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Find the minimal extension field of $\mathbb{F}_2$ such that this extension contains an element of order $21$?


Attempt: I know that such an extension of $\mathbb{F}_2$ is like $\mathbb{F}_{2^s}$ and $2|s$. Such a field has a primitive element, say $\alpha$ that generated the whole field. We know by theory that such a primitive element is such that $\alpha^i =1 <=> 2^s-1|i$

So, $\alpha^{21}=1 <=> 2^s- 1 |21$

So I need to find the minimum $s$ such that $2^s - 1$ divides $21$. $s=3$ is the good candidate ($s=1$ corresponds to $\mathbb{F}_2$ which is the base field).

Therefore, such an extension is $\mathbb{F}_{2^2}=\mathbb{F}_4$

Is it correct?

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  • $\begingroup$ The multiplicative group of $\Bbb{F}_{2^s}$ is cyclic of order $2^s-1$, so you want $21\mid 2^s-1$. Divisibility the other way is silly, because $2^1-1=1$ divides anything. $\endgroup$ – Jyrki Lahtonen May 28 at 20:48
  • $\begingroup$ Put another way, an element of order 21, should have order precisely 21, not just have 21st power 1. $\endgroup$ – Alex J Best May 28 at 20:48
  • $\begingroup$ @JyrkiLahtonen Sorry but I can't really understand why I want $21 | 2^s-1$. I just studied this theorem about primitive elements $\endgroup$ – lukk May 28 at 20:54
  • $\begingroup$ @AlexJBest yes, 21 should be the minimum power such that $a^{21}=1$... but how can I use this? $\endgroup$ – lukk May 28 at 20:55
  • $\begingroup$ Lagrange's theorem from elementary group theory: the order of an element divides the order of the group. Here the group has order $2^s-1$ so if you have an element of order $21$ then you must have $21\mid 2^s-1$. This is a necessary condition. The primitive element theorem implies that it is also sufficient. $\endgroup$ – Jyrki Lahtonen May 28 at 20:56
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No, it's incorrect. You need an element $a$ such that $a^{21}=1$, but $a^k\ne1$ when $0<k<21$.

Since the multiplicative group of a finite field is cyclic, you need to find the least exponent $m$ such that $21\mid(2^m-1)$, which is the reverse of what you're doing.

The group must have order divisible by $21$, and this suffices because the group is cyclic (actually abelian would suffice).

You therefore need $3\mid(2^m-1)$ and $7\mid(2^m-1)$. The former condition yields $m$ even, the latter that $3\mid m$.

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  • $\begingroup$ It's almost everything clear, but I don't understand how you derived $3| (2^m -1)$ and $7 | (2^m-1)$. DId you just factorize it in order to simplify computations, right? $\endgroup$ – lukk May 28 at 21:20
  • $\begingroup$ @lukk That's standard: if $a,b$ are coprime, then $ab$ divides $c$ if and only if both $a$ and $b$ divide $c$. And yes, this simplifies the computation. $\endgroup$ – egreg May 28 at 21:27

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