# Prove/Disprove : Every polynomial with prime degree and coefficients in $[-1,1]$ has galois-group $S_p$

Conjecture :

Let $p$ be a prime number , $f\in \mathbb Z[X]$ an irreducible polynomial with degree $p$ and coefficients in the range $[-1,1]$. Then the galois group of $f$ over $\mathbb Q$ is $S_p$

Can anyone prove or disprove this conjecture ?

The conjecture is true for the primes upto $p=11$. To prove the conjecture it would be sufficient to prove that the galois group of $f$ over $\mathbb Q$ contains a transposition, which is surely the case if there are exactly two non-real roots. But in general, I do not know how this can be shown.

• Do you mean range as in analysis (i.e. the integers $-1,0,1$), or as set $\{-1,1\}$? – ahulpke Feb 9 '16 at 3:35
• How did you prove that this is true until $p=11$? If you can describe your methods, perhaps that would throw some light on why it might be true in general. – Prahlad Vaidyanathan Feb 9 '16 at 11:05
• @ahulpke The coefficients are $-1$ , $0$ or $1$. – Peter Feb 9 '16 at 12:26
• @PrahladVaidyanathan I did not prove it by hand, I simply checked all polynomials with PARI/GP. I wanted to doublecheck it with GAP, but I aborted the calculation for $p=11$ because GAP is very slow in checking whether a polynomial is irreducible and determining the galois group. This is the reason, I did not check the case $p=13$ (which is impossible in PARI/GP, which is limitied to degree $11$). – Peter Feb 9 '16 at 12:28

This is not really a full solution, but it is a bit to unwieldy to put in as a comment:

The property you are conjecturing - Galois group $S_n$ - is known to be very likely. Thus experimental data for small $p$ is not really that convincing.

(Should you want to try degrees larger than 11 in GAP, you might want to look at ProbabilityShapes-- PARI is faster as it does not strictly prove the Galois group type either.)

You cannot expect complex conjugation to have a particularly nice shape. Indeed random examples in degree 17 find a varying number of real roots.

My (maybe naive) guess on a best shot towards proving it is: Prove that the discriminant is not a square. If so (by a classification of transitive groups of prime degree -- e.g. Guralnick's work) the Galois group is either $S_n$ or contained in $AGL_1(p)$. To exclude the latter choice show that there is an element that has at least two fixed points -- i.e. by showing that there must be a prime (not dividing the discriminant) modulo which the polynomial has at least 2 roots but does not split into linear factors.

• PARI does not prove the galois-group-type ? For which degrees is this the case ? – Peter Feb 9 '16 at 16:04
• @Peter To the best of my knowledge PARI uses numerical approximations of the roots to determine the Galois group but does not prove that no rounding errors happen. I think the chance of an error is negligibly small, but so it is for ProbabilityShapes in GAP versus the proven GaloisType. – ahulpke Feb 9 '16 at 16:15