Download A short elementary proof of the Ruffini

A SHORT ELEMENTARY PROOF OF THE RUFFINI-ABEL THEOREM
arXiv:1508.03317v1 [math.HO] 13 Aug 2015
A. Skopenkov
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Abstract. We present a short elementary proof of the well-known Ruffini-Abel theorem
on unsolvability of equations in radicals. This proof is obtained from existing expositions by
stripping it out of material not required for the proof (but presumably required elsewhere). In
particular, we do not use the terms ‘Galois group’ and even ‘group’. However, our presentation is
a good way to learn (or recall) the starting idea of Galois theory: to look at how the symmetry
of a polynomial is decreased when a radical is extracted. So the note provides a bridge (by
showing that there is no gap) between elementary mathematics and Galois theory.
The note is accessible to students familiar with polynomials, complex numbers and permutations; so the note should be interesting easy reading for professional mathematicians (not
necessarily specializing in algebra).
Contents
Statement . . . . . . . . . . . . . . . . . .
Plan of the proof . . . . . . . . . . . . . .
Proof of the Ruffini Theorem (Theorem 3)
Proof of the Abel Theorem (Theorem 4) .
Historical remarks . . . . . . . . . . . . .
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Statement
Motivations for the main definition. We may represent solution of a quadratic equation x2 +
px + q = 0 by a sequence of formulas
f12 = p2 − 4q,
x = (f1 − p)/2.
We may represent solution of a cubic equation x3 + px + q = 0 by a sequence of formulas
p 3 q 2
q
q
2
f1 =
+
, f23 = − + f1 , f33 = − − f1 , x = f2 + f3 .
3
2
2
2
(See comments on the extraction of complex roots in [FT, §5.2].)
We say that the general polynomial equation of degree n is solvable in radicals if there exist
• non-negative integers s, k1, . . . , ks and
• polynomials p0 , p1 , . . . , ps with complex coefficients and in n, n + 1, . . . , n + s variables,
respectively,
such that if a0 , . . . , an−1 , x ∈ C and
xn + an−1 xn−1 + . . . + a1 x + a0 = 0,
then there are f1 , . . . , fs ∈ C for which
f1k1 = p0 (a0 , . . . , an−1 ),
f2k2 = p1 (a0 , . . . , an−1 , f1 ),
1
...
This text is based on the author’s 2015 lectures at summer school ‘Modern Mathematics’, Summer Conference of Tournament of Towns, Kirov region summer school, Moscow ‘olympic’ school, circle ‘Olympiades
and Mathematics’ (MCCME). I am grateful to I. Bogdanov, G. Chelnokov, F. Petrov and V. Volkov for useful
discussions, and to A.B. Sossinsky for correcting English in section ‘Historical remarks’.
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Supported in part by the D. Zimin Dynasty Foundation.
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fsks = ps−1 (a0 , . . . , an−1 , f1 , . . . , fs−1),
...
x = ps (a0 , . . . , an−1 , f1 , . . . , fs ).
Remarks. What we have defined is a property of a number n (not of a specific equation
with given coefficients like in Galois Theorem).
In this definition xn + an−1 xn−1 + . . . + a1 x + a0 = 0 is the equation we are solving and
f1 , . . . , fs are radicals we extracting during the solution.
Example. The general polynomial equation of degree 2 is solvable in radicals: take
s = 1,
k1 = 2,
p0 (y0 , y1 ) = y12 − 4y0,
p1 (y0 , y1 , z1 ) =
z1 − y1
2
and f1 = 2x + a1 ;
check that f12 = p0 (a0 , a1 ) and x = p1 (a0 , a1 , f1 ).
Theorem 1 (Ruffini-Abel). For each n ≥ 5 the general polynomial equation of degree n is not
solvable in radicals. 3
Plan of the proof
The idea is first to prove a weaker insolvability result (the Vieta Lemma 2 and Ruffini Theorem
3 below) and then to deduce from it genuine insolvability (the Abel Theorem 4 below).
