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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?

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This fails if $P$ is allowed to have degree equal to $|G|$, where $G$ is some non-solvable Galois group, since one could take $P$ to be the minimal polynomial of a primitive element of a field extension with Galois group $G$. –  Kevin Ventullo Oct 26 '12 at 6:40
    
@Kevin : Good point, in this case the splitting field of $P$ is even generated by a single root. –  François Brunault Oct 26 '12 at 7:08
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Dear François, see this question mathoverflow.net/questions/24081/… for a proof of Galois's criterion. A purely algebraic application (the determination of the compositum of all separable extensions of given prime degree, with solvable galoisian closure, of a given commutative field), see Section 4 of arxiv.org/abs/1005.2016, specially Lemma 20. This section might furnish a certain number of exercises for your course. –  Chandan Singh Dalawat Oct 26 '12 at 7:18
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Galois himself proved that if a primitive polynomial equation $f(x)=0$ is solvable by radicals, then its degree is a power of a prime number. Definition of primitive according to Galois: Let $mn$ be the degree of $f$, it is not possible that $f$ splits in $m$ irreducible factors of degree $n$ by adjoining all roots of an equation $g$ of degree $m$. For a detailed discussion and relation with the concept of primitivity in group theory, see the paper by Peter Neumann, The Concept of Primitivity in Group Theory and the Second Memoir of Galois (2006). –  ACL Oct 26 '12 at 7:35
    
Dear Antoine, thanks! I always wondered what 'équation primitive' meant in Galois's writings... –  François Brunault Oct 26 '12 at 7:48

1 Answer 1

up vote 22 down vote accepted

The question asks about the relation of the properties 1. and 3., though possibly the intended meaning of 1. was 2.:

  1. The splitting field of $P$ is generated by two roots of $P$.
  2. The splitting field of $P$ is generated by any two roots of $P$.
  3. 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:

If $n=\deg(P)\not\equiv1\pmod{120}$, and the splitting field of $P$ is not generated by a root (see Kevin Ventullo's comment above), then 2. implies 3.

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.

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Thank you very much for your answer. The result you mention is a very nice generalization of Galois's theorem. Just a naive thought : even if $3. \Rightarrow 1.$ (or $3. \Rightarrow 2.$) does not hold, one could try to give an bound on the number of roots needed to generate the splitting field. Could this bound depend only on the prime factorization of the degree? This is a purely group theoretic question on transitive solvable subgroups of $\mathfrak{S}_n$. –  François Brunault Oct 26 '12 at 14:19
    
@François: See the addition to my answer. –  Peter Mueller Oct 26 '12 at 18:39
    
Can you give more details (e.g. references) for the quoted theorems of Thompson and Zassenhaus? Thanks in advance. –  GH from MO Oct 26 '12 at 20:02
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@GH: en.wikipedia.org/wiki/Frobenius_group describes and gives references for these results. Besides the cited book by Passman there, another nice treatment of Thompson's theorem is in Finite Group Theory by Isaacs. A recent and relatively elementary proof of Zassenhaus' theorem was given by Mazurov, see sciencedirect.com/science/article/pii/S0021869303001078 –  Peter Mueller Oct 26 '12 at 21:56
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@GH: Another elementary proof of Zassenhaus' theorem (by Ulrich Meierfrankenfeld) you can find in math.msu.edu/~meier/Preprints/Frobenius/frobenius.pdf –  j.p. Oct 27 '12 at 12:35

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