Is there an easy way to tell how many roots $f(x)$ has in $\Bbb{Q}[x]/(f)$ given the coefficients of the polynomial $f$ in $\Bbb{Q}[x]$?
Is there an easy way to find the roots as rational expressions in $x$?
The easiest example is a pure quadratic: $X^2 + 7$ for instance. If $A$ is a root, then so is $−A$. Good ole $\pm\sqrt{−7}$.
If the Galois group is abelian (like for any quadratic), then all of the roots can be expressed as polynomials in a given root. However, I am not sure how to tell by looking at the polynomial if its Galois group is abelian, and even if it is, I am not sure how to find those rational expressions for the other roots.
It might help to see some non-Abelian (non-Galois) examples:
If $A$ is a root of $X^6 + 2X^4 − 8$, then $−A$ is also a root, but its other $4$ roots cannot be expressed as rational functions of $A$ (assuming I still understand Galois theory).
Is there some easy way (not asking a CAS to calculate the Galois group) to see the other $4$ roots of of $X^6 + 2X^4 − 8$ cannot be expressed as rational functions of $A$?
This one had the nice feature that it was a function of $X^2$, so it was easy to find two roots. For $X^6 − 2X^5 + 3X^3 − 2X − 1$, I still have not found its other root (even using a CAS).
If $A$ is a root of $X^6 − 2X^5 + 3X^3 − 2X − 1$, then what is a rational expression in $A$ for another root?
This all first came up with the polynomial $x^4−4x^2+2$, where several distinct ad hoc arguments each sufficed, but I had no real understanding of how to even tell if my ad hoc arguments were worth trying on other polynomials. If it helps, the roots are $A$, $−A$, $A^3−3A$, and $3A−A^3$.
The context is hand calculations and reasonable degrees (say $\leq 10$), though I am not opposed to having a polynomial evaluation oracle that computes $f(g(x)) \mod f(x)$ in $1$ second (so "try this finite and not too big list of possible roots" is ok).
If someone is interested, I am curious what the normalizer of a point stabilizer in the Galois group actually means in terms of Galois theory. The index of the point stabilizer in its normalizer is the number of roots of $f$ in $\Bbb{Q}[x]/(f)$, but I'm not sure if it really means anything useful.
Best Answer
If $f$ has abelian Galois group and you can find an explicit embedding of its splitting field into $\mathbb{Q}(\zeta_n)$, you get a quotient map $(\mathbb{Z}/n\mathbb{Z})^{\ast} \to \text{Gal}(f)$ which makes the action of the Galois group quite explicit. Applying elements of $(\mathbb{Z}/n\mathbb{Z})^{\ast}$ to a root $a$ of $f$ in $\mathbb{Q}(\zeta_n)$ gives you some explicit expressions in $\mathbb{Q}(\zeta_n)$ which you can then try to express as polynomials in $a$. I don't know how easy this will be to do, though, but in certain cases it is fairly explicit: for example if $f$ is the minimal polynomial of $2 \cos \frac{2 \pi}{n}$ you get Chebyshev polynomials.