MATH2001 Development of Mathematical Ideas
History of Solving Polynomial
Equations
26 April 2012
In Disquisitiones Arithmeticae (1801), Gauss found the first instance
of an important class of equations of arbitrarily high degree that are solvable
by radicals.
Note that xn − 1 can be factored into
xn − 1 = (x − 1)(xn−1 + xn−2 + · · · + x + 1).
Gauss showed that if n is prime, then the second factor is irreducible
over the rational numbers and has n − 1 distinct complex roots.
Gauss also proved that if n − 1 is a product of factors αβγ · · · , then
the cyclotomic equation, xn − 1 = 0 can be solved by solving equations
of degrees α, β, γ . . . . For instance, if n = 17, we have n − 1 = 24, so
x17 − 1 = 0 can be solved by solving four quadratic equations. This implies
that the 17-gon can be constructed with ruler and compass.
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In 1824, the Norwegian mathematician Niels Henrik Abel (1802-1829)
gave a proof that the general equation of degree five was not solvable by
radicals.
Abel’s Theorem (1824). The generic algebraic equation of degree
higher than four is not solvable by radicals, i.e., formulae do not exist
for expressing roots of a generic equation of degree higher than four in
terms of its coefficients by means of operations of addition, subtraction,
multiplication,division, raising to a natural power, and extraction of a
root of natural degree.
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Abel’s Proof
Abel’s idea was that if some finite sequence of rational operations and
root extractions applied to the coefficients produces a root of the equation
x5 − ax4 + bx3 − cx2 + dx − e = 0,
the final result must be expressible in the form
1
m
2
m
x = p + R + p2R + · · · + pm−1R
m−1
m
,
where p, p2, . . . , pm−1, and R are also formed by rational operations
and root extractions applied to the coefficients, m is a prime number,
and R1/m is not expressible as a rational function of the coefficients
a, b, c, d, e, p, p2, . . . , pm−1.
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By straightforward reasoning on a system of linear equations for the
coefficients pj , he was able to show that R is a symmetric function of the
roots, and hence as the roots are permuted, R1/m must assume exactly m
different values, R1/m, αR1/m, α2R1/m, ..., αm−1R1/m.
Moreover, since there are 5! = 120 permutations of the roots and m
must be a (prime) divisor of 120, it follows that m = 2 or m = 5 (the case
m = 3 having been ruled out by Cauchy).
The hypothesis that m = 5 led to certain equation in which the left-hand
side assumed only five values while the right-hand side assumed 120 values
as the roots were permuted.
Then the hypothesis m = 2 led to a similar equation in which one side
assumed 120 values and the other only 10.
Abel concluded that the hypothesis that there exists an algorithm for
solving the equation was incorrect.
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On January 17, 1831, Évariste Galois (1811 − 1832) presented his
Mèmoire sur les conditions de résolubilité des équations par radicaux
(Memoir on the conditions for solvability of equations by radicals) to the
French Academy of Science in Paris.
In less than 12 pages in the first part of the mémoire, entitled Principes,
Galois described his ideas to the then radical new theory that bears his
name today.
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After reviewing the manuscript, S. D. Poisson (1781 − 1840) and S. F.
Lacroix (1765 − 1843) wrote in their report:
Quoi qu’il en soit, nous avons fait tous nos efforts pour comprendre la
démonstration de M. Galois. Les raisonnements ne sont ni assez clairs
ni assez développés pour que nous ayons pu juger de leur exactidude. . .
(We have done everything to understand the proofs of M. Galois. His
reasoning is neither clear nor developed enough to judge its exactness...)
The mémoire was rejected. Galois had already submitted his mémoire
before, in 1829 and 1830, and learned of the recent rejection at the end
of 1831. Just months later, on May 31, 1832, Galois died as a result of a
mysterious duel.
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The duel took place on May 30th, early in the morning, and Galois died
the day after with his brother beside him.
All his papers were taken care of by his friend Chevalier, and it was
Liouville who took the charge to read through Galois’ manuscripts.
