Simple extension of $\mathbb{Z}_3$ by a complex number

Question

So far I've been working with (simple) extensions of the rationals, so this is the first time I've come across a problem of trying to extend another field. I've come across the notion that $\mathbb{Q}(k) = \{a + bk \mid a, b \in \mathbb{Q} \}$ where $k \not \in \mathbb{Q}$ (otherwise you're not extending anything), and think it'd be the same for any other type of simple extension.

So, specifically, if I want to find $[\mathbb{Z}_3(\frac{1}{2}(1 + i\sqrt{7})): \mathbb{Z}_3]$ (a simple extension), i'd want to look at the basis for the set $\{ a + b\frac{1}{2}(1 + i\sqrt{7}) \mid a,b \in \mathbb{Z}_3 \} = \{a + b + i\frac{b}{2}\sqrt{7} \mid a,b \in \mathbb{Z}_3 \} = \{c + i\frac{b}{2}\sqrt{7} \mid c,b \in \mathbb{Z}_3\}$? Which would be $\{1, \sqrt{7}\}$? Is this a way to characterize all simple extensions?

Edit: This is the full problem that I'm working with:

Calculate the degree of $[\mathbb{Z}_3(\alpha) : \mathbb{Z}_3]$ where $\alpha$ is the root of the polynomial $x^3 + x + 2 \in \mathbb{Z}_3[x]$. (I've determined that the 3 roots are $-1$ and $\frac{1}{2}(1 \pm i\sqrt{7})$.

Thanks for reading!

Answer 1

over $\Bbb Z_3$ one root lies in the field, and we have the factorization. $$ x^3 + x + 2 = (x+1)(x^2 + 2x + 2) $$

the quadratic factor is $(x+1)^2 +1^2$ which is irreducible because $3$ is a prime of the form $4n+3$ (Fermat).

if $\alpha \ne 2$ is one root of the quadratic then the other root is $1-\alpha$ since the sum of roots is $1 $ .

hence the extension is $\Bbb Z_3[\alpha]$ of degree 2.

Answer 2

in $\Bbb Z_3$ we have $-1 \equiv 2$ so apparently different forms of an equation can be equivalent over any finite field. note also that $-7 \equiv -4 \equiv -1$ so if you wish to solve using the quadratic formula, then the roots are:

$$ \frac{1 \pm "\sqrt{2}"}2 $$

the quotes indicate that we are not dealing with the $\sqrt{2}$ of real arithmetic, but a root over $\Bbb Z_3$ of the equation $$ x^2 - 2 = 0 $$