Given $f \in \mathbb{Q}[x]$ irreducible. How many and which roots of $f$ are contained in $\mathbb{Q}[x]/(f)$?

Question

It is a fact that struggle me for a while. When working with irreducible polynomial over $\mathbb{Q}$ it is natural to build the extension ${\mathbb{Q}[x]}/{(f)} $ in which "lives " one root of the polynomial. More precisely, $[x]$ is a root of the polynomial in this extension.

BUT

Take for example $f= x^3-2$ which roots are $ 2^{1/3} , \ w2^{1/3} , \ w^2 2^{1/3}$ where $w \in RPU(3)$. Doing the extension (the same as above) I build a field where $[x]$ is a root of $f$. Where are the other two roots? I don't think they are here, just for a matter of dimension as vector space over the field of rationals. I calculate the size of the splitting field of $f$ and have as a result 6. But my extension have dimension 3 over Q. So something is missing. Why? And if this reasoning is correct, given an irreducible polynomial $g$, which root(s) I found in this way?

Hope it is clear :), thanks in advance :)

Answer 1

The other two roots are in an extension field of the field you have found.

As to which root you find, unmask the question. There is no way, from an algebraic point of view, to distinguish among the roots of an irreducible polynomial.

Answer 2

At least one, and 'which' has no sense in this setting, as this root can be taken into any other root by a field automorphism of a (perhaps bigger) field over which f factors to linear components.

However, I think, in this generality this is not an easy question.

But 'a priori' only one root enters into the quotient $\Bbb Q[x]/(f)$. But! If the field operations yield other roots from this one, they also must be present. (I.e. in case of quadratic polynomial.)