Cyclic extension of degree $2$ of $\mathbb{F_2}$

Question

I want to find a cyclic extension of degree $2$ of $\mathbb{F_2}$

The degree of its minimal polynomial is $2$, i.e. a quadratic polynomial.

The only cyclic field of degree $2$ which I can think of is $\mathbb{Z_2}$. Is this ok? I am not sure if I misunderstand the concept of "cyclic field extension"

Thank you for your help

Answer 1

A cyclic extension is a finite extension for which the Galois group of the extension is cyclic. It has nothing to do with "cyclic fields." Let the original field be $\Bbb{F}_p$ and the field extension be $K$. If K is a cyclic extension, then there is an automorphism of $K$ that fixes $\Bbb{F}_p$ such that if you apply that automorphism multiple times, you will eventually loop through all of the automorphisms in the Galois group until you get to the identity. However, any finite extension of a finite field is cyclic, so as long as we can find a finite extension of $\Bbb{F}_2$ of degree 2, we are guaranteed that is is cyclic.

Now that we have that out of the way, we can completely ignore the cyclic part of the question. Now, we want to find a quadratic polynomial in $\Bbb{F}_2$ such that it is irreducible, meaning it has no roots. Let's do some guessing and checking now:

• Does $x^2$ have a root? Yes, $0^2=0$.
• Does $x^2+1$ have a root? Yes, $1^1+1=0$.
• Does $x^2+x$ have a root? Yes, $1^1+1=0$.
• Does $x^2+x+1$ have a root? No, $0^2+0+1=1$ and $1^1+1+1=1$.

Thus, the polynomial we want to use is $x^2+x+1$. Because this is a field extension of degree $2$ over $\Bbb{F}_2$, the field we get is of order $2*2=4$, meaning we now have $\Bbb{F}_4$. Finally, we want to generate the multiplicative group of this field using our polynomial. We can do this by starting with $\alpha$, which represents our new found root to the polynomial $x^2+x+1$, and then multiplying it by itself multiple times.

• $\alpha^1=\alpha$
• $\alpha^2=(\alpha^2+\alpha+1)-\alpha-1=\alpha+1$ (Here, I use the fact that $\alpha^2+\alpha+1=0$ and the fact that $-1=1$ in fields of characteristic $2$.)
• $\alpha^3=\alpha*\alpha^2=\alpha(\alpha+1)=\alpha^2+\alpha=(\alpha^2+\alpha+1)-1=1$

Once we have found the identity, we stop. Now we know that $\alpha^2=\alpha+1$ and that $\alpha^3=1$ which shows the structure of the multiplicative group in this representation of $\Bbb{F}_4$.