Let $F$ be a field and $f(x) \in F[x]$ s.t. $f(x)$ is irreducible and of degree $n \ge 1$. I've seen it proved that there exists an extension field $E$ of $F$ s.t. that there is a root $\alpha \in E$ of $f(x)$ and the degree of $E$ over $F$ is $n$. In particular, the construction works like this.

1. Consider $(f(x)) \subset F[x]$ and let $V = (f(x))$.
2. Then $F[x]/V$ is a field with basis $\{1 + V, x + V, \ldots , x^{n-1} + V\}$.
3. The polynomial $f(x) + V$ has root $x + V$.

Now this seems to be skirting the issue somewhat to me. We originally wanted to extend $F$, but instead we have moved from the setting of $F$ and $F[x]$ to some other setting.

It's been pointed out that if $\overline{F} = \{f\_i + V : f\_i \in F\}$, then $F \cong \overline{F}$. In particular, this is done via the isomorphic mapping $\psi(f\_i) = f\_i + V$. But I'm still not clear on how this guarantees us that there exists some $E$ over $F$ of degree $n$ with a root of $f(x)$. Presumably that $E$ is isomorphic to $F[x]/V$. But what is that $E$?

So to sum up my question: we have moved from the setting of $F$ and $f(x) \in F[x]$ to the new setting of $\overline{F}$ and $f(x) + V \in F[x] / V = \overline{F}[x]$. We've shown some analogous result in this **new setting**, but how exactly does this net us the desired result in our **original setting**?