Where does one Evaluate the Roots of a Polynomial? Apologies in advance about this question; it is kind of hard to explain what I am trying to ask. 
Suppose you have a polynomial in some ring $F[x]$ where $F$ is a field. Now suppose I take some $\alpha \in R$ where $R$ is some other field. If I say $\alpha$ is the root of some polynomial $m(x) \in F[x]$, how does that make sense as a statement. Reading the wikipedia and other definitions seem to imply what I have just said is not a sensible thing to say since a root of a polynomial in $F[x]$ must itself be in $F$. For ex. wikipedia says:

In mathematics, a univariate polynomial is an expression of the form $a_0 + a_1 + \cdots + a_nx^n$ with $a_n \neq 0$ where $a_i$ belong to some field (in this case $\mathbb{C}$) ... A root of this polynomial is some $a$ in $\mathbb{C}$...

Note the part where it says $a$ must be in $\mathbb{C}$.
The reason I ask is because of this example from finite fields. Consider the field of $9$ elements given by $\mathbb{F}_3[x]/(x^2 + 1)$. In this field we have the elements 
$$0,1,2,x,(x+1),(x+2),2x,(2x+1),(2x+2)$$
Now let us take any $\alpha$ in this field, say let $\alpha$ be the element $x+1$. Then since $\mathbb{F}_3[x]/(x^2 + 1)$ is a vector space over $\mathbb{F}_3$ of dimension $2$, we can find a non-trivial solution to $c_0 + c_1\alpha + c_2 \alpha^2 = 0$ and this implies that there is a minimum polynomial in $\mathbb{F}_3[y]$ (different variable name to avoid confision) for which $\alpha$ is a root. You can read about this here for example.
My question is, if a root of a polynomial in $\mathbb{F}_3[y]$ must itself be in $\mathbb{F}_3$ (as per the wikipedia for ex), how can our $\alpha \in \mathbb{F}_3[x]/(x^2 + 1)$, which recall was $x+1$, be a root of a polynomial in $\mathbb{F}_3[y]$. I can tell you by computation that the minimum polynomial for 
our specific $\alpha$ is $m(y) = y^2 + y + 1$. But $m(\alpha) = x^2 + 1$ which is definitely not zero is $\mathbb{F}_3[y]$. It is a root of the polynomial if it is evaluated in the original field it came from (where $x^2 + 1$ is $0$), but not in $\mathbb{F}_3[y]$. Can someone help clear my confusion?
 A: It's certainly not true that roots of polynomials must belong to the field (ring) of their coefficients.  In fact, this is the whole reason field extensions exist.  For example, there is no $\sqrt{2}$ in $\mathbb{Q}$, but there's a very good geometric reason to have one.
That's the reason for (as I said before) field extensions.  If you have a field, you can look at a subset as a smaller field, a subfield.  You can think of polynomials with coefficients in that subfield as standalone polynomials, but it's perfectly possible to have roots outside that subfield but inside the overall field.  $\mathbb{R}$ can be embedded inside $\mathbb{C}$, and $x^2+1$ is a polynomial with real coefficients that has only nonreal complex roots.
The reason $\mathbb{C}$ is different is because there is no field above it that has roots of polynomials in $\mathbb{C}[x]$ that $\mathbb{C}$ doesn't.  That is, $\mathbb{C}$ is algebraically closed.  That's what the Wikipedia article is describing.  But $\mathbb{F}_3$ is not algebraically closed; you may add a root of the polynomial $x^2+1$, and as long as you know how it interacts with everything, it's valid to do.  In this case, you know how it interacts because it's a single element $\alpha$ and you know what $\alpha^2$ is.  And that's basically all that you need to know; it's encompassed in the statement $\mathbb{F}_3[\alpha]=\mathbb{F}_3[x]/(x^2+x+2)$, because you know how to work with the right side.