The first time Galois’ works were published was in 1846 in Journal de
Mathématiques pures et appliqués de Liouville, which contains the original
exposition of Galois’ theory. Almost all of Galois’ mathematical writings
(with English translation) can be found in the following book:
Peter M. Neumann, The mathematical writings of Evariste Galois,
European Mathematical Society, 2011.
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In order to understand the work of Galois, we need to consider different
subfields of the complex numbers C other than the rational number field Q
and the real number field R.
Recall that a subfield K of C is a subset of C that is closed under
addition and multiplication, so that if α ∈ K, then so is −α. Moreover, if
α ̸= 0, then 1/α ∈ K.
According to Galois, it is important to specify not just the equation
under consideration, but also the field of numbers that one considers as
known, and over which the equation is irreducible.
Galois called the elements of the ground field K ”rationnelles” (rational)
or ”connues” (known).
To enlarge the field of numbers over which we are considering the
equation, we need to adjoining some new numbers to the equation.
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Galois then explained how to ”adjoindre á l’équation” (adjoin to the
equation): for any numerical value or abstract quantity t, K(t) consists of
all algebraic expressions in K and t , using addition, multiplication, and
division.
Galois emphasized the importance of predetermining which values are
considered known when attempting to solve a given equation.
[Galois: One sees that the properties and difficulties of an equation can be
completely different depending on the adjoined quantities. For example,
an irreducible equation can become reducible.]
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2
Example
1.
The
equation
x
− 2 is irreducible over Q. However, if we
√
2
2
adjoin
2
to
the
equation
x
−
2
=
0,
then
x
− 2 becomes reducible over
√
√
b ∈ Q} and x2 − 2 decomposes into linear factors
Q( 2) = {a + b 2|a, √
over the larger field Q( 2),
√
√
2
x − 2 = (x − 2)(x + 2).
√
Thus Q( 2) has the property that it is the smallest subfield of C
which contains all the zeros of the equation x2 − 2 = 0. Such a field is
called a splitting field of the polynomial, since over this field, it splits into
a product of linear factors.
In general, if f (x) is a polynomial with coefficients in K ⊂ C, and the
zeros of f are a, b, c, . . ., then a splitting field for f is the smallest subfield
of C that contains all the zeros a, b, c, . . . and is denoted by K(a, b, c, . . .).
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One can show that every element of K(a, b, c, . . .) is a rational
combination of the zeros a, b, c, . . . with coefficients in K, where a rational
combination is formed from elements of K and the a, b, c, . . . using sums,
products, additive inversions and multiplicative inversions (except divisions
by zero). It can be proved that every polynomial equation f (x) = 0 has a
splitting field, something that Galois seems to have taken for granted.
Example 2. The equation x3 − 1 = 0 is reducible over Q as
x3 − 1 = (x − 1)(x2 + x + 1).
If we adjoin ω to the equation x3 − 1 = 0, where ω is a cube root of unity
and ω ̸= 1, Then x3 − 1 = (x − 1)(x − ω)(x − ω 2)). Since ω 2 + ω + 1 = 0,
the splitting field for the equation x3 − 1 = 0 is the field
Q(ω) = {a + bω | a, b ∈ Q}.
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3
Example 3. Consider
the
equation
x
− 2 = 0 which is irreducible over Q.
√
If we now adjoin 3 2 to the equation, then we can factor it as
√
√
√
3
3
3
2
3
x − 2 = (x − 2)(x + 2x + 22).
So x3 − 2 becomes reducible over
√
√
√
3
3
3
Q( 2) = {a + b 2 + c 22|a, b, c ∈ Q}.
√
√
3
3
2
One can check that x + √
2x + 22 has imaginary roots and is therefore
irreducible over the field Q( 3 2), which contains only real numbers.
