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With a little more effort we can go the whole hog. Abel’s proof contains one fur-ther idea, which lets us delete the word ‘Ruffini’ from Theorem 8.10. This section is an optional extra, and nothing later depends on it. We continue to work with the gen-eral polynomial, so throughout this section L = C(t1, . . . ,t_n) and K = C(s1, . . . , s_n), where the s_jare the elementary symmetric polynomials in the t_j.

To delete ‘Ruffini’ we need:

Definition 8.12. An extension L : K in C is radical if L = K(α1, . . . , αm) where for each j = 1, . . . , m there exists an integer n_jsuch that

αⁿ_j^j ∈ K(α1, . . . , α_j−1) ( j ≥ 2)

The elements αjform a radical sequence for L : K. The radical degree of the radical α_jis n_j.

The essential point is:

Theorem 8.13. If the general polynomial equation F(t) = 0 can be solved by radi-cals, then it can be solved by Ruffini radicals.

Corollary 8.14. The general polynomial equation F(t) = 0 is insoluble by radicals if n≥ 5.

To prove the above, all we need is the so-called ‘Theorem on Natural Irrational-ities’, which states that extraneous radicals like√⁵

s1cannot help in the solution of F(t) = 0. More precisely:

Theorem 8.15 (Natural Irrationalities). If L contains an element x that lies in some radical extension R of K, then there exists a radical extension R⁰of K with x∈ R⁰and R⁰⊆ L.

Once we have proved Theorem 8.15, any solution of F(t) = 0 by radicals can be converted into one by Ruffini radicals. Theorem 8.13 and Corollary 8.14 are then immediate.

It remains to prove Theorem 8.15. A proof using Galois theory is straightforward, see Exercise 15.11. With what we know at the moment, we have to work a little harder—but, following Abel’s strategic insights, not much harder. We need several lemmas, and a technical definition.

Definition 8.16. Let R : K be a radical extension. The height of R : K is the smallest integer h such that there exist elements α1, . . . , αh∈ R and primes p₁, . . . , p_h such that R= K(α1, . . . , αh) and

α^p_j^j∈ K(α1, . . . , αj−1) 1 ≤ j ≤ h where when j= 1 we interpret K(α1, . . . , α_j−1) as K.

Proposition 8.9 shows that the height of every radical extension is defined.

We prove Theorem 8.15 by induction on the height of a radical extension R that contains x. The key step is extensions of height 1, and this is where all the work is put in.

Lemma 8.17. Let M be a subfield of L such that K ⊆ M, and let a ∈ M, where a is not a pth power in M. Then

(1) a^kis not a pth power in M for k= 1, 2, . . . , p − 1.

(2) The polynomial m(t) = t^p− a is irreducible over M.

Proof. (1) Since k is prime to p there exist integers q, l such that qp + lk = 1. If a^k= b^pwith b ∈ M, then

(a^qb^l)^p= a^qpb^{l p}= a^qpa^kl= a contrary to a not being a pth power in M.

(2) Assume for a contradiction that t^p− a is reducible over M. Suppose that P(t) is a monic irreducible factor of m(t) = t^p− a over M. For 0 ≤ j ≤ p − 1 let P_j(t) = P(ζ^jt), where ζ ∈ C ⊆ K ⊆ M is a primitive pth root of unity. Then P0= P, and Pjis irreducible for all j, for if P(ζ^jt) = g(t)h(t) then P(t) = g(ζ^{− j}t)h(ζ^{− j}t). Moreover, m(ζ^jt) = (ζ^jt)^p− a = t^p− a = m(t), so P_j divides m for all j = 0, . . . , p − 1 by Lemma 5.6.

We claim that P_k and P_j are coprime whenever 0 ≤ j < k ≤ p − 1. If not, by irreducibility

P_j(t) = cP_k(t) c∈ M Let

P(t) = p₀+ p₁t+ · · · + p_r−1t^r−1+ t^r where r ≤ p. By irreducibility, p₀6= 0. Then

P_j(t) = p₀+ p₁ζ^jt+ · · · + p_r−1ζ^j(r−1)t^r−1+ ζ^jrt^r P_k(t) = p₀+ p₁ζ^kt+ · · · + p_r−1ζ^k(r−1)t^r−1+ ζ^krt^r

so c = ζ^{( j−k)r}from the coefficient of t^r. But then p₀= ζ^{( j−k)r}p₀. Since p₀6= 0, we must have ζ^{( j−k)r}= 1, so r = p. But this implies that ∂ P = ∂ m, so m is irreducible over M.

Thus we may assume that the P_jare pairwise coprime. We know that P_j|m for all j, so

P0P1. . . P_p−1| m

Natural Irrationalities 123 Since ∂ p = r, it follows that pr ≤ p, so r = 1. Thus P is linear, so there exists b ∈ M such that (t − b)|m(t). But this implies that b^p= a, contradicting the assumption that ais not a pth power. Thus t^p− a is irreducible.

