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I want to know how to find a polynomial $f(x)$ of degree $6$ in $\mathbb{Q}[x]$ with Galois groups $G_f=\mathbb{S}_6$. I have a criterion to find a polynomial wich Galois group is $\mathbb{S}_p$ with $p$ prime, but i don´t now if it work in this case

Jyrki Lahtonen
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Dimitri
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  • Probably the same method as in the case of $S_5$ will work. Find an irreducible polynomial with $4$ real roots and $2$ complex roots. Keith Conrad has some notes on recognizing $S_n$ and $A_n$ as Galois groups you might also find useful. – JSchlather Feb 18 '13 at 20:20
  • Ok, the criterion i find say $p$ prime so i am not sure, the idea of the $2$ complx root was that $\mathbb{S}_p$ is generated by a transposition (the complex conjugation of those 2 roots) and a $p$-cycle given by the irreducibility of the polynomial. But i don't know if is the same for $\mathbb{S}_n$ in general – Dimitri Feb 18 '13 at 20:32
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    almost all the polynomials of degree $n$ will have Galois group $S_n$. so potentially any random polynomial with rational coefficients will do the job ... of course this begs the question of a truly random number generator, so is not much useful – magguu Mar 06 '13 at 15:09
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    The method of @JSchlather isn't guaranteed to work, as there's no guarantee that there will be a 6-cycle. That's what's different about the prime case. – Gerry Myerson Oct 01 '14 at 07:28

1 Answers1

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Consider the polynomial $$ f(x)=x^6-8x^4-5x^3+4x^2+5x-12. $$ We view its Galois group $G$ as a subgroup of $S_6$ as permutations of its roots. I shall be using the fact that if a polynomial in $\mathbb{Z}[x]$ only has simple factors modulo a prime $p$, then the degrees of the irreducible factors give lengths of cycles in an element of a Galois group.

Modulo $p=2$ it factors as $f(x)\equiv x(x^5+x^2+1)$, so $G$ contains a 5-cycle. Furthermore we know that either $f(x)$ is irreducible or it has a rational integer as a root. The latter case can be excluded with the high school level rational root test, so we know that $f(x)$ is irreducible and $G$ is a transitive subgroup of $S_6$. (The presence of that 5-cycle proves that a point stabilizer of $G$ is also transitive, so we could also already conclude that $G$ is doubly transitive.)

Modulo $p=3$ it factors as $f(x)\equiv (x^3-x)(x^3-x+1)=x(x-1)(x+1)(x^3-x+1)$ so the Galois group has a 3-cycle.

Modulo $p=5$ it factors as $f(x)\equiv (x^4-1)(x^2+2)\equiv (x+1)(x+2)(x-1)(x-2)(x^2+2)$ so we can conclude that there is also a 2-cycle in $G$.

By transitivity of $G$ we can conclude that the point stabilizer $G_1$ of a fixed root $x_1$ of $f(x)$ in $G$ also contains a 5-cycle, a 3-cycle and a 2-cycle. (If the permutations produced above don't give one right away, then conjugate them by a suitable element of $G$). Therefore we can conclude that $2\cdot3\cdot5\mid |G_1|$. We can naturally view $G_1$ as a subgroup of $S_5$. The presence of a 2-cycle tells that $G_1$ is not a subgroup of $A_5$. The index of $G_1\cap A_5$ in $A_5$ is thus at most 4. But $A_5$ cannot have a proper subgroup of index $\le 4$, because that would give rise to a non-trivial homomorphism from $A_5$ to $S_k, k\le 4$, violating the known fact that $A_5$ is simple. Thus $G_1$ must contain all of $A_5$. Again, the presence of a 2-cycle in $G_1$ tells us that we must have $G_1=S_5$. From this it follows that $G=S_6$, because the only transitive subgroup of $S_6$ that has full $S_5$ as a point stabilizer is obviously $S_6$.

This was a bit ad hoc. At this point it is probably best that I admit having constructed $f(x)$ by prescribing those three factorizations modulo $p=2,3,5,$ and then using the Chinese Remainder Theorem. The rest was just wishfully making deductions hoping that they would take me to the finish line. The thing guiding me was that I knew enough about subgroups of $S_5$ to hope that the presence of those short cycles would force the group to be quite big.

Jyrki Lahtonen
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