√
√
√
√
3 2
3
3
2
2 3
Since the zeros of x + 2x + 2 are ω 2 and ω 2, where ω is a
cube root of unity and ω ̸= 1, one can show that the splitting field for the
equation x3 − 2 = 0 is the field
√
√
√
√
√
3 2
3
3
3
3
Q( 2, ω) = {a + b 2 + c 2 + dω + eω 2 + f ω 22 | a, b, c, d, e, f ∈ Q}.
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Meaning II:
Suppose we can solve the equation xn = c (i.e. taking roots), then try to
express the the roots of a degree n polynomial using only the usual algebraic
operations (addition, subtraction, multiplication, division) and application
of taking roots.
Let K be a subfield of C and f (x) ∈ K[x] be a polynomial of degree n
with zeros a1, . . . , an. Let E = K(a1, . . . , an) be the splitting field of f (x).
Then the polynomial f is solvable by radicals if there exists a chain of
fields
K = K0 ⊂ K1 ⊂ · · · ⊂ KN −1 ⊂ KN = E,
where Km+1 is obtained from Km by adjoining a zero of the polynomial
X pm − Am ∈ Km[X] for m = 0, ..., N − 1. Here, the pm can be taken to
be primes .
13
Let K be a subfield of C and f (x) ∈ K[x] be a polynomial of degree n
with zeros a1, . . . , an. Let E = K(a1, . . . , an) be the splitting field of f (x).
Consider a linear polynomial
T = Au + Bv + Cw + · · ·
in n variables u, v, w, . . . with integer coefficients A, B, C, . . ..
In the splitting field E, the polynomial f (x) has n zeros, and there are
n! ways of substituting those zeros for the variables u, v, w, . . ..
For each such substitution we obtain an element of E. In general,
these n! elements may not be distinct. However, Galois showed that if the
zeros a1, . . . , an are all distinct, then one can make a choice of coefficients
A, B, C, · · · such that T takes on n! distinct values and in this case we call
T a Galois resolvent of f (x) = 0 over K.
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Galois then proved that if t is one of these n! distinct values in E,
then the field E = K(a, b, c, . . .) is of the form K(t), so that every root
of f (x) = 0 can be expressed as a rational function of t, that is, as the
quotient of two polynomials in t with coefficients in K.
Now suppose that t1, t2, . . . , tn! are all the distinct values of T .
Then one can show that after multiplying out the product
F (X) = (X − t1)(X − t2) · · · (X − tn!),
the polynomial in F (X) has coefficients lie in K.
Now decompose F (X) into factors that are irreducible over K,
F (X) = G1(X) · · · Gs(X).
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The zeros of each Gi are a subset of the zeros t1, . . . , tn! of F (X).
Galois showed that if t is any zero of any of the Gi, then the splitting
field E for f (x) is equal to K(t), i.e.,
K(a1, . . . , an) = K(t).
Nowadays, this is called the Primitive Element Theorem.
Try to solve Exercise 5.20,5.21,5.22.
Lagrange vs Galois
Recall that given a proposed equation, Lagrange wanted to find an
associated reduced equation whose zeros could be expressed as linear
polynomials in the zeros of the proposed equation, with roots of unity as
coefficients, so that in turn, the zeros of the proposed equation were rational
functions of the zeros of this reduced equation.
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Galois first produced a candidate t for the zero of an as yet undetermined
reduced equation, namely the value of a Galois resolvent T under any
substitution of the zeros of f (x).
Since t is an element of the splitting field E = K(a1, . . . , an), it can be
written as a rational combination of the zeros a1, . . . , an, with coefficients
in K.
On the other hand, since E = K(t), we know the zeros a1, . . . , an are
rational functions of t even though we may not know the explicit form of
these rational functions.
To get a reduced equation, we can simply take any of the Gi above.
Thus, Galois has shown that this part of Lagrange’s program can be
carried out in principle, even though he did not indicate a procedure that
one could use for a specific polynomial f (x).
17
Galois then associated to the equation f (x) = 0 a group of
permutations of the zeros a1, . . . , an.
