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If $\beta$ is a primitive element in extension field GF$(2^m)$ does $\beta$ and $\beta^3$ are always non-conjugate elements?

In my textbook, I saw they assume they are, but I couldn't understand it for all $m$. It seems reasonable, but as I thought it seems reasonable that $\beta^3$ and $\beta^5$ are non-conjugate elements, I found that for $m=3$, they are conjugate to each other.

Mr.O
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1 Answers1

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I assume that to you a primitive element is a generator of the multiplicative group. This is standard meaning of primitive in the context of finite fields.

The conjugates of $\beta$ in $GF(2^m)$ are $$ \beta,\beta^2,\beta^4,\ldots,\beta^{2^{m-1}}.\qquad(*) $$ Because $\beta$ is primitive the exponent $\ell$ identifies the power $\beta^\ell$ uniquely while it stays in the range $0\le \ell<2^m-1$.

The exponents in $(*)$ are all in this range, so $\beta^3$ is not a conjugate simply because $3$ is not in that list of exponents.

More generally, we need to look at the exponents modulo $2^m-1$. So $\beta^3$ and $\beta^5$ are conjugates if and only if $$ 3\cdot2^i\equiv 5\pmod{2^m-1} $$ for some $i$. Indeed, $3\cdot2^2\equiv 5\pmod7$, $7=2^3-1$. Thus $\beta^3$ and $\beta^5$ are conjugates when $\beta$ is a primitive element of $GF(8)$.

With $i$ ranging over the interval $0\le i \le m-1$ we see that $3\cdot2^i$ "overflows" $2^m-1$ only when $i=m-1$ exactly. We have $$ 3\cdot2^{m-1}=2^m+2^{m-1}\equiv2^{m-1}+1\pmod{2^m-1}. $$ So $\beta^3$ and $\beta^5$ can be conjugates only when $2^{m-1}+1=5$, a case we already handled.

Jyrki Lahtonen
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  • Right. I didn't think about this simple way. Writing all the conjugates till $\beta^{2^{m-1}}$ (knowing they are distinct) , and use the fact that $(\beta^{2^{m-1}})^2=\beta$ since it is a primitive element. And thank you for the extended explanation for $\beta^5$ and $\beta^3$ – Mr.O Jan 07 '21 at 12:42