I've read this statement in a presentation slide, but it isn't obvious to me on why this is true:


Forgetting about BCH codes, the question is: if an alement $\beta$ has even order ($2k$ is always even for $k \in \mathbb{N}$), why can't such an element exist in a field with characteristic 2 (e.g. in $\mathbb{F}_2$)? I know that the characteristic of a field is defined as

$$char(F) = k \Leftrightarrow \underbrace{1 + 1 + ... + 1}_{k\text{ times}} = 0 $$ and the $ord(\beta)$ is defined as the smallest $a$ for which the statement $\beta^a = 1$ holds true, but I can't seem to wrap my head around why the upper statement must be true. An explanation of this is highly appreciated.

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    $\begingroup$ The group of units in a finite field of characteristic $2$ has order $2^n - 1$, which is odd. $\endgroup$ – Qiaochu Yuan Feb 6 '16 at 18:44
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    $\begingroup$ Maximilian: In a field, @QiaochuYuan's unit is just a fancy way of saying non-zero element. $\endgroup$ – TonyK Feb 6 '16 at 19:55

In a finite group, the order of an element divides the order of the group. A finite field of characteristic $2$ has $2^n$ elements for some positive integer $n$, so its multiplicative group has odd order.

  • $\begingroup$ I see that the multiplicative group of $\mathbb{F}$ is $\mathbb{F}^* = \mathbb{F} - \left\lbrace0\right\rbrace$, and that $ord(\beta) | ord(\mathbb{F})$ which is equivalent to $2k | 2^n$, which should be possible for some value of $k$ for a fixed $n$. I also see that $ord(\mathbb{F}^*)=2^n-1$, and that it's impossible that $2k | 2^n-1$, right? So $\beta$ with order $2k$ cannot be an element of $\mathbb{F}^*$ since the order of $\beta$ doesn't divide the oder of that group, and since $\beta$ can't be $0$, it follows that $\beta \notin \mathbb{F}$, is that reasoning correct? $\endgroup$ – Maximilian Gerhardt Feb 6 '16 at 20:32
  • $\begingroup$ When we ralk about the order of an element of a finite field, it is always the multiplicative order, for the additive order of a non-zero element is always the characteristic $p$. So the relevant group is $F^\ast$, and it has order $2^n-1$, odd. $\endgroup$ – André Nicolas Feb 6 '16 at 20:37

To see this directly: If there is an element of order $2k$, then there is an element of order $2$.

But $x^2-1 = (x-1)^2$ in a field of characteristic $2$, so there are no nontrivial square roots of $1$.

  • $\begingroup$ I think I partly understood that answer, but not enterily. When we have a $\beta$ with order $2k$ then $\beta^{2k} = 1$, then the element $\beta^k$ has order 2, because $\beta^{2k} = 1 \Leftrightarrow \left(\beta^k\right)^2 = 1$. I also see that $(x-1)^2=x^2-2x+1=x^2+1=x^2-2$ over the a field with characteristic $2$, but from that, how does it follow that there are no nontrivial square roots of 1, and why does it follow then that there didn't exist an element with order $2k$ in the first place? $\endgroup$ – Maximilian Gerhardt Feb 6 '16 at 20:32
  • $\begingroup$ Since $x=\beta^k$ satisfies $x^2=1$, it satisfies $(x-1)^2=0$, so $x-1=0$. But $x=1$ does not have order $2$, contradiction. $\endgroup$ – Slade Feb 6 '16 at 20:57

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