Lemma 9.20.7. Let $L/K$ be a finite extension of fields. The following are equivalent:

1. $L/K$ is separable,

2. $\text{Trace}_{L/K}$ is not identically zero, and

3. 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. $\square$

Comment #1772 by Carl on

The first sentence of the proof is missing a "not" somewhere.

In the last paragraph, "If \alpha\notin K, then K' is not maximal" should be \alpha\notin K'.

Comment #3226 by Fan on

The last paragraph actually proves the same thing as the the previous ones: (2) implies (1), so (1) implies (2) is still missing. Maybe the standard proof using $\det(Tr(a_ia_j))=\det(\sigma_i(a_j))^2$ and linear independence of characters will do.

Comment #3329 by on

Oops! Good catch. This is fixed here in a slightly different manner than you suggest.

Comment #3377 by on

Actually, rather you should look here since my previous attempt was erroneous.

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