# Showing that Trace($ax$) = $0$ for all $x$ in field $F$ of order $2^N$ implies $a = 0$.

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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Is $a$ a scalar? Is $x$ an endomorphism? Is the dimension of the space $x$ is an endomorphism on finite? –  Sh4pe Oct 30 '12 at 14:06
It is not a good idea to rely on the title alone to express the assumption that $Tr(ax)=0$ is supposed to hold for all the elements $x$ of the finite field. That is the only interpretation that makes sense, so I answered accordingly. You should list all the assumptions in the question body - even when they were mentioned in the title. Not doing that confuses many people, and you may miss an answer because of that (or the answer may come only later). –  Jyrki Lahtonen Oct 31 '12 at 9:08
Oops. I just noticed (after Martin's edit) that you had written out trace incorrectly. The exponents are supposed to be consecutive powers of two, starting from $1=2^0$ up to $2^{N-1}$. –  Jyrki Lahtonen Nov 2 '12 at 10:09

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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