If $G(x)=x^2+x+1$ and $H(x)=x^3-x+1$ where $r_G$ and $r_H$ are roots of $G(x)$ and $H(x)$ respectively, what do the elements of (I). $\mathbb{F}_2(r_G)$, (II). $\mathbb{F}_2(r_H)$, (III). $\mathbb{F}_3(r_G)$, and (IV). $\mathbb{F}_3(r_H)$ look like? How many elements does each have?
I guess I'm confused, because for (I), I feel like the elements should look like $a+br_G$ where $a,b\in\{0,1\}$ which would be 4 elements. Similarly, for (II), I thought the elements should be of the form $a+br_H$ where $a,b\in\{0,1\}$ which is also four elements. Continuing, (III)'s elements would look like $a+br_G$ where $a,b\in\{0,1,2\}$ which is 6 elements and (IV)'s would look like $a+br_H$ where $a,b\in\{0,1,2\}$, also with 6 elements.
I've instead tried saying that since G has 2 roots, that (I) $\mathbb{F_2}$ with the roots of $H(x)$ adjoined should look something like $\mathbb{F_2}(r_1,r_2)$ which would be elements of the form $a+b{r_1}+c{r_2}+d{r_1}{r_1}$ where $a,b,c,d \in \{0,1\}$, but that can make 8 elements (I suppose some of these would be equivalent). Similarly for (IV), that since $H(x)$ has 3 roots that when we adjoin $\mathbb{F_3}$ with $H(x)$ that we would get elements that look like $a+br_1+cr_2+dr_3+er_1r_2+fr_2r_3+gr_1r_3+hr_1r_2r_3$ where $a,b,c,d,e,f,g,h \in \{0,1,2\}$ which could yield up to 24 elements. (Again, I suppose these could contain copies). Continuing with this trend (II) and (III) would have 16 and 12 elements respectively.
Obviously, I'm very confused, and am missing something critical about what adjoin means. As it turns out, $\mathbb{F}_2(r_G)$ has 4 elements, $\mathbb{F}_3(r_G)$ has 9, $\mathbb{F}_2(r_H)$ has 8, and $\mathbb{F}_3(r_H)$ has 27. This makes a lot more sense (as fields should have orders of prime powers), but yeah. halp.
For a field $K$, an irreducible polynomial $f$ of degree $n$ and a root $\alpha$ of $f$, we can adjoin $\alpha$ to $K$. Then $K(\alpha)$ is the smallest field containing $K$ and $\alpha$. More concretely, $K(\alpha)$ consists precisely of elements of the form $k_0 + k_1 \alpha + \ldots + k_{n-1} \alpha^{n-1}$ with $k_i \in K$ (these elements form a field isomorphic to $K[X]/(f)$)
So, for example the elements of $\mathbb{F}_2(r_H)$ are of the form $a+br_H +cr_H^2$ with $a,b,c \in \mathbb{F}_2$ since $H$ is a polynomial of degree $3$. The elements $a+br_H$ are not enough: $\mathbb{F}_2(r_H)$ is a field and therefore must also contain $r_H^2$ which is not of the form $a+br_H$.
The elements of $\mathbb{F}_3(r_G)$ are indeed of the form $a+br_G$ with $a,b \in \mathbb{F}_3$. There are $9$ of them, not $6$.
Also, adjoining one root is not the same as adjoining all roots: take for example $\mathbb{Q}(\sqrt[3]{2})$. The elements are of the form $a +b \sqrt[3]{2} + c (\sqrt[3]{2})^2$ with $a,b,c \in \mathbb{Q}$ and those are all real. But the polynomial $X^3-2$ has two complex roots which can't be in $\mathbb{Q}(\sqrt[3]{2})$.