Denote the elementary symmetric polynomials in variables x1 , . . . , xn by
σ1 (x1 , . . . , xn ) := x1 + . . . + xn ,
...,
σn (x1 , . . . , xn ) = x1 · . . . · xn .
A radical n-formula is a collection of
• non-negative integers s, k1, . . . , ks ,
• polynomials p0 , p1 , . . . , ps with complex coefficients and in n, n + 1, . . . , n + s variables,
respectively, and
• functions f1 , . . . , fs : Cn → C (which are not assumed to be continuous)
such that
f1k1 = p0 (σ1 , . . . , σn ), f2k2 = p1 (σ1 , . . . , σn , f1 ), . . .
...
fsks = ps−1 (σ1 , . . . , σn , f1 , . . . , fs−1 ),
x1 = ps (σ1 , . . . , σn , f1 , . . . , fs ).
In this formula we omit the argument (x1 , . . . , xn ) of polynomials σ1 , . . . , σn and functions
f1 , . . . , fs ; the equalities are equalities of functions.
Example. There is a radical 2-formula: take
s = 1,
k1 = 2,
f1 (x1 , x2 ) = x1 − x2 ,
p0 (y1 , y2 ) = y12 − 4y2
and p1 (y1 , y2, z1 ) =
y1 + z1
;
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check that f12 (x1 , x2 ) = p0 (x1 + x2 , x1 x2 ) and x1 = p1 (x1 + x2 , x1 x2 , f1 (x1 , x2 )).
Lemma 2 (Vieta Lemma). If the general polynomial equation of degree n is solvable in radicals,
then there is a radical n-formula. 4
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In the definition we can replace the polynomials p0 , p1 , . . . , ps by rational functions (i.e., ratios of polynomials), and assume that for each j = 0, 1, . . . , s the value pj (a0 , . . . , an−1 , f1 , . . . , fj ) is defined whenever used in
the formulas. The obtained analogue of the Ruffini-Abel Theorem is true, see comments to the proof below.
4
Clearly, the converse statements of Lemma 2 and Theorem 4 below hold.
These results are named after Vieta and Abel only for clarity of exposition. It is unlikely that Vieta stated
Lemma 2, but he was able to prove it. It is likely that Abel proved Theorem 4, but I am not sure about that.
2
Proof. Take integers s, k1 , . . . , ks and polynomials p0 , . . . , ps from the definition of the solvability
in radicals of the general polynomial equation of degree n. Take any x1 , . . . , xn ∈ C. Let aj :=
(−1)n−j σn−j (x1 , . . . , xn ) and x := x1 . Then by Vieta Theorem xn +an−1 xn−1 +. . .+a1 x+a0 = 0.
So we can choose any f1 , . . . , fs ∈ C from the definition and name them the values of functions
f1 , . . . , fs : Cn → C at (x1 , . . . , xn ).
Clearly, s, k1 , . . . , ks , p0 , . . . , ps and f1 , . . . , fs form a radical n-formula.
A polynomial (or Ruffini) radical n-formula is defined in the same way as a radical n-formula,
except that ‘functions f1 , . . . , fs : Cn → C (which are not assumed to be continuous)’ is replaced
by ‘polynomials f1 , . . . , fs with complex coefficients’; and ‘equalities of functions’ is replaced by
‘equalities of polynomials’.
E.g. the previous example of a radical 2-formula is a radical polynomial 2-formula.
The Ruffini-Abel Theorem (Theorem 1) is implied by the Vieta Lemma 2 and the following
two results.
Theorem 3 (Ruffini). For each n ≥ 5 there is no polynomial radical n-formula.
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Theorem 4 (Abel). If there is a radical n-formula, then there is a polynomial radical n-formula.
Denote
εq := cos
2π
2π
+ i sin .
q
q
Proof of the Ruffini Theorem (Theorem 3)
A permutation α is even if it is a composition of an even number of transpositions. Denote by
An the set of all even permutations of n-element set.
In this section equality signs involving f mean equality of polynomials. For a permutation
α denote
x := (x1 , . . . , xn ) and xα := (xα(1) , . . . , xα(n) ).
