In a book on abstract algebra I'm reading it is said that if $f(x)=x^2+x+2$ is a polynomial over $\mathbb{Z}_3$ then $\mathbb{Z}_3[i]$ is a splitting field for $f(x)$ over $\mathbb{Z}_3$, which I understand. It is also said that the element $x+\langle x^2+x+2\rangle$ of $F:=(\mathbb{Z}_3/\langle x^2+x+2\rangle)[x]$ is a zero of $f(x)$. It is in my understanding that $F$ is an extension field of $\mathbb{Z}_3$. Now, the author of the book adds that, since $f(x)$ is of degree 2, then the Factor Theorem implies that the other zero of $f(x)$ is in $F$. Now, the Factor Theorem says that if $F$ is a field, $a\in F$, and $f(x)\in F[x]$, then $a$ is a zero of $f(x)$ if and only if $x-a$ is a factor of $f(x)$ - plain and simple.
My questions are:
(1) $F$ is an extension field of $\mathbb{Z}_3$, but it may not be the splitting field of the polynomial above. Is this correct?
(2) How does the fact that $\deg f(x)=2$ combined with the Factor Theorem imply that if one of the roots of $f(x)$ is in $F$ then the other root is also in $F$?
(3) The author says that $F$ is a splitting field of for $f(x)$ (this part is clear), and that because $F$ is a two-dimensional vector space over $\mathbb{Z}_3$, $F$ is also a splitting field of $f(x)$ over $\mathbb{Z}_3$. Isn't this statement extremely confusing? Because:
$\text{ }$(i) How come $F$ is a two-dimensional vector field if in fact it appears to be three-dimensional? That is, its basis appears to be $\{1, x, x^2\}$ because it contains polynomials of order two.
$\text{ }$(ii) If we know that $F$ is a splitting field for $f(x)$, then, just by looking at how $F$ is defined we can readily tell that $F$ is a splitting field over $\mathbb{Z}_3$! I mean how can $F$ be a splitting field over something else than $\mathbb{Z}_3$?!
Concerning question (2), if we consider the polynomial of degree two $x^2+2$ over $\mathbb{Q}[x]$ with its root $\alpha$ in some extension field $E$, and the other root $\beta$ in some extension field, then it can be factored as $(x-\alpha)(x-\beta)$. We know that $(x-\alpha)\in E[x]$ and that $(x-\alpha)(x-\beta)\in \mathbb{Q}[x]$. But how does this imply that $\beta\in E$? Why does its degree being two matter here? Consider a polynomial $a+bx+cx^2+cx^3=(x-\alpha)(x-\beta)(x-\gamma)$, with roots $\alpha, \beta, \gamma$. If we know that $\alpha\in E_1$, then, by the same token described in the book, wouldn't this imply that $\beta$ and $\gamma$ are in $E_1$?
I'm quite puzzled, would appreciate your explanation. Sorry for my question being to long.
To answer both (1) and (2): No, $F$ must be a splitting field for the polynomial. You've already explained why - if $a$ is a root of $x^2 + x + 2$ in $F$, then $x - a$ is a factor of $x^2 + x + 2$ over $F$. So $x^2 + x + 2 = (x - a)p(x)$ for some polynomial $p$. $p(x)$ has degree exactly $1$, otherwise $x^2 + x + 2$ wouldn't be able to have degree $2$. So $p(x)$ is linear - but linear polynomials have solutions over any field, so $F$ has the solution to $p$ as well. This idea is also the answer to your question in the last paragraph.
To address (3)(i): $F$ does not properly include $x^2$. In $F$, $x^2 + x + 2 = 0$ (because that's what the quotient does). So $x^2 = -x - 2$. Which means that $x^2$ is not independent of $x$ and $1$.
As for (3)(ii): I'm not actually sure - but I suspect you're misreading the author, and that the author may be drawing a distinction between a splitting field and the splitting field. For it to be a splitting field, it just has to have the roots of the polynomial; for it to be the splitting field, it must also be minimal. Because $F$ has dimension $2$ - and the splitting field of a quadratic has dimension $2$, so can't fit properly inside - $F$ must be the minimal splitting field.
To re-address your last paragraph: No, in the cubic case the additional roots need not be present. Suppose, for example, that in $\mathbb{Q}(\alpha)$ the cubic $ax^3 + bx^2 + cx + d$ factors as $(x - \alpha)(a'x^2 + b'x + c')$. The latter quadratic might be irreducible in $\mathbb{Q}(\alpha)$, so $\mathbb{Q}(\alpha)$ doesn't have to include the other two roots. The issue is that with a quadratic, the piece that's leftover must be linear, and a linear polynomial over a field $K$ always has a root in $K$.