Show that $\mathbb{F}_{2^2}=\mathbb{Z}_2(a)$

Question

How could we 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??

Answer 1

This might help, depending on the definition you're using: $\mathbb Z_2$ is a field and $\mathbb F_{2^2}$ is a vector space over $\mathbb Z_2$. What is its dimension?

Answer 2

Let $K$ be a field with four elements (call it $\mathbb{F}_{2^2}$ if you want). Observe that $K$ contains $0$, $1$, and two other elements that we will call $a$ and $b$.

Observe that $1+1=0$. Indeed, the subgroup of $(K,+)$ generated by $1$ has to have order that divides $4$, so either $1$, $2$ or $4$. And $1$ is not possible since $1\not=0$, and $4$ is not possible otherwise $2\cdot 2=0$.

Hence, $K$ contains $\mathbb{F}_2=\{0,1\}$ as a subfield.

You can then check that $1+0=1$, $1+1=0$, so $1+a=b$ and $1+b=a$.

Observe then that $a^2\not=1$, because otherwise $(a+1)^2=0$. We also have $a^2\not=a$, as $a\notin\{0,1\}$. This implies that $a^2=b$.

Hence, $a^2=a+1$, so $a^2+a+1=0$. This yields $K=\mathbb{F}_2[a]/(a^2+a+1)$.

Answer 3

There's only one irreducible polynomial of degree $2$ over $\mathbf F_2$, namely $x^2+x+1$, hence only one quadratic extension of $\mathbf F_2$. Moreover, over any field, an extension of prime degree is necessarily simple.