Understanding Kronecker's Theorem The way I understand Kronecker's theorem (as I learnt it) is as follows.
We have a field $F$ and it's assosciated polynomial ring $F[x]$. If there's an irreducible polynomial $f(x) \in F[x]$ then $f(x)$ has no roots in $F$.
The main part of Kronecker's theorem is the following. We can take $F$ and embed it in the field extension $F[x]/f(x)F[x]$ through an embedding $h$. That is $F \cong h[F] \subseteq F[x]/f(x)F[x]$
Then the induced polynimial $\bar{f}(x)$ (which is induced by $f(x)$) in the polynomial ring $(F[x]/f(x)F[x])[x]$ has a root in $F[x]/f(x)F[x]$ (which is a field extension of $F$)
However I've seen Kronecker's theorem states as follows :

Let $F$ be a field and let $f(x)$ be a nonconstant polynomial in $F[x]$. Then there is an extension field $E$ of $F$ in which $f(x)$ has a root.

However in light of what I said above, it isn't $f(x)$ which has a root in the extension field, it's the induced polynomial $\bar{f}(x)$ which has a root in the extension field.
Since it isn't really our polynomial that we started off with that has a root in this larger field that we embedded our original field into, why is Kronecker's theorem so important?
I'll take a concrete example to show why I ask this. Let $\mathbb{R}$ bne our field and consider $f(x) = x^2 + 1 \in \mathbb{R}[x]$. We know that $f(x)$ is irreducible. I've seen authors claim that $\mathbb{R}[x]/(x^2 + 1) \cong \mathbb{C}$ 
(In this notation I'm assuming the authors mean that $\mathbb{R}[x]/(x^2 + 1)$ is just a shorthand notation for $\mathbb{R}[x]/(x^2 + 1)\mathbb{R}[x]$)
The induced polynomial is given by $\bar{f}(x) = (1 + (x^2+1)\mathbb{R}[x]) + (1 + (x^2 + 1)\mathbb{R}[x])x^2$ and it's root in $\mathbb{R}[x]/(x^2+1)\mathbb{R}[x]$ is $x + (x^2 +1)\mathbb{R}[x]$. But what I've seen people prequently mention is that $f(x) = x^2 +1$ has a root in $\mathbb{C}$ (which is isomorphic to $\mathbb{R}[x]/(x^2+1)\mathbb{R}[x]$) which is incorrect (at least it seems to me at this point) because it is in fact the induced polynomial $\bar{f}(x)$ which has a root in $\mathbb{C}$ I believe.
Kronecker's theorem seems very different to me compared to (the concept) if we took the rational field $\mathbb{Q}$ and an irreducible polynomial $x^2 - 2 \in \mathbb{Q}[x]$, and then embedded $\mathbb{Q}$ into $\mathbb{R}$ and then took the polynomial $x^2-2 \in \mathbb{R}[x]$ which is irreducible, since it has a root $\sqrt{2} \in \mathbb{R}$. Note that in this last example I may have overlooked subtleties so it's a very informal example I would think.
Is the last example I gave any different to the concept of Kronecker's theorem or are they both doing the same thing?
 A: The point is that $\overline{f}$ has coefficients in the image of $F$ under the embedding $F\to F[X]/(f)$. 
Hence if you modify $F[X]/(f)$ in such a way that we have an actual inclusion $F\subset \tilde{F[X]/(f)}$, then $\overline{f}$ becomes precisely $f$: $\tilde{\overline{f}}=f$; and so $\tilde{F[X]/(f)}$ really is an extension where $f$ has a root.
Now what's $\tilde{F[X]/(f)}$ ? Well it's a process that mathematicians do all the time when they have an embedding : they replace the image of this embedding by the embedded object so that the embedding becomes an inclusion.
In our situation, say you have a field morphism $i : k\to K$ (necessarily injective, hence an embedding), and you want to see $k$ as an actual subfield of $K$. Firstly, make sure that $k$ and $K$ are disjoint. This can be done for instance by putting $K' = K\times\{k\}$ and the obvious field structure on $K'$. Note that $K'$ is then isomorphic to $K$, with isomorphism $f$,and so you still have an embedding $j: k\to K'$ such that $f\circ i = j$.
Then define $\tilde{K'} := K'\setminus j(k) \cup k$; and you can then define a field structure on $\tilde{K'}$ such that : (i) $k$ is a subfield of $\tilde{K'}$ and (ii) : the map $\tilde{K'}\to K'$ defined by $x\mapsto j(x)$ if $x\in k$, $x$ otherwise is a field isomorphism. 
Hence we have that for any embedding of fields $i: k\to K$, there is a field $\tilde{K}$  and an isomorphism $m: K\to K'$ such that (i) $k$ is a subfield of $\tilde{K}$, and (ii) $m\circ i$ is the inclusion.
Applying this to our situation : $k= F$, $K= F[X]/(f)$ yields $\tilde{K}$ of whom $k$ is a subfield, and $m^*(\overline{f})= m^*(i*(f) ) = (m\circ i)^*(f) = f$ (where for a field morphism $j: L\to L'$, I let $j^*$ denote the induced morphism $L[X]\to L'[X]$); and thus, since $\overline{f}$ has a root in $K$, $m^*(\overline{f})=f$ has a root in $\tilde{K}$. 
Thus we have a field extension where $f$ has a root.
Now this is all set-theoretic and really "uninteresting" to the eyes of an algebraist : $\overline{f}$ is really essentially $f$ and if it has a root it's the same as saying that $f$ does in an extension; this is what the above argument shows. 
Note also that what I described above is in no way specific to fields : you can do this for algebraic structures whenever you have an embedding, and that's why algebraists tend to treat embeddings as inclusions (though there are some contexts where one wants to remember the embeddings as such) and embedded structures as substructures.
