# 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.

• Without the context this is false. There are three indistinguishable third roots of $2$ over $\mathbb Q$, and their minimal polynomial $x^3 - 2$ has degree $3$, so the left hand side is a $3$-dimensional $\mathbb Q$-vector space, whereas the right hand side is $6$-dimensional. This is assuming that your minimal polynomial is the minimal polynomial of some element of an algebraic field extension over $\mathbb Q$. May 26 '15 at 9:29
• so it is specific to just one element in such a case? May 26 '15 at 9:32
• In fact, the statement is true if we only write $\mathbb Q(\alpha)$ on the right, if $f$ is the minimal polynomial of $\alpha$. May 26 '15 at 9:33

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 :

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

• Could you provide a proof for that at some source perhaps? May 26 '15 at 9:34
• I don't know at which level you are but this paper from Milne might do the job... jmilne.org/math/CourseNotes/FT.pdf May 26 '15 at 9:36
• Right level certainly, I just want it because it's somewhere and I just need to brush up on the details, could you say which page it is on? May 26 '15 at 9:36
• @ZelosMalum page 27 : splitting fields (i.e. the field generated by the roots of a polynomial), and page 17 stem fields (i.e. some field generated by a root of a polynomial). What darij grinberg and I are saying is that stem fields are not always splitting fields... May 26 '15 at 9:41
• Fair enough, thank you for the information it helped! May 26 '15 at 9:42

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...

• Aha that's the distinction I am missing then May 26 '15 at 9:41
• If $n=2$, then $\mathbb{Q}/\langle f\rangle=\mathbb{Q}[\alpha_1,\alpha_2]$. May 26 '15 at 13:47