5) Q(i) Q(i
√ 5)
Extensive but routine calculations (Exercise 8.2) show that these, together with K, are the only subfields of L. So in this case the Galois correspondence is bijective.
8.7 Diet Galois
To provide further motivation, we now pursue a modernised version of La-grange’s train of thought in his memoir of 1770-1771, which paved the way for Galois. Indeed we will follow a line of argument that is very close to the work of Ruffini and Abel, and prove that the general quintic is not soluble by radicals. Why, then, does the rest of this book exist? Because ‘general’ has a paradoxically special meaning in this context, and we have to place a very strong restriction on the kind of radical that is permitted. A major feature of Galois theory is that it does not as-sume this restriction. However, quadratics, cubics, and quartics are soluble by these restricted types of radical, so the discission here does have some intrinsic merit. It could profitably be included as an application in a first course of group theory, or a digression in a course on rings and fields.
We have already encountered the symmetric group Sn, which comprises all per-mutations of the set {1, 2, . . . , n}. Its order is n!. When n ≥ 2, Snhas a subgroup of index 2 (that is, of order n!/2); namely, the alternating group An,F which consists of all products of an even number of transpositions (ab). The elements of Anare the even permutations. The group Anis a normal subgroup of Sn. It is well known that Anis generated by all 3-cycles (abc): see Exercise 8.7. The group A5holds the secret of the quintic, as we now explain.
Introduce the polynomial ring C[t1, . . . ,tn] in n indeterminates. Let its field of fractions be C(t1, . . . ,tn), consisting of rational expressions in the tj. Consider the polynomial
F(t) = (t − t1) . . . (t − tn)
over C(t1, . . . ,tn), whose zeros are t1, . . . ,tn. Expanding and using induction, we see that
F(t) = tn− s1tn−1+ s2tn−2+ · · · + (−1)nsn (8.5)
Diet Galois 117 where the sjare the elementary symmetric polynomials
s1 = t1+ · · · + tn
s2 = t1t2+ t1t3+ · · · + tn−1tn . . .
sn = t1. . .tn Here sris the sum of all products of r distinct tj.
The symmetric group Snacts as symmetries of C(t1, . . . ,tn):
σ f (t1, . . . ,tn) = f (tσ (1), . . . ,tσ (n))
for f ∈ C(t1, . . . ,tn). The fixed field K of Snconsists, by definition, of all symmetric rational functions in the tj, which is known to be generated over C by the n elemen-tary symmetric polynomials in the tj. That is, K = C(s1, . . . , sn). Moreover, the sj
satisfy no nontrivial polynomial relation: they are independent. There is a classical proof of these facts based on induction, using ‘symmetrised monomials’
t1a1t2a2· · ·tnan+ all permutations thereof
and the so-called ‘lexicographic ordering’ of the list of exponents a1, . . . , an. See Exercise 8.5. A more modern but less constructive proof is given in Chapter 18.
Assuming that the sjgenerate the fixed field, we consider the extension C(t1, . . . ,tn) : C(s1, . . . , sn)
We know that in C(t1, . . . ,tn) the polynomial F(t) in (8.5) factorises completely as F(t) = (t − t1) . . . (t − tn)
Since the sj are independent indeterminates, F(t) is traditionally called the general polynomial of degree n. The reason for this name is that this polynomial has a uni-versal property. If we can solve F(t) = 0 by radicals, then we can solve any specific complex polynomial equation of degree n by radicals. Just substitute specific num-bers for the coefficients sj. The converse, however, is not obvious. We might be able to solve every specific complex polynomial equation of degree n by radicals, but using a different formula each time. Then we would not be able to deduce a radi-cal expression to solve F(t) = 0. So the adjective ‘general’ is somewhat misleading;
‘generic’ would be better, and is sometimes used.
The next definition is not standard, but its name is justified because it reflects the assumptions made by Ruffini in his attempted proof that the quintic is insoluble.
Definition 8.8. The general polynomial equation F(t) = 0 is soluble by Ruffini rad-icalsif there exists a finite tower of subfields
C(s1, . . . , sn) = K0⊆ K1⊆ · · · ⊆ Kr= C(t1, . . . ,tn) (8.6) such that for j = 1, . . . , r,
Kj= Kj−1(αj) and αnjj ∈ Kj for nj≥ 2, nj∈ N
The aim of this definition is to exclude possibilities like the√
−121 in Cardano’s solution (1.10) of the quartic equation t4− 15t − 4 = 0, which does not lie in the field generated by the roots, but is used to express them by radicals.
