How can we consider $\Bbb Z_3[x]/ \langle x^2+x+2\rangle$ as splitting field for $x^2+x+2$ over $\Bbb Z_3$ My book introduce two kind of splitting field. One of them is $\Bbb Z_3(i)=\{a+bi : a, b \in \Bbb Z_3\}$. The other is  $\Bbb Z_3[x]/\langle x^2+x+2\rangle$. But, I am confused how the latter case could be splitting field. As follows my book, since $F=\Bbb Z_3[x]/\langle x^2+x+2\rangle$ is field and by fundamental Theorem of Field, $\beta = x+\langle x^2+x+1\rangle$ is one root of $f(x)=x^2+x+1$ in $F$. So by factor theorem, $f(x)$ splits in $F$. And 
because $F$ is two-dimensional vector space over $\Bbb Z_3$, we know that $F$ is also a splitting field of $f(x)$ over $\Bbb Z_3$
My curious part is here. 
First, how can the fact that $F$ is 2-dim vector space induce consequent conclusion. 
Second, If the statement is true, by definition of splitting field $F=\Bbb Z_3(\alpha, \beta)$ where $\alpha$ is the other root of $f(x)$ in $F$. But rigorously the equality is not true because each side of elements are different. For example, right side contain 2, but 2 is not contained in left side. I guess just identifying $a$ in $\Bbb Z_3$ with $a+\langle x^2+x+1\rangle$ may induce the equality. But even if it is the case, components of two sets are different!  So pleas help me with more additional explanation. Thanks!
 A: Firstly note that $x^2+x+2$ is an irreducible polynomial over $\Bbb{Z}_3$ because it has no linear factors (since it is degree $2$, so enough to check linear factors). So we consider the quotient ring 
$$\Bbb{F}_9=\Bbb{Z}_3[x]/\langle f(x) \rangle=\Bbb{Z}_3[x]/\langle x^2+x+2 \rangle=\{ax+b \, | \, a,b \in \Bbb{Z}_3 \text{ and } x^2+x+2=0\}.$$
 One can use the irreducibility of $f(x)$ (hence the ideal $\langle f(x) \rangle$ is maximal etc.) to claim that $\Bbb{F}_9$ is indeed a field.
By the definition of this quotient ring, the element $\alpha=x+\langle f(x) \rangle$ is a root of the polynomial $f(x)$. Suppose $\beta$ is the other root, then sum of roots=-coefficient of $x$ term gives 
$$\alpha + \beta =-1=2.$$ 
So
$$\beta=2-\alpha=2-[x+\langle f(x) \rangle]=(2+2x)+\langle f(x) \rangle.$$
Observe that this is also an element of $\Bbb{F}_9$ defined as above. Thus the polynomial $f(x)$ can be factored into linear factors in $\Bbb{F}_9$. This means it splits completely in $\Bbb{F}_9$. 
Note: In case you want to view $\Bbb{F}_9$ as a vector space over $\Bbb{Z}_3$ with a basis 
$$\{\mathbf{1}+\langle f(x) \rangle,\mathbf{x}+\langle f(x) \rangle\}.$$ 
Then consider the vector $\beta=2(\mathbf{1}+\langle f(x) \rangle)+2(\mathbf{x}+\langle f(x) \rangle)=\mathbf{2+2x}+\langle f(x) \rangle \in \Bbb{F}_9$ lying in this vector space. Observe that this $\beta$ is the other root of $f(x)$. Hence $f(x)$ splits completely.
