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Let F be a field extension of K, then F over K is a vector space, and for each a in F define f:F-->F as f(x)=ax, this is a linear transformation, define trace of a as trace of this linear transformation. If K has characteristic zero then clear 1 has trace n, which is non zero. Are there fields F and K as above, such that every element of F has zero trace.

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    $\begingroup$ As you've noted the only place to look is degree p (or a multiple) extensions in characteristic p. Looking up some results on purely inseparable extensions might help. $\endgroup$ – Alex J Best Feb 24 at 20:15
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Let $F=\Bbb{F}_p(x)$ and $K=\Bbb{F}_p(x^p)$. A basis of the extension $F/K$ consists of $1,x,\ldots,x^{p-1}$. The element $1$ has $tr^F_K(1)=0$ because the $p\times p$ identity matrix has trace zero. The element $x^i, i=1,2,\ldots,p-1$, has trace zero because its minimal polynomial $m_i(T)$ over $K$ is $$(T-x^i)^p=T^p-(x^p)^i\in K[T],$$ and the coefficient of the degree $p-1$ term is manifestly zero.

The trace $tr^F_K$ is $K$-linear, so if it vanishes on a basis it vanishes everywhere.

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  • $\begingroup$ It seems the converse - finding all finite extensions whose trace vanishes identically - is not obvious as the problem of relating the trace polynomials with the elementary symmetric polynomials $\endgroup$ – reuns Feb 25 at 6:20

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