Ruffini tacitly assumed that if F(t) = 0 is soluble by radicals, then those radicals are all expressible as rational functions of the roots t1, . . . ,tn. Indeed, this was the sit-uation studied by his predecessor Lagrange in his deep but inconclusive researches on the quintic. So Lagrange and Ruffini considered only solubility by Ruffini cals. However, this is a strong assumption. It is conceivable that a solution by radi-cals might exist, for which some of the αj constructed along the way do not lie in C(t1, . . . ,tn), but in some extension of C(t1, . . . ,tn). For example,√5
s1might be use-ful. (It is useful to solve t5− s1= 0, for instance, but the solutions of this equation do not belong to C(t1, . . . ,tn).) However, the more we think about this possibility, the less likely it seems. Abel thought about it very hard, and proved that if F(t) = 0 is soluble by radicals, then those radicals are all expressible in terms of rational func-tions of the roots—they are Ruffini radicals after all. This step, historically called
‘Abel’s Theorem’, is more commonly referred to as the ‘Theorem on Natural Irra-tionalities’. From today’s perspective, it is the main difficulty in the impossibility proof. So, following Lagrange and Ruffini, we start by defining the main difficulty away. In compensation, we gain excellent motivation for the remainder of this book.
For completeness, we prove the Theorem on Natural Irrationalities in Section 8.8, using classical (pre-Galois) methods. As preparation for all of the above, we need:
Proposition 8.9. If there is a finite tower of subfields (8.6), then it can be refined (if necessary increasing its length) to make all njprime.
Proof. For fixed j write nj= p1. . . pkwhere the plare prime. Let βl= αpjl+1...pk, for 0 ≤ l ≤ k. Then β0∈ Kjand βlpl ∈ Kj(βl−1), and the rest is easy.
For the remainder of this chapter we assume that this refinement has been per-formed, and write pjfor njas a reminder. With this preliminary step completed, we will prove:
Theorem 8.10. The general polynomial equation F(t) = 0 is insoluble by Ruffini radicals if n≥ 5.
All we need is a simple group-theoretic lemma.
Lemma 8.11. (1) The symmetric group Snhas a cyclic quotient group of prime order p if and only if p= 2 and n ≥ 2, in which case the kernel is the alternating group An. (2) The alternating group Anhas a cyclic quotient group of prime order p if and only if p= 3 and n = 3, 4.
Proof. (1) We may assume n ≥ 3 since there is nothing to prove when n = 1, 2.
Suppose that N is a normal subgroup of Snand Sn/N ∼= Zp. Then Sn/N is abelian, so N contains every commutator ghg−1h−1for g, h ∈ Sn. To see why, let ¯gdenote the image of g ∈ Snin the quotient group Sn/N. Since Sn/N is abelian, ¯g¯hg¯−1¯h−1= ¯1 in Sn/N; that is, ghg−1h−1∈ N.
Diet Galois 119 Let g, h be 2-cycles of the form g = (ab), h = (ac) where a, b, c are distinct. Then
ghg−1h−1= (bca)
is a 3-cycle, and all possible 3-cycles can be obtained in this way. Therefore N con-tains all 3-cycles. But the 3-cycles generate An, so N ⊇ An. Therefore p = 2 since
|Sn/An| = 2.
(2) Suppose that N is a normal subgroup of Anand An/N ∼= Zp. Again, N contains every commutator. If n = 2 then Anis trivial. When n = 3 we know that An∼= Z3.
Suppose first that n = 4. Consider the commutator ghg−1h−1 where g = (abc), h = (abd) for a, b, c, d distinct. Computation shows that
ghg−1h−1= (ab)(cd)
so N must contain (12)(34), (13)(24), and (14)(23). It also contains the identity. But these four elements form a group V. Thus V ⊆ N. Since V is a normal subgroup of A4and A4/V ∼= Z3, we are done.
The symbol V comes from Klein’s term Vierergruppe, or ‘fours-group’. Nowa-days it is usually called the Klein four-group.
