Let $L$ be a finite field extension of $K$. I have the following question:

Show that $L/K$ is separable if and only if the bilinear trace form $\text{Tr}_{L/K}:L\times L \to K$ is non-degenerate.

A proof or reference will be great. Thank you!


This is a quite well-known fact, the proof can be find in every textbook on algebraic number theory but I would like to tell a proof which is different from the standard one but more elementary.

The fact that $L/K$ is separable means that for every $a\in L$ its minimal polynomial $f$ over $K$ does not have multiple roots. Since $f$ is irreducible it is equivalent to $f'\not\equiv 0$. Consider the logarifmic derivative of $f$(the formal one) $$\frac{f'}{f}=(\log(x-a_1)\dots(x-a_n))'=\frac{1}{x-a_1}+\dots+\frac{1}{x-a_n}$$ there $a_1\dots a_n$ are the roots of the $f$(possibly not lying in $L$). Now we expand this as a formal series $$\frac{1}{x}\sum\limits_{k=0}^{\infty}(\frac{a_1^k}{x^k}+\dots+\frac{a_n^k}{x^k})=\frac{1}{x}\sum\limits_{k=0}^{\infty}\frac{tr_{K(a)/K}(a^k)}{x^k}$$ so $f'\equiv 0$ iff $tr(a^k)=0$ for every $k$. (EDIT $tr_{K(a)/K}(a^k)$ is equal to ${a_1^k}+\dots+{a_n^k}$ because the characteristic polnomial of $a^k$ is equal to $\pm(x-a_1^k)\dots(x-a_n^k)$ since it actually annihilates $a^k$ and has degree equal to the degree of the extension)

If $a$ is an inseparable element of $L$ then $tr_{K(a)/K}\equiv 0$ so $tr_{L/K}=tr_{L/K(a)}\circ tr_{K(a)/K}\equiv 0$. Conversely, if $tr_{L/K}\equiv 0$ then obviously there exists an inseparable element.

Finally, the form $Tr$ is degenerate iff the covector $tr$ is identically zero.

  • $\begingroup$ $\operatorname{tr}\left(a^k\right)$ needs not be $a_1^k+\cdots+a_n^k$. We do not know if $a_1, a_2, \ldots, a_n$ span $L$. $\endgroup$ May 28 '15 at 22:49
  • $\begingroup$ @darijgrinberg Yes we should consider the extension generated by $a$, thank you. $\endgroup$
    – SashaP
    May 28 '15 at 22:52
  • $\begingroup$ I'm not sure about that. Do we even know that $a_1, a_2, \ldots, a_n$ belong to $L$ ? It's not like $L/K$ is normal. $\endgroup$ May 28 '15 at 22:55
  • 1
    $\begingroup$ Note that you do not need to invoke the logarithmic derivative. An alternative way of showing $f' / f = \sum_i 1/\left(x-a_i\right)$ is to write $f = c \left(x-a_1\right) \left(x-a_2\right) \cdots \left(x-a_n\right)$ and apply the Leibniz identity. $\endgroup$ May 28 '15 at 23:03
  • 1
    $\begingroup$ Nice answer! Why is it true that if $tr_{L/K} \equiv 0$ then obviously there exists an inseparable element? $\endgroup$
    – Watson
    Dec 4 '16 at 16:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.