Now suppose that R is a radical extension of height 1 over M. Then R = M(α) where α^p∈ M, α 6∈ M. Therefore every x ∈ R \ M is uniquely expressible as

x= x₀+ x₁α + x2α²+ · · · x_p−1α^p−1 (8.7) where the x_j∈ M.This follows since [M(α) : M] = p by irreducibility of m. We want to put x into a more convenient form, and for this we need the following result:

Lemma 8.18. Let L ⊆ M be fields, and let p be a prime such that L contains a primitive pth root of unity ζ . Suppose that α, x0, . . . , x_p−1∈ M with α 6= 0, and L contains all of the elements

X_r= x₀+ (ζ^rα )x₁+ (ζ^rα )²x₂+ · · · + (ζ^rα )^p−1x_p−1 (8.8) for0 ≤ r ≤ p − 1. Then each of the elements x₀, αx1, α²x₂, . . . , α^p−1x_p−1also lies in L. Hence, if x₁= 1, then α and each xj(0 ≤ j ≤ p − 1) lies in L.

Proof. For any m with 0 ≤ m ≤ p − 1, consider the sum

X₀+ ζ^−mX₁+ ζ^−2mX₂+ · · · + ζ^−(p−1)mX_p−1

Since 1 + ζ + ζ²+ · · · + ζ^p−1= 0, all terms vanish except for those in which the power of ζ is zero. These terms sum to pα^mx_m. Therefore pα^mx_m∈ L, so α^mx_m∈ L.

If x1= 1 then the case m = 1 shows that α ∈ L, so now x_m∈ L for all m with 0 ≤ m ≤ p − 1.

We can also prove:

Lemma 8.19. With the above notation, for a given x ∈ R, there exist β ∈ M(α) and b∈ M with b = β^p, such that b is not the pth power of an element of M, and

x= y₀+ β + y2β²+ · · · y_p−1β^p−1 where the yj∈ M.

Proof. We know that x 6∈ M, so in (8.7) some x_s6= 0 for 1 ≤ s ≤ p − 1. Let β = xsα^s, and let b = β^p. Then b = x^p_sα^sp= x_s^pa^s, and if b is a pth power of an element of M then a^sis a pth power of an element of M, contrary to Lemma 8.17(2). Therefore b is not the pth power of an element of M.

Now s is prime to p, and the additive group Zpis cyclic of prime order p, so s generates Zp. Therefore, up to multiplication by nonzero elements of M, the powers β^jof β run through the powers of α precisely once as j runs from 0 to p − 1. Since β⁰= 1, β¹= x_sα^s, we have

x= y₀+ β + y2β²+ · · · + y_p−1β^p−1 for suitable y_j∈ M, where in fact y₀= x₀.

Lemma 8.20. Let q ∈ L. Then the minimal polynomial of q over K splits into linear

has q as a zero. Symmetry under Snimplies that f_q(t) ∈ K[t]. The minimal polyno-mial m_qof q over K divides f_q, and f_qis a product of linear factors; therefore m_qis the product of some subset of those linear factors.

We are now ready for the climax of Diet Galois:

Proof of Theorem 8.15. We prove the theorem by induction on the height h of R.

If h = 0 then the theorem is obvious.

Suppose that h ≥ 1. Then R = R₁(α) where R1 is a radical extension of K of height h − 1, and α^p∈ R₁, α 6∈ R1, with p prime. Let α^p= a ∈ R₁.

By Lemma 8.19 we may assume without loss of generality that x= x₀+ α + x₂α²+ · · · + x_p−1α^p−1

where the x_j∈ R₁. (Replace α by β as in the lemma, and then change notation back to α.) The mimimum polynomial m(t) of x over K splits into linear factors in L by Lemma 8.20. In particular, x is a zero of m(t), while all zeros of m(t) lie in L.

Take the equation m(α) = 0, write x as above in terms of powers of α with coefficients in R₁, and consider the result as an equation satisfied by α. The equation has the form f (α) = 0 where f (t) ∈ R1[t]. Therefore f (t) is divisible by the minimal polynomial of α, which is t^p− a. Hence all the roots of that equation, namely ζ^rα for 0 ≤ r ≤ p − 1, are also roots of f (t). Therefore all the elements X_r in (8.8) are roots of m(t), so they lie in L. Lemma 8.18 now shows that α, x0, x₂, . . . x_p−1∈ L.

Also, α^p, x₀, x₂, . . . x_p−1∈ R₁. The height of R₁is h − 1, so by induction, each of these elements lies in some radical extension of K that is contained in L. The subfield Jgenerated by all of these radical extensions is clearly radical (Exercise 8.12), and contains α^p, x₀, x₂, . . . x_p−1. Then x ∈ J(α) ⊆ L, and J(α) is radical. This completes the induction step, and with it, the proof.

