# Minimal extension field of $\mathbb{F}_2$ such that

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?

• 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. – Jyrki Lahtonen May 28 at 20:48
• Put another way, an element of order 21, should have order precisely 21, not just have 21st power 1. – Alex J Best May 28 at 20:48
• @JyrkiLahtonen Sorry but I can't really understand why I want $21 | 2^s-1$. I just studied this theorem about primitive elements – lukk May 28 at 20:54
• @AlexJBest yes, 21 should be the minimum power such that $a^{21}=1$... but how can I use this? – lukk May 28 at 20:55
• 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. – 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 $$0.
Since the multiplicative group of a finite field is cyclic, you need to find the least exponent $$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$$.
• 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? – lukk May 28 at 21:20
• @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. – egreg May 28 at 21:27