I am currently teaching Galois theory and this week, I mentioned the following theorem of Galois :
Let $P(x) \in \mathbf{Q}[x]$ be an irreducible polynomial of prime degree. Then $P$ is solvable by radicals if and only if the splitting field of $P$ is generated by any two roots of $P$.
I was asked by a student whether this theorem can be generalized to polynomials whose degree is composite, maybe allowing the splitting field to be generated by more than two roots. I know that the proof of Galois's theorem relies on determining the solvable subgroups of $\mathfrak{S}_p$, but I don't know enough group theory to tell what can be proved in the case where the degree of the polynomial is composite, say $pq$ where $p$ and $q$ are (possibly equal) primes.
Does such a generalization of Galois's theorem exist? Or is there a conceptual reason why such a generalization cannot hold? In the latter case, do there already exist generalizations of Galois's theorem, possibly in different directions?
Best Answer
The question asks about the relation of the properties 1. and 3., though possibly the intended meaning of 1. was 2.:
- The splitting field of $P$ is generated by two roots of $P$.
- The splitting field of $P$ is generated by any two roots of $P$.
- The Galois group $G$ of $P$ is solvable.
In the prime degree case, all three properties are equivalent.For arbitrary degrees, 1. is a weak condition and so doesn't tell much about solvability of $G$. Condition 1. is also much weaker than 2. Despite its weakness, 3. does not imply 1. In fact, for each non-prime degree $n\ge6$, there is a solvable group $G$ of degree $n$ for which 1. does not hold. Indeed, it's easy to construct explicit examples for all such degrees: Let $n=rs$ with $r\ge3$, $s\ge2$. Then for suitable rational $a,b$, the polynomial $P(X)=(X^s-b)^r-a$ doesn't fulfill 1.
So let's forget about 1. Also, this example shows that 3. is far from implying 2.
The question remains whether 2. implies 3. Indeed, it comes close:
The reason for this is as follows: 2. says that $G$ is a Frobenius group. By Frobenius' Theorem, $G$ is a semidirect product of a regular normal subgroup $N$ and a point stabilizer $H$, called the Frobenius complement. By Thompson's Theorem, $N$ is nilpotent, so in particular solvable. What about $H$? By an old result of Zassenhaus, $H$ is either solvable, or its series of derived subgroups terminates in $\text{SL}(2,5)$, a group of order $120$. As $n=\lvert N\rvert$, and $H$ has regular orbits on $N\setminus\{e\}$, we get what I claimed above.
As to the excluded degrees: The smallest candidate of degree $121$ exists group theoretically: $\text{SL}(2,5)$ has a regular action on the nonzero elements of $\mathbb F_{11}^2$, yielding a Frobenius group $\mathbb F_{11}^2\rtimes\text{SL}(2,5)$. By a result of Jack Sonn (see here), every finite Frobenius group is a Galois group over the rationals. Thus there is an irreducible $P(X)$ of degree $121$ for which 2. holds, but 3. does not.
Added (Answering François' question in his comment below): This minimal number of roots which generate the splitting field is the size of a so-called minimal base of the permutation group $G$. A base of a permutation group is a subset of the points the group acts on whose elementwise stabilizer is trivial. If you take the wreath product $G=C_2\rtimes C_m$, in its natural action on $2m$ points, this number is $m$. So even if the degree is a product of two primes, the minimal base size can be arbitrarily large.
Things are better if $G$ is primitive and solvable: Then the minimal base size is at most $4$. See here for more results on minimal bases.