So much for the general quintic. We have used virtually everything that led up to Galois theory, but instead of thinking of a group of automorphisms of a field exten-sion, we have used a group of permutations of the roots of a polynomial. Indeed, we have used only the group Sn, which permutes the roots t_jof the general polynomial F(t). It would be possible to stop here, with a splendid application of group theory to the insolubility of the ‘general’ quintic. But for Galois, and for us, there is much more to do. The general quintic is not general enough, and it would be nice to find out why the various tricks used above actually work. At the moment, they seem to be fortunate accidents. In fact, they conceal an elegant theory (which, in particular, makes the Theorem on Natural Irrationalities entirely obvious; so much so that we can ignore it altogether). That theory is, of course, Galois theory. Now motivated up to the hilt, we can start to develop it in earnest.

Exercises 125

EXERCISES

8.1 Prove that in Section 8.2, the permutations R and S of equations (8.1, 8.2) pre-serve every valid polynomial equation over Q relating α, β , γ, and δ . (Hint:

The permutation R has the same effect as complex conjugation. For the permu-tation S, observe that any polynomial equation in α, β , γ, δ can be expressed as

pγ + qδ = 0 where p, q ∈ Q(i). Substitute γ =√

5, δ = −√

5 to derive a condition on p and q. Show that this condition also implies that the equation holds if we change the values so that γ = −√

5, δ =√ 5.) 8.2 Show that the only subfields of Q(i,√

5) are Q, Q(i), Q(√

5), Q(i√ 5), and Q(i,

√5).

8.3 Express the following in terms of elementary symmetric polynomials of α , β , γ .

(a) α²+ β²+ γ² (b) α³+ β³+ γ³

(c) α²β + α²γ + β²α + β²γ + γ²α + γ²β (d) (α − β )²+ (β − γ)²+ (γ − α)²

8.4 Prove that every symmetric polynomial p(x, y) ∈ Q[x, y] can be written as a polynomial in xy and x + y, as follows. If p contains a term axⁱy^j, with i 6= j ∈ N

Hence show that p is a sum of terms that are polynomials in x + y, xy.

8.5* This exercise generalises Exercise 8.3 to n variables. Suppose that p(t₁, . . .,t_n) ∈ K[t₁, . . .,t_n] is symmetric and let the s_i be the elementary sym-metric polynomials in the t_j. Define the rank of a monomial t₁^a1t₂^a2. . .t_n^an to be a1+ 2a2+ · · · na_n. Define the rank of p to be the maximum of the ranks of all monomials that occur in p, and let its part of highest rank be the sum of the terms whose ranks attain this maximum value. Find a polynomial q composed of terms of the form ks^b₁¹s^b₂². . . s^b_nⁿ, where k ∈ K, such that the part of q of high-est rank equals that of p. Observe that p − q has smaller rank than p, and use induction on the rank to prove that p is a polynomial in the s_i.

8.6 Suppose that f (t) = a_ntⁿ+ · · · + a₀∈ K[t], and suppose that in some subfield Lof C such that K ⊂ L we can factorise f as

f(t) = a_n(t − α1) . . . (t − αn) Define

λ_j= α₁^j+ · · · + α_n^j Prove Newton’s identities

a_n−1+ a_nλ1 = 0 2a_n−2+ a_n−1λ₁+ a_nλ₂ = 0

. . . na0+ a₁λ1+ · · · + a_n−1λ_n−1+ a_nλ_n = 0

. . .

a0λ_k+ a₁λk+1+ · · · + a_n−1λk+n−1+ a_nλk+n = 0 (k ≥ 1) Show how to use these identities inductively to obtain formulas for the λj. 8.7 Prove that the alternating group Anis generated by 3-cycles.

8.8 Prove that every element of A5is the product of two 5-cycles. Deduce that A5

is simple.

8.9 Solve the general quadratic by Ruffini radicals. (Hint: If the roots are α1, α2, show that α1− α2is a Ruffini radical.)

8.10 Solve the general cubic by Ruffini radicals. (Hint: If the roots are α1, α₂, α₃, show that α1+ ωα₂+ ω²α₃and α1+ ω²α₂+ ωα₃are Ruffini radicals.) 8.11 Suppose that I ⊆ J are subfields of C(t1, . . . ,t_n) (that is, subsets closed under

the operations +, −, ×, ÷), and J is generated by J₁, . . . , J_r where I ⊆ J_j⊆ J for each j and J_j: I is radical. By induction on r, prove that J : I is radical.

8.12 Mark the following true or false.

(a) The K-automorphisms of a field extension L : K form a subfield of C.

(b) The K-automorphisms of a field extension L : K form a group.

(c) The fixed field of the Galois group of any finite extension L : K contains K.

(d) The fixed field of the Galois group of any finite extension L : K equals K.

(e) The alternating group A5has a normal subgroup H with quotient isomor-phic to Z5.

(f) The alternating group A5has a normal subgroup H with quotient isomor-phic to Z3.

Exercises 127 (g) The alternating group A5has a normal subgroup H with quotient

isomor-phic to Z2.

(h) The general quintic equation can be solved using radicals, but it cannot be solved using Ruffini radicals.

Chapter 9