Find an irreducible polynomial $f$ with degree $2$ over $\mathbb{Z}_2$. Assign a root $u$ of $f$ to $\mathbb{Z}_2$ to get the field $\mathbb{Z}_2 (u)$ of order 4. Use the same method to construct a field of order 8.
My attempt:
$f$ is reducible if and only if $f=gh$ with $g,h$ non constant polynomial. Hence $2=deg(f)=deg(h)+deg(g)$. From here we have that $1=deg(g)=deg(h)$.
We know that $\mathbb{Z}_2=\lbrace [0], [1]\rbrace$, the polynomials of degree 1 are $x,x+1$ and the polynomials of degree 2 are $x^2,x^2+1,x^2+x,x^2+x+1$. Hence, the only irreducible polynomial is $f(x)=x^2+x+1$ with roots $-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $-\frac{1}{2}-\frac{\sqrt{3}}{2}i$.
The difficulty I have is that I don't know which root to attach to $ \mathbb {Z} _2 $, how to check that the resulting field has order 4 and how to do the same to obtain a field of order 8.
Thanks in advance.
Best Answer
We have that $f(x)=x^2+x+1$ is an irreducible polynomial. If $u$ is a root of $f$ we have that $\mathbb{Z}_2(u)=\lbrace a+bu : a,b \in \mathbb{Z}_2 \rbrace$.
In a similar way, we can find that the irreducible polynomials of degree 3 are $x^3+x+1$ and $x^3+x^2+1$. Then if we take $u$ as a root of any of these polynomials, we have that $$\mathbb{Z}_2(u)= \lbrace a+bu+cu^2 : a,b,c \in \mathbb{Z}_2 \rbrace $$
And clearly these fields have 4 and 8 elements, resp.