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t
¦n (X) = (X ’ ζn ).
t=1,...,n’1
gcd(t,n)=1

Proof. We will give a reformulation and proof of this in Theorem 6.2.
Example 1.43. For n = 6 we have

1 3
ζ6 = e2πi/6 = eπi/3 =+ i.
2 2
Then •(6) = 2 and
¦6 (X) = X 2 ’ X + 1 = (X ’ ζ6 )(X ’ ζ6 ).
5


It is also worth recording a related general result on cyclic groups.
Proposition 1.44. Let n 1 and C = g be a cyclic group of order n and a generator g.
Then an element g r ∈ C is a generator if and only if gcd(r, n) = 1; the number of such elements
of C is •(n).
This leads to a useful group theoretic result.
Lemma 1.45. Let G be a ¬nite group satisfying the following condition:
• For each n 1, there are at most n solutions of xn = ι in G.
Then G is cyclic and in particular is abelian.
1.4. FINDING ROOTS OF COMPLEX POLYNOMIALS OF SMALL DEGREE 13


Proof. Let θG (d) denote the number of elements in G of order d. By Lagrange™s
™™¦

Theorem, θG (d) = 0 unless d divides |G|. Since
G= {g ∈ G : |g| = d},
d||G|

we have
|G| = θG (d).
d||G|
Recall the Euler •-function satis¬es Equation (1.5), hence
|G| = •(d).
d||G|

Combining these we obtain
(1.11) θG (d) = •(d).
d||G| d||G|

Let d be a divisor of |G|. By Proposition 1.44, for each element g ∈ G of order d, the cyclic
subgroup g G has •(d) generators, each of order d. As there are at most d such elements g
in G, this gives θG (d) •(d). So
θG (d) •(d).
d||G| d||G|

Now if θG (d) < •(d) for some d, we would have a strict inequality in place of Equation (1.11).
Hence θG (d) = •(d) for all d. In particular, there are •(|G|) elements of order |G|, hence there
must be an element of order |G|, so G is cyclic.
The above results for polynomials over Q and Z have analogues over the ¬eld of fractions
k(T ) and polynomial ring k[T ], where k is a ¬eld.
A polynomial f (X) ∈ k[T ][X] is an element of k(T )[X]. If R = k[T ] or k(T ), we say that
f (X) has a proper factorisation over R if f (X) = g(X)h(X) for some g(X), h(X) ∈ R[X] with
deg g(X) > 0 and deg h(X) > 0.
Proposition 1.46 (Gauss™s Lemma). Let f (X) ∈ k[T ][X]. Then f (X) has a proper fac-
torisation over k[T ] if and only it has a proper factorisation over k(T ).
Proposition 1.47 (Eisenstein Test). Let f (X) ∈ k[T ][X] and s ∈ k. Choose ai ∈ k[T ] so
that
f (X) = a0 + a1 (X ’ s) + · · · + ad’1 (X ’ s)d’1 + ad (X ’ s)d ,
where d = deg f (X). Suppose that p(T ) ∈ k[T ] is an irreducible for which the following condi-
tions hold:
• ak ≡ 0 (mod p(T )) for k = 0, . . . , d ’ 1;
• a0 ≡ 0 (mod p(T )2 );
• ad ≡ 0 (mod p(T )).
Then f (X) is irreducible in k(T )[X] and hence also in k[T ][X].
Example 1.48. Let k be a ¬eld. Then the polynomial X n ’ T is irreducible in k(T )[X].

1.4. Finding roots of complex polynomials of small degree

™™¦

In this section we work within the complex numbers and take k ⊆ C; in practice we will
have k = R or k = C.
For monic linear (degree 1) or quadratic (degree 2) polynomials, methods of ¬nding roots are
very familiar. Let us consider the cases of cubic (degree 3) and quartic (degree 4) polynomials.
14 1. INTEGRAL DOMAINS, FIELDS AND POLYNOMIAL RINGS

Cubic polynomials: Cardan™s method. The following 16th century method of ¬nding
roots of cubics is due to Jerˆme Cardan who seems to have obtained some preliminary versions
o
from Niccol` Tartaglia by somewhat disreputable means! For historical details see [2, 3].
a
A monic cubic
f (X) = X 3 + a2 X 2 + a1 X + a0 ∈ C[X]
can be transformed into one with no quadratic term by a change of variables X ’’ X ’ a2 /3
giving
a1 a2 2a3
123
+2
g(X) = f (X ’ a2 /3) = X ’ a1 ’ a2 X ’ a0 + ∈ C[X].
3 3 27
Clearly ¬nding the roots of f (X) is equivalent to ¬nding those of g(X), so we may as well
assume that we want to ¬nd the complex roots of
f (X) = X 3 + pX + q ∈ C[X].
Suppose that x ∈ C is a root of f (X), i.e.,
x3 + px + q = 0.
(1.12)
Introduce u ∈ C for which
p
x=u’
3u
Then
p p
3
u’ +p u’ +q =0
3u 3u
and so
p3
3
u’ + q = 0,
27u3
hence
p3
6 3
u + qu ’ = 0.
27
Solving for u3 we obtain
4p3
qq
3
q2
u =’ ± + ,
22 27
4p3
q2
where + denotes one of the complex square roots of the discriminant of the quadratic
27
equation
p3 2
U + qU ’ = 0.
27
Now if we take u to be a cube root of one of the complex numbers
4p3
qq 2+
’± q
22 27
we obtain the desired root of f (X) as x = u ’ p/3u. Notice that we have a choice of 2 values
for u3 and for each of these a choice of 3 values for u, di¬ering by factors of the form ω r for
r = 0, 1, 2 where ω = e2πi/3 is a primitive cube root of 1. However, since
4p3
4p3 q2
’q ’ +
q2 +
’q ’ 27
1 27
= = ’27 ,
q2 (q 2 4p3 /27) 4p3
’ +
4p3
q2 +
’q +
27
it is easy to verify that there are in fact only 3 choices of the root x which we can write
symbolically as

