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?

  • $\begingroup$ 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

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 $3\mid m$.

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.