A polynomial f ∈ C[x1 , . . . , xn ] is even-symmetric if f (x) = f (xα ) for each even permutation α.
Lemma 5 (Keeping Symmetry Through Extraction of Radical). If n ≥ 5 and f is a polynomial
such that f q is even-symmetric for some positive integer q, then f is even-symmetric.
Proof of the Ruffini Theorem modulo Lemma 5. By induction on k we obtain that fk is evensymmetric for each k = 1, . . . , s. So x1 ∈ C[x1 , . . . , xn ] is even-symmetric, which is a contradiction because even permutation (13)(12) = (123) carries x1 to x2 6= x1 .
Proof of Lemma 5. We may assume that q is a prime and f 6= 0. Take an even permutation α.
Since
q−1
Y
q
q
(f (x) − εm
f (x) = f (xα ), we have
q f (xα )) = 0.
m=0
Since f is a non-zero polynomial, there is a unique
χ(α) = χf (α) ∈ Zq := {εm
q : m ∈ Z} such that f (x) = χ(α)f (xα ).
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In the definition of the polynomial radical n-formula we can replace polynomials p0 , p1 , . . . , ps by rational
functions and assume that the equalities are equalities of rational functions. The obtained analogue of the
Ruffini Theorem is true with the same proof; the only change is to replace in the proof of Lemma 5 ‘Since f
is a polynomial’ by ‘Since f is a rational function’. Proof of the Abel Theorem in fact requires proof of such a
‘rational’ analogue.
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So we have a map χ : An → Zq . Clearly, this map carries the composition to the product:
χ(αβ) = χ(α)χ(β) for each α, β ∈ An (i.e. χ is a homomorphism, or a character). Now the
Lemma is implied by the following Proposition. 6
Proposition 6. For each integer n ≥ 5 and prime q any map χ : An → Zq carrying the
composition to the product is identically one.
Proof. In this proof i, j, k, l, m are pairwise distinct integers between 1 and n. Any even permutation is a composition of an even number of transpositions. So it suffices to prove that
χ((ij)(kl)) = 1 and χ((ij)(jk)) = 1 for each i, j, k, l. Since (ij)(jk) = (ijk) and (ij)(kl) =
(ijk)(jkl), it suffices to prove that χ((ijk)) = 1 for each i, j, k.
For q 6= 3 this holds because 1 = χ((ijk)3 ) = χ((ijk))3 . For q = 3 we have
1 = χ((ijklm)5 ) = χ((ijklm))5 ⇒ χ((ijklm)) = 1 ⇒ χ((ijk)) = χ((mlkji)(ikjlm)) = 1.
Proof of the Abel Theorem (Theorem 4)
Proof of the Abel Theorem is interesting because it illustrates one of the main ideas of Galois’
work: if an equation is solvable in radicals at all, then it is solvable in radicals using Lagrange
resolutions [S]. See more general remarks in [S].
A formal radical formula is a collection of primes k1 , . . . , ks and rational functions p0 , p1 , . . . , ps ,
pj ∈ C(σ1 , . . . , σn )[y1 , . . . , yj ]. Given a formal radical formula, define
F0 := C(σ1 , . . . , σn ) and Fj :=
Fj−1 [yj ]
k
yj j − pj−1
for each j = 1, . . . , s.
Here 7 and below the omitted argument of pk and p′k is (σ1 , . . . , σn , y1 , . . . , yk ).
For f ∈ C[x1 , . . . , xn ] an f -formula is a formal radical formula for which [f ] = [ps ] ∈ Fs .
If we replace some pj by a rational function representing an equivalent element of Fj , we
obtain an f -formula. So we may assume that pj ∈ Fj for each j.
Lemma 7 (Last Radical). Assume that there is an f -formula. Among f -formulas take an f formula with minimal s. Then there are u ∈ Fs−1 − {0} and an integer l not divisible by ks such
that ysl u ∈ Fs is represented by a polynomial of x1 , . . . , xn .
