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How can I show that Trace $(ax) = 0$ implies that $a = 0$ in a field $F$ (over F2) of order $2^N$? I get that I can something like the following:

Trace($ax_1)=$ Trace($ax_2$) $\implies$ $a(x_1-x_2)+a^2(x_1-x_2)^2+...+a^{2^{N-1}}(x_1-x_2)^{2^{N-1}} = 0,$

but I don't see where to go from here or why this is useful.

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The claim results from the following statement:

(*) If $E/F$ is a degree $n$ separable extension, then the trace $$ t(E/F):E\to F $$ of $E/F$ is nonzero.

To prove (*), denote by $$ t(A/K):A\to K $$ the trace of $A$ over $K$, whenever $K$ is a field and $A$ a finite dimensional $K$-algebra.

As $$ t((K\otimes E)/K)=K\otimes t(E/K): K\otimes E\to K $$ for any extension $K/F$, the tensor products being taken over $F$, it suffices to check that $K/F$ can be chosen so that $t((K\otimes E)/K)\ne0$.

Thus, it only remains to observe that if $K/F$ splits the minimal polynomial of a generator of $E/F$, then the $K$-algebra $K\otimes E$ is isomorphic to $K^n$ by the Chinese Remainder Theorem.

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Let $F$ be the field of $2^N$ elements. If $Tr(ax)=0$ for all $x\in F$, then the polynomial $$ p(x)=Tr(ax)=ax+a^2x^2+a^4x^4+\cdots+a^{2^{N-1}}x^{2^{N-1}} $$ has (at least) $2^N$ distinct zeros in the field $F$, namely all its elements. If $a\neq0$, then this polynomial has degree $2^{N-1}$. Therefore $\ldots$

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