Distribution of Trace values

Question

I try to prove that ${2^{n-1}}$ elements of the field $\mathbf{F}_{2^{n}}$ have a Trace with value 1, while the other ${2^{n-1}}$ elements have a Trace with value 0.

I started to show that Trace(1) = 1, and I tried to use the additivity of the Trace but I wasn't successful. Any advice ?

"elements based..." ? Are you talking of the elements in that field? And thus their trace is just the sum of their values over that extensions Galois group? — Timbuc, Feb 14 '15 at 15:12

score 1 · Answer 1 · answered Feb 14 '15 at 18:07

1

Because $tr(x)$ is a polynomial of degree $2^{n-1}$ it can take the value zero at most $2^{n-1}$ times. Because it is linear it takes the value zero at least $2^{n-1}$ times (rank-nullity tells that the kernel has dimension $\ge n-1$). Because $tr(x)$ is either zero or $1$, it has to take both values equally often.

answered Feb 14 '15 at 18:07

Jyrki Lahtonen

133,153

Thanks for the explanation. Is this explanation independent if n is even or odd ? – MathPowerUser Feb 25 '15 at 21:36
In the script here is a hint to prove it by additivity of the trace function. Do you have any advice how this can be used ? Thank you. – MathPowerUser Feb 25 '15 at 21:37

score 0 · Answer 2 · answered Feb 14 '15 at 16:11

This is just linear algebra: the trace map is a linear functional $\;\Bbb F_{2^n}\to\Bbb F_2\;$, and since the extension $\;\Bbb F_{2^n}/\Bbb F_2\;$ is separable it is not the zero functional (or just show there's some element with trace different from zero), from where it follows that it is onto (this much is true for any linear functional over any vector space) and thus $\;\dim\ker\,Tr.=n-1\;$, and this means that

$$|\ker Tr.|=|\Bbb F_{2^{n-1}}|=2^{n-1}$$

and we're done.

Distribution of Trace values

2 Answers2