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A continuous radical n-formula is defined in the same way as a polynomial radical n-formula, except that
‘polynomials f1 , . . . , fs with complex coefficients’ are replaced by ‘continuous functions f1 , . . . , fs : Cn → C’ and
the equalities are equalities of functions. Any continuous radical n-formula is a polynomial radical n-formula,
because a continuous function Cn → C whose power is a rational function has to be a polynomial (since it
is one-valued and defined on Cn ). The analogue of the Ruffini Theorem for continuous radical n-formulas
either follows from this fact or is proved analogously. A continuous function f : Cn → C is even-symmetric if
f (x1 , . . . , xn ) = f (xα(1) , . . . , xα(n) ) for each even permutation α and (x1 , x2 , . . . , xn ) ∈ Cn . In Lemma 5 we can
replace ‘polynomial’ by ‘a function fj from a continuous radical n-formula’; the proof is the same except that
we replace ‘Since f is a polynomial, there is’ by ‘Since f is a function fj from the continuous radical n-formula,
by the following assertion there is’.
Assertion. Let x, y ∈ Cn and fj : Cn → C be a function from the continuous radical n-formula such that
fj (x)fj (y) 6= 0. Then there is a path between x and y avoiding the zeroes of fj (i.e., there is a continuous map
l : [0, 1] → fj−1 (C − {0}) such that l(0) = x and l(1) = y).
k
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Recall the definition of Fj . Two polynomials a, b ∈ Fj−1 [yj ] are congruent modulo yj j − pj−1 if a − b is
k
divisible by yj j − pj−1 . Let Fj be the set of congruence classes. The addition and the multiplication on Fj−1
give an addition and a multiplication on Fj in an obvious way.
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Lemma 8. Let k be a prime, F a field 8 containing εk , and ρ ∈ F not a k-th power of an
F [y]
element from F and F ′ := k
.
y −ρ
(a) The polynomial tk − ρ is irreducible over F .
(b) If u ∈ F ′ − {0}, then there is v ∈ F ′ such that uv = 1 ∈ F ′ .
(c) (Conjugation) If Q ∈ F [t], r = [y] ∈ F ′ and Q(r) = 0, then Q(rεjk ) = 0 for each
j = 1, . . . , k − 1.
Proof of (a). All the roots of the polynomial tk − ρ are r, rεk , rε2k , . . . , rεkk−1. Then the free
coefficient of a factor of tk − ρ is the product of some q of these roots. Then r q ∈ F . For a
proper factor, if it existed, 0 < q < k. Since k is prime, ka + qb = 1 for some integers a, b. Then
r = (r k )a (r q )b = ρa (r q )b ∈ F . A contradiction.
Proof of (b). Since u ∈ F ′ −{0} and ρ is not a k-th power of an element of F , by (a) any element
u0 ∈ F [y] representing u is relatively prime to y k − ρ in F [y]. Therefore there are v0 , w0 ∈ F [y]
such that u0 v0 + (y k − ρ)w0 = 1 ∈ F [y]. Then uv = 1 ∈ F ′ for v = [v0 ] ∈ F ′ .
Proof of (c). Let R(t) be the remainder of division of Q(t) by tk − ρ in F . Since Q(r) = 0 and
r k = ρ, we have R(r) = 0. By (a) and deg R < deg(tk − ρ) we have R = 0. (Alternatively,
r = [y] ∈ F [y]/(y k − ρ) and Q(r) = 0 means that Q(t) is divisible by tk − ρ.) Take any
j = 0, 1, . . . , k − 1. Since (rεjk )k = ρ, we obtain Q(rεjk ) = 0.
Proof of the Abel Theorem modulo the Last Radical Lemma 7. A radical n-formula gives an x1 formula. We may assume that our x1 -formula has minimal s among x1 -formulas. Let ys+1 :=
[x1 ] ∈ Fs . It suffices to prove that
for each j = 1, . . . , s + 1 there is an x1 -formula which has minimal s among x1 -formulas,
and for which yj , . . . , ys+1 ∈ Fs are represented by polynomials of x1 , . . . , xn .
