Let $K$ be a finite field, and $F$ be the finite dimensional extension field over $K$. Prove that the trace map $\operatorname{tr}_K^F: F\to K$ is surjective.

I consider the problem as follows.

Since $F$ is finite dimensional over finite field $K$, $\operatorname{Aut}_KF$ is finite and cyclic. Suppose that Aut$_KF=\langle \sigma \rangle$ of order $n$. Then $\operatorname{tr}K^F(u)=u+\sigma(u)+\cdots+\sigma^{n-1}(u)$ and $\operatorname{tr}_K^F$ is $K$-linear. We know that $\operatorname{tr}_K^F(u)\in K$, $\forall u\in F$, $\operatorname{tr}_K^F$ is not trivial, and $K$ is a 1-dimensional vector space over $K$, $\operatorname{tr}_K^F$ is surjective.

I don't know the last step is right. If it is right, is the proposition still true for any field of characteristic $p\ne 0$? Can anyone give me help? Thank you.

  • 2
    $\begingroup$ As long as the extension is separable $\;\iff\;$ the trace bilinear form isn't degenerate, it is. $\endgroup$ – DonAntonio Nov 28 '16 at 8:49
  • $\begingroup$ I understand, thank you very much. $\endgroup$ – Edelweiss Ntu Nov 28 '16 at 9:21

It is not true for any field of characteristic $p \ne 0$. As noted in the comments, if $K/F$ is not separable, the result can fail.

For your particular case of finite fields, note that we can take $\sigma(x)=x^q$ for all $x \in F$, where $q=|K|$. Then trace($u$) is a polynomial in $u$ of degree $q^{n-1}$ and therefore can't have more than $q^{n-1}$ roots in $F$. But $|F|=q^n$. Thus the trace map is nonzero, hence surjective, and your argument works.

  • $\begingroup$ Here $u$ is any non-zero element in $F$ or the $u$ as in the primitive element theorem $F=K(u)$ ? $\endgroup$ – reflexive Apr 23 '19 at 6:52
  • $\begingroup$ $u$ is an arbitrary element of $F$. $\endgroup$ – Ted Apr 24 '19 at 3:51

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