Proof for quotient polynomial rings equivalent to field extension I am predominantely looking for a proof, I have seen in my books and around but seem to have a hard time finding that if we let $\alpha_1,\alpha_2,...,\alpha_n$ be the roots of the minimal polynomial $f$, then we have that
$$\mathbb{Q}[x]/\langle f\rangle\cong \mathbb{Q}[\alpha_1,\ldots,\alpha_n]$$
As said I am after finding the proof somewhere, I just have not been able to find it anywhere.

As noted elsewhere in the thread, if $f$ is irreducible with roots $\{a_1, a_2, ..., a_n\}$, it is definitely not the case that $\mathbb{Q}[x]/\langle f(x) \rangle \cong \mathbb{Q}[a_1, a_2, ..., a_n]$.  It is the case, however, that $\mathbb{Q}[x]/ \langle f(x) \rangle \cong \mathbb{Q}[a_j]$ for any $1 \leq j \leq n$.
More generally, this result holds for any field $F$ and any irreducible polynomial $f(x) \in F[x]$.  We can prove this with the isomorphism theorem.  Let $\alpha$ be any root of $f$ and consider the evaluation homomorphism $ev_{\alpha}:F[x] \rightarrow F[\alpha]$ defined where $g(x) \mapsto g(\alpha)$.  You'll want to prove that this is indeed a homomorphism.  
Now certainly $\ker(\phi) = \langle f(x) \rangle$.  This is because $F[x]$ is a PID and $f(x)$ is irreducible over $F$ with $\alpha$ as a root.  We also know that $ev_\alpha$ maps surjectively to $F[\alpha]$.  
And so...