Let G(X) be an irreducible factor of degree r of the polynomial F (X)
above, and let t1, t2, . . . , tr be all of its zeros.
Then
E = K(a1, . . . , an) = K(ti),
for any zero ti.
Suppose that f (x) has n distinct zeros, which we will label a1, . . . , an.
Then each zero can be expressed as a rational function of t, that is, there
exist rational functions h1(X), . . . , hn(X) such that
ai = hi(t),
for i = 1, . . . , n.
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The same holds true if we replace t by any other of the tj . Therefore, for
each j = 1, . . . , r, the elements h1(tj ), ..., hn(tj ) are the roots a1, a2, . . . , an
of f (x) in some order.
Therefore, we obtain a permutation σj of the zeros of f (x) by setting
σj (a1) = h1(tj ), σj (a2) = h2(tj ), . . . , σj (an) = hn(tj )
for every j = 1, . . . , r.
The Galois group G(E/K) of f (x) = 0 over K is the group (under
composition) of the r permutations obtained in this way.
One now needs to make sure that this Galois group G(E/K) is
independent of the choices of the irreducible factor G and its root t
and this is in fact the case.
19
If the field K is extended to a larger field K ′, and one views the
equation f (x) = 0 as defined over K ′, then the Galois group G(E/K ′) of
the equation over K ′ is expected to be a smaller group (why ?).
It is because F may split into more irreducible factors over a larger field
K ′. In fact, Galois showed that G(E/K ′) is a subgroup of the G(E/K).
We shall see that the subgroup structure of G(E/K) can tell us whether
f can be solvable by radicals or not.
20
In fact, Galois discovered that there is a one-to-one correspondence
between subgroups of G(E/K) and subfields M of E such that K ⊂ M ⊂
E.
The Fundamental Theorem of Galois theory:
The intermediate fields K ′, that is, the fields K ′ such that
K ⊂ K ′ ⊂ E = K(a, b, . . .),
are in one-to-one correspondence with the subgroups H of the Galois group
G = G(E/K).
21
E = K(a, b, . . .)
∪
..
{id}
∩
..
∪
∩
K′
1:1
←→
H = G(E/K ′)
∪
..
∩
..
∪
∩
K
G = G(E/K)
22
Recall that a polynomial f is solvable by radicals if there exists a chain
of fields
K = K0 ⊂ K1 ⊂ · · · ⊂ KN −1 ⊂ KN = E,
where Km+1 is obtained from Km by adjoining a zero of the polynomial
X pm − Am ∈ Km[X] for m = 0, ..., N − 1. Here, the pm can be taken to
be primes .
Note that the adjunction of a radical, i.e., a root of an equation
X pm − Am leads to a reduction of the Galois group !.
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It follows that an equation f (x) = 0 is solvable by radicals if and only if
there exists a sequence of subgroups
G ⊃ H1 ⊃ H2 ⊃ · · · ⊃ HN = {id},
such that every Hk is a normal subgroup of the preceding Hk−1 or G, while
all indices are prime. If this is the case we say that the group G is solvable.
Galois then showed the Galois group of the general quintic equation is
not solvable, hence this equation cannot be solved by radicals. Thus, Abel’s
result follows from the theory of Galois. Nowadays, one shows that there
exists some polynomial whose Galois group is S5 which is not solvable.
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Papers on history of Galois theory
B. M. Kiernan, The development of Galois theory from Lagrange to
Artin. Arch. History Exact Sci. 8 (1971), no. 1-2, 40–154.
I. Radloff, Évariste Galois: principles and applications. Historia Math.
29 (2002), no. 2, 114–137.
Books on Galois theory which emphasis on the historical background.
H.M. Edwards, Galois theory. Graduate Texts in Mathematics, 101.
Springer-Verlag, New York, 1984.
J. Bewersdorff, Galois theory for beginners. A historical perspective.
Translated from the second German (2004) edition by David Kramer.
Student Mathematical Library, 35.
American Mathematical Society,
Providence, RI, 2006.
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