MHB Are the Fields $\mathbb{F}_{2^2}$ and $\mathbb{Z}_2(a)$ Equal?

[mathmari]

Hey!

I have to show that $\mathbb{F}_{2^2}=\mathbb{Z}_2(a)$, where $a \in \mathbb{F}_{2^2}$ is of degree $2$ over $\mathbb{Z}_2$.

Could you give me some hints how I could show that?? (Wondering)

 

[Fallen Angel]

Hi,

Every two finite fields $F,K$ with $|F|=|K|$ are isomorphic.

 

[Fernando Revilla]

Explicitly $\mathbb{F}_{2^2}=\mathbb{Z}_2/\langle x^2+x+1\rangle=\{0,1,\xi,\xi +1\}$ ($\xi^2+\xi+1=0$). That is, $\left|\mathbb{F}_{2^2}\right|=4.$ On the other hand, $\left[\mathbb{Z}_2(a):\mathbb{Z}_2\right]=2$ and $\left|\mathbb{Z}_2\right|=2,$ so $\left|\mathbb{Z}_2(a)\right|=4.$