The case j = 1 proves the Abel Theorem (even if p0 , p1 , . . . , ps are rational functions). Let
us prove this assertion by downward induction on j. The base j = s + 1 is trivial.
In order to prove the inductive step suppose that the assertion holds for j replaced by j + 1.
Take u ∈ Fj−1 − {0} and l given by the Last Radical Lemma 7 for s := j. Since k := kj is
a prime, there are integers a and b such that ak + bl = 1. Since s is minimal, pj−2 ∈ Fj−2
is not a k-th power of an element of Fj−2. Thus by Lemma 8.b there is v ∈ Fj−1 such that
uv = 1 ∈ Fj−1 . 9 Informally, we change yj to zj := yjl u, then yj = zjb u−b paj−1 = zjb v b paj−1
and zjk = uk plj−1, so we extend this change to a change of the ‘after j’ part of the x1 -formula.
Formally, define
p′m = pm
for each m = 1, . . . , j − 2,
p′j−1 := uk plj−1
and
p′m := pm (σ1 , . . . , σn , y1 , . . . , yj−1, yjb v b paj−1, yj+1, . . . , ym ) for each m = j, . . . , s.
The obtained x1 -formula satisfies the required properties. The inductive step is proved.
Proof of the Last Radical Lemma 7. Denote k := ks−1 and F := Fs−1 . Then ps ∈ Fs =
F [ys ]/(ysk − ps−1 ) is represented by a polynomial p ∈ F [ys ] of degree less than k.
Since s is minimal, there is l > 0 such that the coefficient u ∈ F of ysl in p is non-zero. Since
k is a prime, there are integers a and b such that ak + bl = 1. Denote z := uysl . Then
ys = ysak+bl = (ysk )a u−b z b = pas−1 u−b z b ∈ Fs ,
where pas−1 u−b ∈ F.
To understand the idea a reader can consider the case F = Q[εk ] := {a0 + a1 εk + a2 ε2k + · · · + ak−2 εkk−2 |
| a0 , . . . , ak−2 ∈ Q}. A reader not familiar with the notion of a field may assume that F = Fs for some f -formula.
9
This explains why we have to work with rational functions p0 , . . . , ps of σ1 , . . . , σn , even if initially we had
polynomials p0 , . . . , ps of σ1 , . . . , σn .
8
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Define q(z) := p(pas−1 u−b z b ). Thus q ∈ F [z] is a polynomial of degree less than k and whose
degree 1 coefficient isQ1.
(z − f (xα(1) , . . . , xα(n) )). The coefficients of R as a polynomial of z are
Define R(z) :=
α∈Sn
symmetric polynomials of x1 , . . . , xn , so R ∈ F [z]. We have f = p(ys ) = q(z) ∈ F [z], so
R(q(z)) = 0. By the Conjugation Lemma 8.c R(q(zεjk )) = 0 for each j = 1, . . . , k − 1. Thus
q(zεjk ) ∈ C[x1 , . . . , xn ]. Since q ∈ F [z] is a polynomial of degree less than k whose degree 1
coefficient is 1, we have that
z=
−2
1−k
k−1
2
q(z) + ε−1
k q(zεk ) + εk q(zεk ) . . . + εk q(zεk )
k
is a polynomial of x1 , . . . , xn .
Historical remarks
Our exposition of the Ruffini Theorem is based on, but different from, [K, L]. Our exposition
of the Abel Theorem is based on, but different from, [PS]. See the exposition containing much
introductory material in [AB].
Our exposition is different from [A, FT, S11] (and is ‘dual’: in our approach the symmetry
group of a polynomial is decreased after the extraction of a radical, while in [A, FT, S11] the
group of permutations of roots generated by paths is increased after the extraction of a radical).
The elementary exposition of [FT, L, PS, K] was more useful to the author (in spite of some
drawbacks, mentioned below) than formal expositions which start with numerous definitions
whose role in the proof of the main theorem is not clear at the moment of their formulations.
Let me explain in more detail the relation between the present exposition and the previous
(available to me) elementary expositions of the Ruffini-Abel Theorem.
