# The Stacks Project

## Tag 0BIL

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 $\text{Trace}_{L/K}$ is not identically zero. Let $L/K'/K$ be a maximal subfield separable over $K$. Then by Lemma 9.20.5 we see that $\text{Trace}_{L/K'}$ is not identically zero. Then we pick $\alpha \in L$ such that $\text{Trace}_{L/K'}(\alpha) \not = 0$. Then by Lemma 9.20.3 we see that $\alpha$ is separable over $K'$. If $\alpha \not \in K'$ then $K'$ is not maximal. If $\alpha \in K'$ then Lemma 9.20.3 shows that the characteristic of $K$ does not divide $[L : K']$ which implies that $L/K'$ is separable (as the inseparable degree of $L/K'$ is forced to be $1$, see Definition 9.14.7) and hence trivial. $\square$

The code snippet corresponding to this tag is a part of the file fields.tex and is located in lines 2423–2431 (see updates for more information).

\begin{lemma}
\label{lemma-separable-trace-pairing}
Let $L/K$ be a finite extension of fields. The following are equivalent:
\begin{enumerate}
\item $L/K$ is separable,
\item $\text{Trace}_{L/K}$ is not identically zero, and
\item the trace pairing $Q_{L/K}$ is nondegenerate.
\end{enumerate}
\end{lemma}

\begin{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).

\medskip\noindent
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 \ref{lemma-trace-and-norm-from-minimal-polynomial}.
Hence for this kind of purely inseparable degree $p$ extension
we see that $\text{Trace}_{L/K}$ is identically zero.

\medskip\noindent
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 \ref{lemma-separable-first} and \ref{lemma-finite-purely-inseparable}.
Hence by Lemma \ref{lemma-trace-and-norm-tower}
we see that $\text{Trace}_{L/K}$ is identically zero.

\medskip\noindent
Assume on the other hand that $\text{Trace}_{L/K}$ is not identically zero.
Let $L/K'/K$ be a maximal subfield separable over $K$. Then by
Lemma \ref{lemma-trace-and-norm-tower}
we see that $\text{Trace}_{L/K'}$ is not identically zero.
Then we pick $\alpha \in L$ such that
$\text{Trace}_{L/K'}(\alpha) \not = 0$.
Then by Lemma \ref{lemma-trace-and-norm-from-minimal-polynomial}
we see that $\alpha$ is separable over $K'$. If $\alpha \not \in K'$
then $K'$ is not maximal. If $\alpha \in K'$ then
Lemma \ref{lemma-trace-and-norm-from-minimal-polynomial}
shows that the characteristic of $K$ does not divide $[L : K']$
which implies that $L/K'$ is separable
(as the inseparable degree of $L/K'$ is forced to be $1$,
see Definition \ref{definition-insep-degree})
and hence trivial.
\end{proof}

Comment #1772 by Carl on December 21, 2015 a 7:36 pm UTC

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'.

There is also 1 comment on Section 9.20: Fields.

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