# Semialgebraic conditions that convey properties of Galois group

Let $f \in \mathbb{Z}[x]$ be a polynomial of degree $n$ with integer coefficients and let $G_f$ be the Galois group of $f$ over $\mathbb{Q}$.

I am trying to collect results that convey some information about $G_f$ that only involve the coefficients of $f$ in semialgebraic conditions. Let me explain what I mean by also giving the few such results that I have found.

Example 1

When trying to determine $G_f$, the first thing to check is if $f$ is irreducible over $\mathbb{Q}$ since this holds if and only if $G_f$ is transitive. However, the condition "$f$ is irreducible over $\mathbb{Q}$" is not a semialgebraic condition of the coefficients of $f$. There are many popular methods to show that a polynomial is irreducible, including Eisenstein's criterion (a special case of Dumas's criterion, which uses Newton diagrams, see Section 2.2.1 in Polynomials), irreducibility over $\mathbb{F}_p$ for some prime $p$, and Pólya-type criteria (also see Section 2.2.2 in Polynomials). However, these criteria are still not semialgebraic conditions of the coefficients of $f$.

Instead, an irreducibility criterion that only involves the coefficients in a semialgebraic expression is Perron's criterion:

Let $f(x) = \sum_{i=0}^n a_i x^i$ be a polynomial with integer coefficients such that $a_n = 1$ and $a_0 \ne 0$.

a) If $|a_{n-1}| > 1 + |a_{n-2}| + \dotsb + a_0$, then f is irreducible.

b) If $|a_{n-1}| \ge 1 + |a_{n-2}| + \dotsb + a_0$ and $f(\pm 1) \ne 0$, then f is irreducible.

The only other results of this kind that I know of are by Brauer (either Theorem 2.2.6 in Polynomials or results in On the Irreducibility of Polynomials with Large Third Coefficient).

Example 2

The method of resolvents for Galois groups seems quite strong to me. If I understand it correctly, for each subgroup of $H \le S_n$, there exists a polynomial $\varphi_H$ such that $G_f \le H$ if and only if $\operatorname{Res}(f,\varphi_H)$ contains a rational root. The simplest example of this is when $H = A_n$ is the alternating group, in which case $\varphi_H = f'$ is the derivative of $f$ and $\operatorname{Res}(f,\varphi_H) = x^2 - \Delta(f)$, where $\Delta(f)$ is the discriminant of $f$. Once again, the condition that some polynomial does (or doesn't) contain a rational root is not a semialgebraic condition of the roots of $f$.

For this simplest case of Galois resolvent, it is easy to obtain the kind of results that I am looking for.

If $\Delta(f) < 0$, then $G_f \not\le A_n$.

In this case (as well as in the stronger iff condition), there are an odd number of nonreal complex conjugate pairs, so the Galois group contains an odd permutation of order 2, namely complex conjugation.

Question

To explicitly state my question:

What other results are known that convey information about the Galois group of $f$ but only involve the coefficients of $f$ in semialgebraic conditions?

• I would hardly call that “algebraic conditions”. Inequalities are more “analytic” in my view. And “irreducibility” is defined by multiplication, and is thus very “algebraic” in a sense. I find your terminology confusing. – Ewan Delanoy Nov 24 '13 at 17:06
• @EwanDelanoy Yes, thanks for your comments. I wasn't sure if "algebraic" was the right word. I have changed to using "analytic". – Tyson Williams Nov 24 '13 at 20:58
• To tell the truth, I don’t think “analytic” is better than “algebraic” in this context. Perhaps you could just say “simple polynomial conditions”. – Ewan Delanoy Nov 24 '13 at 21:01
• @EwanDelanoy They don't have to be polynomial conditions though. Other functions, like exponentials, would also be fine. – Tyson Williams Nov 24 '13 at 21:42
• The word is semi-algebraic (if I understand OP correctly) – Grigory M Jan 1 '14 at 20:05

I don't know if this is sufficiently "analytic", but let's give it a try:

If $f\in\mathbb Q[x]$ is irreducible of prime degree $p$ such that $f$ has exactly $p-2$ zeros in $\mathbb R$, then the Galois group of $f$ is $S_p$.

Example: $f = X^5 - 4X + 2$ is of prime degree $5$, and it is irreducible by Eisenstein. The derivative $f' = 5X^4 - 4$ has exactly $2$ zeros, so $f$ has at most $3$ zeroes in $\Bbb R$. Furthermore, $f(-2) < 0$, $f(0) > 0$, $f(1) < 0$, $f(2) >0$, so by the intermediate value theorem there are three zeroes in $\mathbb R$. By the criterion, the Galois group of $f$ is isomorphic to $S_5$.

A typical application of this criterion is to show that non-solvable polynomial equations actually do exist. Here, a polynomial with a non-solvable Galois group is needed. The above example provides a such a polynomial (since the group $S_5$ is not solvable).

• I am happy with all of your assumptions except the condition that $f$ is irreducible. How can you determine if a polynomial is irreducible using only analytic conditions? In my question, I mentioned (or linked to) all such results of which I am aware. – Tyson Williams Dec 2 '13 at 19:58
• @TysonWilliams: So just replace the assumption "$f$ irreducible" by one of the "analytic" criterions for irreducibility in your question! – azimut Dec 2 '13 at 20:09
• Sure, but then your answer doesn't add anything I didn't already mention in my question or is basic abstract algebra. If $f$ has exactly two nonreal zeros, then complex conjugation is an element of $G_f$ as I stated. In this case, it is a transposition. If $f$ is irreducible, then $G_f$ is transitive, so it contains an element of order $p$. Since $p$ is prime, the only such element is a $p$-cycle. Then a transposition and large cycle (length more than half the degree of $f$) generate $S_n$. The most important part of this argument is that $f$ is irreducible. – Tyson Williams Dec 3 '13 at 14:39
• Do you know of any analytic way to show that $f$ is irreducible that I didn't already mention? – Tyson Williams Dec 3 '13 at 14:40
• @TysonWilliams Sorry for bothering your with "basic abstract algebra". – azimut Dec 3 '13 at 15:00