Recall that $\mathbb{C}$ was constructed from $\mathbb{R}$ by `formally' introducing a root $i$ of $p(x)=x^2+1$. In the field $F=\mathbb{Z}_2$ with 2 elements, the polynomial $x^2+1$ has a root since $p(1)=1^2+1=1+1=0$. On the other hand the polynomial $$q(x)=x^2+x+1$$ has no root: Indeed, $q(0)=q(1)=1$. So, following the strategy for complex numbers let us `formally' introduce an element $j$ with the property $$(\ast)\ \ \ j^2+j+1=0.$$ More precisely, let $F=\mathbb{Z}_2\times \mathbb{Z}_2$, with elements written in the form $$(a,b)=a+ j b $$ where $j$ is some "formal symbol". Define addition component-wise, and multiplication by the rule $(\ast)$. For example, $$(1+j)+j=1+2j=1$$ since $2=0$ in $ \mathbb{Z}_2$, and $$ (1+j)j=j+j^2=-1=1$$ since $-1=1$ in $\mathbb{Z}_2$.
Prove: $F$ is a field.
This is the field with 4 elements!