4p3 4p3
q1 q1
3 3
q2 q2
(1.13) x= ’+ + + ’’ +
22 27 22 27
1.4. FINDING ROOTS OF COMPLEX POLYNOMIALS OF SMALL DEGREE 15

or more precisely as

4p3
q1 p
3
q2
(1.14) x= ’+ + ’ .
22 27
4p3
q1
3
q2
3 ’+ +
22 27
Example 1.49. Find the complex roots of the polynomial
f (X) = X 3 + 3X ’ 10 ∈ R[X].
Solution. Applying the method above, we reduce to the quadratic equation
U 2 ’ 10U ’ 1 = 0
√ √ √
whose roots are 5 ± 26 ∈ R. Notice that 5 + 26 > 0 and 5 ’ 26 < 0; we also have
√ ’1
√.
5 ’ 26 =
5 + 26

Now 5 + 26 has the complex cube roots
√ √ √
3 3 3
5 + 26 ω 2 .
5 + 26, 5 + 26 ω,
Here we have x = u ’ 1/u, so the 3 complex roots of f (X) are
√ 1
3
ωr
5+ 26 ’ (r = 0, 1, 2).

3
5 + 26
Notice that one of these is real, namely
√ 2
3
5+ 26 ’ 1
√ 1
3
5+ 26 ’ √= .

3 3
5 + 26 5 + 26
Quartic polynomials: Ferrari™s method. The following method of ¬nding roots of quar-
tics was publicised by Cardan who attributed it to his student Lodovicio Ferrari.
A general monic quartic polynomial
f (X) = X 3 + a3 X 3 + a2 X 2 + a1 X + a0 ∈ C[X]
can be transformed into one with no cubic term by a change of variables X ’’ X ’ a2 /3 giving
g(X) = f (X ’ a3 /4) =
3 13 1 1 341
Y 4 + a2 ’ a2 Y 2 + a2 a2 ’
a ’ a2 a3 + a1 Y ’ a + a1 a3 + a0 ∈ C[X].
83 83 2 3
256 3 4
16
Clearly ¬nding the roots of f (X) is equivalent to ¬nding those of g(X), so we may as well
assume that we want to ¬nd the complex roots of
f (X) = X 4 + pX 2 + qX + r ∈ C[X].
Suppose that x is a root and introduce numbers y, z such that z = x2 + y (we will ¬x the
values of these later). Then
z 2 = x4 + 2x2 y + y 2
= ’px2 ’ qx ’ r + 2xy + y 2
= (2y ’ p)x2 ’ qx + y 2 ’ r.
Now choose y to make the last quadratic expression in x a square,
(2y ’ p)x2 ’ qx + (y 2 ’ r) = (Ax + B)2 .
(1.15)
This can be done by requiring the vanishing of the discriminant
q 2 ’ 4(2y ’ p)(y 2 ’ r) = 0.
(1.16)
16 1. INTEGRAL DOMAINS, FIELDS AND POLYNOMIAL RINGS

Notice that if y = p/2 then we would require q = 0 and then
f (X) = X 4 + pX 2 + r = (X 2 )2 + p(X 2 ) + r = 0
can be solved by solving
Z 2 + pZ + r = 0.
Since Equation (1.16) is a cubic in y, we can use the method of solution of cubics to ¬nd a root
y = t say. Then for Equation (1.15) we have
(x2 + t) = (Ax + B)2 ,
whence
x2 = ’t ± (Ax + B).
Thus taking the two square roots of the right hand side we obtain 4 values for x, which we
symbolically write as
x = ± ’t ± (Ax + B).
Remark 1.50. In the case of cubic and quartic polynomials over C we can obtain all the
roots by repeatedly taking square or cube roots (or radicals). Consequently such polynomials are
said to be solvable by radicals. Later we will see that this is not true in general for polynomials
of degree at least 5; this is one of the great early successes of this theory.

1.5. Automorphisms of rings and ¬elds
Definition 1.51. Let R be a ring and R0 ⊆ R a subring.
• An automorphism of R is a ring isomorphism ± : R ’’ R. The set of all such auto-
morphisms is denoted Aut(R).
• An automorphism of R over R0 is a ring isomorphism ± : R ’’ R for which ±(r) = r
whenever r ∈ R0 . The set of all automorphisms of R over R0 is denoted AutR0 (R).
Proposition 1.52. For a ring R with a subring R0 ⊆ R, Aut(R) and AutR0 (R) form groups
under composition of functions.
Proof. The composition ±—¦β of two automorphisms ±, β : R ’’ R is also an automorphism

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