I was surprised not to find a rigorous statement of the Ruffini-Abel Theorem in Wikipedia
(English, French, German, Russian, Italian, in summer 2015). The statement in this note
follows [FT, 5.2, 5.3 and Theorem 5.1]. The statement in [A, p.6] is heuristic rather than
rigorous because it is not defined what ‘formulas for expressing roots’ are, i.e. what functions
fj appear in the definition of this note. In that proof [A, 2.10, 2.11], it is assumed that they
are multi-valued functions having the monodromy property. This additional assumption is very
natural, just as Ruffini’s additional assumption. This additional assumption is used in [FT, 5.9,
S11] in sentences like ‘variation of a along a path takes x1 to...’ (not explicitly defined there).
A professional can recover the rigorous statement from the proof, and observe that in [A, FT,
S11] a weaker result than the Ruffini-Abel Theorem is proved.
The proof of the Ruffini Theorem in [K] is hidden by material which is very interesting but
is not part of the proof (so I was not able to understand the proof before I rediscovered it).
In [S, T] a stronger Kronecker’s Theorem is proved; the proof is more complicated.
The exposition of [L] does not claim to be rigorous (and is not).
In [PS] there is some confusion between numbers and variables. There is no ‘equation (5.1)’;
presumably equation in p. 215 is meant. It is not defined what is meant by ‘the coefficients are
considered as independent variables over Q’. It is not defined what is meant by ‘the roots of a
polynomial (5.1) with variable coefficients’, so definition of ∆(F ) is not clear. Presumably ∆ =
Q(σ1 , . . . , σn ) and ∆(F ) = Q(α1 , . . . , αn ) or, in the notation of this note, ∆(F ) = Q(x1 , . . . , xn ).
√
But then it is not defined what is meant by s1 a1 , where a1 ∈ Q(x1 , . . . , xn ). So the main notion
of solvability in radicals is not defined (p. 215). Presumably the definition from the first section
of this note is meant. The solvability in radicals of the polynomial G used in the proof of
Theorem 4 (p. 219) is not defined (and it is defined differently). So instead of Theorem 3 and
6
the first part of the proof of Theorem 4 one needs to use more general results. These results
are not stated. The exact meaning of ‘we are dealing with a general polynomial of degree n’ is
not clear (p. 220, after the formula for ρ1 ).
Hopefully the present note covers all these gaps. It would not have been possible to write it
if the above-mentioned books and articles did not exist.
References
[A] V. Alexeev, Abel’s theorem in problems and solutions, Kluwer A.P., New York, 2004.
http://www.maths.ed.ac.uk/~ aar/papers/abel.pdf
[AB] Solving equations using one radical,
presented by D. Akhtyamov,
I.
Bogdanov,
A. Glebov,
A. Skopenkov,
E. Streltsova and A. Zykin,
http://www.turgor.ru/lktg/2015/4/index.htm.
[FT] D.
Fuchs,
S.
Tabachnikov,
Mathematical
http://www.math.psu.edu/tabachni/Books/taba.pdf.
Omnibus.
AMS,
2007.
[K] V. Kolosov, Theorems and Problems of Algebra, Combinatorics and Number Theory (in
Russian), Gelios, Moscow, 2001.
[L] L. Lerner, Galois Theory without abstract algebra, http://arxiv.org/abs/1108.4593.
[PS] V. Prasolov and Yu. Solovyov, Elliptic Functions and Elliptic Integrals, AMS, 1997. Russian version: www.mccme.ru/~ prasolov.
[S11] A. Skopenkov, A simple proof of the Abel-Ruffini theorem, Mat. Prosveschenie, 15 (2011)
113-126, http://arxiv.org/abs/1102.2100.
[S] A. Skopenkov, Some more proofs from the Book: solvability and insolvability of equations
in radicals, submitted, http://arxiv.org/abs/0804.4357 v6.
[T] V.
Tikhomirov,
Abel
and
his
great
theorem,
http://kvant.mccme.ru/pdf/2003/01/kv0103abel.pdf.
7
Kvant,
2003,
N1.