following are equivalent:
L/K is separable,
\text{Trace}_{L/K} is not identically zero, and
the trace pairing Q_{L/K} is nondegenerate.
Lemma 9.20.7. Let L/K be a finite extension of fields. The following are equivalent:
L/K is separable,
\text{Trace}_{L/K} is not identically zero, and
the trace pairing Q_{L/K} is nondegenerate.
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
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