One can see by brute force that $x^3+2x+2$ has no roots in GF(3). So it is irreducible and hence the minimal polynomial of $\alpha$.
My question is what is $\alpha$, how can I think about it? I determined that $GF(3)(\alpha) \cong GF(3^3)=GF(27)$. I then wrote a script to check $1,2,3,\dots,26$ to see if they are a root $\mod 3$ and $\mod 27$, and none of those are... Do I have a misconception about $\alpha$?
On
Careful, $\operatorname{GF}(27) \neq \mathbb Z/27\mathbb Z$ !
The latter is a ring of order $27$ which has zero-divisors ($3 \cdot 9 = 0$) and so not a field, and whose elements are $0$ up to $26$.
The first is a field of order $27$ of characteristic $3$, in which $\{0,1,\ldots, 26\} = \{0, 1, 2\}$ because $4=1$ etc.
On
$GF(27)$ is the set of all elements of the form $pa^2+qa +r$, where $p,q,r \in \mathbb{Z_3}$. Thus accounting for 27 elements. With the additional constraint/relation that $a^3+2a^2+2=0$.
You may want to think of it as working with the expressions of the form $pa^2+qa +r$ modulo $3$ as well as modulo $a^3+2a^2+2$.
On
You “add” an element $\alpha$ subject to $\alpha^3+2\alpha+2=0$, as explained in other answers.
But wait! How do I know that the system we find, that is, the set of linear combinations of the form $p+q\alpha+r\alpha^2$ is consistent?
There are two ways. The first one is to consider the polynomial ring $\mathbb{F}_3[x]$ and quotient it by the ideal $I$ generated by $x^3+2x+2$. Since this polynomial is irreducible, the quotient ring $\mathbb{F}_3[x]/I$ is a field.
It does not contain $\mathbb{F}_3$, but it has an isomorphic copy into it, namely the elements of the form $\bar{p}=p+I$, for $p\in\mathbb{F}_3$. As it's usually done in algebra, we can identify $\mathbb{F}_3$ with its isomorphic copy. Here $\overline{f(x)}=f(x)+I$ denotes the image of $f(x)$ in the quotient ring.
Now consider $\alpha=\bar{x}=x+I$. Then $$ \alpha^3+2\alpha+2=\bar{x}^2+2\bar{x}+2=\overline{x^3+2x+2}=\overline{0} $$ so we do have found a suitable $\alpha$.
The second approach is to consider that the extension field has to be a $3$-dimensional vector space. Call $\{1,\alpha,\alpha^2\}$ a basis, where $\alpha$ and $\alpha^2$ are just formal names.
The action of multiplying by $\alpha$ should be a linear transformation and we can write down the matrix relative to the basis: $1\mapsto\alpha$, $\alpha\mapsto\alpha^2$ and $\alpha^2\mapsto \alpha^3=-2\alpha-2=\alpha+1$ (the formal names are useful to guess what the linear transformation should do); the matrix is $$ A=\begin{bmatrix} 0 & 0 & 1 \\ 1 & 0 & 1 \\ 0 & 1 & 0 \end{bmatrix} $$ Guess what? The characteristic polynomial of $A$ is $x^3+2x+2$ (up to the sign) and so $A^3+2A+2I=0$. The subring of the ring of $3\times 3$ matrices generated by $A$ is the extension field we are looking for (the base field $\mathbb{F}_3$ is identified with the scalar multiples of the identity matrix).
We so have found a field $K$ that (up to an identification of the base field with a subfield of $K$) satisfies the request. Hence, we can also do the “formal enlargement” we started with, because it can be proved that
given an irreducible polynomial $g(x)$ there is an extension field in which $g$ has a root $\alpha$ and the minimal subfield containing $\alpha$ is uniquely determined up to isomorphism.
The existence part has been outlined above (and works for arbitrary irreducible polynomials over arbitrary fields). The uniqueness up to isomorphism is not difficult to show. Hence, no matter how we add the root, we end up with “the same thing”, as far as algebra is concerned.
You can think of this field by analogy with how you first thought of the complex numbers: as the real numbers with an extra element $i$ such that $i^2+1 = 0$.
In this case you start with $GF(3)$ and "add" an extra element $\alpha$ that satisfies the given polynomial. Then every element is a combination $$ x + y\alpha + z\alpha^2 . $$
The defining equation tells you how to express $\alpha^3$ this way, by analogy with $i^2 = -1$.