The Stacks Project


Tag 0BIL

Chapter 9: Fields > Section 9.20: Trace and norm

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}

    Comments (1)

    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.

    Add a comment on tag 0BIL

    Your email address will not be published. Required fields are marked.

    In your comment you can use Markdown and LaTeX style mathematics (enclose it like $\pi$). A preview option is available if you wish to see how it works out (just click on the eye in the lower-right corner).

    All contributions are licensed under the GNU Free Documentation License.




    In order to prevent bots from posting comments, we would like you to prove that you are human. You can do this by filling in the name of the current tag in the following box. So in case this where tag 0321 you just have to write 0321. Beware of the difference between the letter 'O' and the digit 0.

    This captcha seems more appropriate than the usual illegible gibberish, right?