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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.

Stahl
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Jack Schmidt
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2 Answers2

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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.

Qiaochu Yuan
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I haven't found a stellar way to do this by hand, but it is now easy to do with Pari/GP. The basic idea is you just factor f over Q[x]/(f).

Often this is easy to do: find some prime p in Q such that {x,x^p,x^(p^2),…} has exactly deg(f) distinct residues mod (f,p). Choose p larger than the twice the largest coefficient of the ones in the (unknown) answer. Replace a mod p with the integer of smallest absolute value congruent to a mod p for each of the coefficients of x^(p^i) mod (f,p). Check that the formula works. I had to take p=31 in the particularly 6th degree case, so this is not exactly great for by hand exams.

There are more refined versions of factoring using Hensel lifting or combining several primes both of which allow smaller primes to be used (and for it to work in general). The details of one (or two) algorithms are in section 3.6.2 of Henri Cohen's textbook for a Course in Computational Algebraic Number Theory, and some others are also in the source code to Pari/GP (with references).

Jack Schmidt
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