Question About a Classical Root Appending Extension Field Proof 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.


*

*Consider $(f(x)) \subset F[x]$ and let $V = (f(x))$.

*Then $F[x]/V$ is a field with basis $\{1 + V, x + V, \ldots , x^{n-1} + V\}$.

*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?
 A: Consider $(F[x]/V\,\setminus \{\,a\cdot x^0+V\mid a\in F\,\})\cup F$.
Define addition as
$$ a+b=\begin{cases}a+ b&\text{if } a,b\in F\\
(a\cdot x^0+V)+b&\text{if }a \in F, b\notin F\\
a+(b\cdot x^0+V)&\text{if }a \notin F, b\in F\\
a+b&\text{if }a,b\notin F\end{cases}$$
and similarly for multiplication. Now go through the hell of showing that this givs us a field. Or please don't, work with $E$ and note that there is a canonical embedding $F\to E$.
This is essentially thge same "trick" we use when saying that $\mathbb Z\subseteq \mathbb Q$, although $\mathbb Q$ is constructed as a set of equivalence classes of pairs of integers and does not contain a single integer.
Edit: In fact, those canonical maps are so natual that I forgot to explicitly mention $F\to F[x]$, $a\mapsto a\cdot x^0$. If you don't trust that one may view $F$ as a subset of $F[x]/V$, you shuld not believe that $F\subseteq F[x]$ either: $F[x]$ is a ring together with a ring morphism $i\colon F\to F[x]$ and a map $\iota\colon\{x\}\to F[x]$ with the universal property that for each ring morhism $f\colon F\to Y$ and map $\phi\colon\{x\}\to  Y$ there exists a unique ring morphism $h\colon F[x]\to Y$ with $h\circ i=f$ and $h\circ\iota=\phi$. A priori, $F$ need not be a subring of $F[x]$; and yet we identify $F$ with $i(F)$ and $x$ with $\iota(x)$. - And even in this excursion I ought to add a handful of forgetful functors :)
A: You are correct. In the most general and careful scenario we can only say there exists another field $E$ with a subfield isomorphic to $F$ for which the corresponding polynomial has a root. HOWEVER, at this point we are used to, for example,  totally identifying $F$ with the constants in $F[x]$. Then, since $F \cap V = 0$, we can identify $F$ with its image in the quotient. But yes, this is after a series of identifications, though they are all natural.
