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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 $Bbb Z_3[x]/langle x^2+x+2rangle$. But, I am confused how the latter case could be splitting field. As follows my book, since $F=Bbb Z_3[x]/langle x^2+x+2rangle$ is field and by fundamental Theorem of Field, $beta = x+langle x^2+x+1rangle$ 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 $a+langle x^2+x+1rangle$ 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!

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.