Finally, assume that n ≥ 5. The same argument shows that N contains all per-mutations of the form (ab)(cd). If a, b, c, d, e are all distinct (which is why the case n= 4 is special) then
(ab)(cd) · (ab)(ce) = (ced)
so N contains all 3-cycles. But the 3-cycles generate An, so this case cannot occur.
As our final preparatory step, we recall the expression (1.13) δ =
n
∏
j<k(tj− tk)
It is not a symmetric polynomial in the tj, but its square ∆ = δ2is, because
∆ = (−1)n(n−1)/2
n
∏
j6=k(tj− tk)
The expression ∆, mentioned in passing in Section 1.4, is called the discriminant of F(t). If σ ∈ Sn, then the action of σ sends δ to ±δ . The even permutations (those in An) fix δ , and the odd ones map δ to −δ . Indeed, this is a standard way to define odd and even permutations.
We are now ready for the:
Proof of Theorem 8.10
Assume that F(t) = 0 is soluble by Ruffini radicals, with a tower (8.6) of sub-fields Kjin which all nj= pjare prime. Let K = C(s1, . . . , sn) and L = C(t1, . . . ,tn).
Consider the first step in the tower,
K⊆ K1⊆ L
where K1= K(α1), α1p∈ K, α16∈ K, and p = p1is prime.
Since α1∈ L we can act on it by Sn, and since every σ ∈ Snfixes K we have (σ (α1))p= α1p
Therefore σ (α1) = ζj(σ )α1, for ζ a primitive pth root of unity and j(σ ) an integer between 0 and p − 1. The set of all pth roots of unity in C is a group under multipli-cation, and this group is cyclic, isomorphic to Zp. Indeed ζaζb= ζa+bwhere a + b is taken modulo p.
Clearly the map
j: Sn → Zp
σ 7→ j(σ )
is a group homomorphism. Since α16∈ K, some σ (α1) 6= α1, so j is nontrivial. Since Zphas prime order, hence no nontrivial proper subgroups, j must be onto. Therefore Snhas a homomorphic image that is cyclic of order p. By Lemma 8.11, p = 2 and the kernel is An. Therefore α1is fixed by An.
We claim that this implies that α1∈ K(δ ). Since p = 2, the relation α1p∈ K becomes α12∈ K, so α1is a zero of t2− α12∈ K[t]. The images of α1under Snmust all be zeros of this, namely ±α1. Now α1is fixed by An but not by Sn, so some permutation σ ∈ Sn\ Ansatisifes σ (α1) = −α1. Then δ α1is fixed by both Anand σ , hence by Sn. So δ α1∈ K and α1∈ K(δ ).
If n = 2 we are finished. Otherwise consider the second step in the tower K(δ ) ⊆ K2= K(δ )(α2)
By a similar argument, α2defines a group homomorphism j : An→ Zp, which again must be onto. By Lemma 8.11, p = 3 and n = 3, 4. In particular, no tower of Ruffini radicals exists when n ≥ 5.
It is plausible that any tower of radicals that leads from C(s1, . . . , sn) to a subfield containing C(t1, . . . ,tn) must give rise to a tower of Ruffini radicals. However, it is not at all clear how to prove this, and in fact, this is where the main difficulty of the problem really lies, once the role of permutations is understood. Ruffini appeared not to notice that this needed proof. Abel tackled the obstacle head on.
Galois worked his way round it, by way of the Galois group—an extremely el-egant solution. The actual details of his work differ considerably from the modern presentation, see Neumann (2011), both notationally and strategically. However, the underlying idea of studying what we now interpret as the symmetry group of the polynomial, and deriving properties related to solubility by radicals, is central to Ga-lois’s approach. His method also went much further: it applies not just to the general polynomial F(t), but to any polynomial whatsoever. And it provides necessary and sufficientconditions for solutions by radicals to exist.
Exercises 8.9-8.11 provide enough hints for you to show that when n = 2, 3, 4 the equation F(t) = 0, where F is defined by (8.5), can be solved by Ruffini radi-cals. Therefore, despite the special nature of Ruffini radicals, we see that the quintic
Natural Irrationalities 121 equation differs (radically) from the quadratic, cubic, and quartic equations. We also appreciate the significant role of group theory and symmetries of the roots of a poly-nomial for the existence—or not—of a solution by radicals. This will serve us in good stead when the going gets tougher.