Proof for quotient polynomial rings equivalent to field extension

Question

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.

Answer 1

This might not be true. See for instance $f(X)=X^3-2$ (irreducible by Eisenstein) then :

$$\dim_{\mathbb{Q}}(\mathbb{Q}[x]/(f(x))=\deg(f)=3 $$

However the roots of $f$ are $\sqrt[3]{2}$, $\sqrt[3]{2}j$ and $\sqrt[3]{2}j^2$. It is easy to see that $F:=\mathbb{Q}[{\sqrt[3]{2}},{\sqrt[3]{2}}j,{\sqrt[3]{2}}j^2]$ cannot be of dimension $3$ over $\mathbb{Q}$ because $F':=\mathbb{Q}[{\sqrt[3]{2}}]$ is a subfield of dimension $3$ of $F$ over $\mathbb{Q}$ and $F'$ is a proper subfield of $F$...

What is true is that $\mathbb{Q}[\alpha_1,\dots,\alpha_n]$ is always isomorphic to any splitting field of $f$ (i.e. minimal field such that $f$ is split in it).

What is true is that for any $1\leq i\leq n$ you will have :

$$\mathbb{Q}[x]/(f)\text{ is isomorphic to } \mathbb{Q}[\alpha_i] $$

Answer 2

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...