confused about $\Bbb F_3(\alpha) \cong \Bbb F_3/\langle x^3+2x+2\rangle$

Question

What is the form of, and what method can be used to find $\Bbb F_3(\alpha)$

if $\Bbb F_3(\alpha) \cong \Bbb F_3[x]/\langle x^3+2x+2\rangle$ ?

I know how to do this for more simple polynomials, namely using this method trying to understand the relation between quotient rings and Field extensions. However I'm confused as to what approach to take for higher degrees.

Answer 1

What you can prove is that any congruence class in $\mathbf F_3[x]/(x^3+2x+2)$ has a unique representative as the congruence class of a polynomial of degree $\le 2$, since by Euclidean division, for any polynomial $p(x)$, there exists a unique pair of polynomials $q(x)$, $r(x)$, such that $$p(x)=(x)(x^3+2x+2)+r(x),\qquad \deg r(x)\le 2.$$

Thus if we denote $\alpha$ the congruence class of $x$ mod. $x^3+2x+2$, we have $$p(\alpha)=r(\alpha)= a+b\alpha+c\alpha^2\quad\text{for some }\enspace a, b,c\in \mathbf F_3.$$

Answer 2

Hint: $R = \Bbb{F}_3[x]/\langle x^3 + 2x + 2\rangle$ is a vector space over $\Bbb{F}_3$ of dimension $3$, generated by (the images in $R$ of) $1$, $x$, $x^2$. To see that this is closed under multiplication, note that the relation $x^3 = -2x - 2$ holds in $R$ and use this relation to simplify any polynomial in $x$ to a polynomial of degree at most $2$. To see that $1$, $x$ and $x^2$ are linearly independent, note that a polynomial $ax^2 + bx + c$ cannot have the form $p(x)(x^ 3 + 2x + 2)$ unless $a = b = c = 0$ (and $p(x) = 0$). If we write $\alpha$ for (the image in $R$ of) $x$, then as $x^3 + 2x + 2$ happens to be irreducible over $\Bbb{F}_3$, we use the traditional notation $\Bbb{F}(\alpha)$ for $R$ viewed as a field generated by the element $\alpha$.