Normat 59:3-4, 117–130 (2011) 117
How to describe all cubic Galois extensions?
Juliusz BrzeziÒski and Ulf Persson.
Mathematical Sciences
Chalmers and Gothenburg University
S–41296 Göteborg, Sweden
jub@chalmers.se, ulfp@chalmers.se
1 Introduction
Every quadratic field extension of the rational numbers is a field Q(
Ô
d),whered is
a square-free integer different from 1. Moreover, the fields corresponding to different
d are different, that is, they are not isomorphic as fields. Of course, the number
Ô
d generating the e xtension is a zero of the polynomial X
2
≠ d.Thiswell-known
description of the quadratic fields over the rational numbers is very satisfactory. Is
it possible to describe in a similar way all cubic extensions of the rational numbers?
Trying to find such a description, we note that there are two types of cubic exten-
sions K of Q. If the degree [K : Q]=3, then the automorphism group G(K/Q)
whose order must divide 3, consists of either 1 or 3 elements. In the context of
Galois theory, it is more interesting to look at cubic Galois extensions, that is,
cubic extensions which are splitting fields of cubic polynomials. This will be our
purpose. We want to describe all cubic Galois extensions, that is, the fields K such
that |G(K/Q)| =[K : Q]=3. Usually, such extensions are called cubic cyclic
fields, since the Galois group G(K/Q) is, of course, cyclic of order 3.
We would like to have a similar description as in the case of quadratic extensions,
which means that we want to construct a list of some simple cubic polynomials in
such a way that different polynomials define different (non-isomorphic) cubic Galois
fields and every such field can be obtained as a splitting field of a cubic polynomial
belonging to our list. Moreover, when a cubic Galois field is given as a splitting
field of a cubic polynomial, we would like to have a possibility to find a polynomial
belonging to our list, which defines this field. Let us notice that there exist different
solutions of this problem and rather vast list of refere nces (see [C], 6.4.2 and the
references), so our purpose is to present the topic in an easily accessible way.
Before we start, let us note that we can not expect an equally simple answer as in
the case of quadratic fields. If we take K = Q(
3
Ô
d),whered is a cube-free integer
different from ±1,thenK is not a Galois extension of Q,sinceX
3
≠ d,whichis
the minimal polynomial of
3
Ô
d has zeros Á
i
3
Ô
d,whereÁ =
≠1+
Ô
≠3
2
is the primitive
3rd root of 1 and i =0, 1, 2. Two of these numbers are non-real and, consequently,
not in the field K. In the next section, we discuss how to construct all cyclic cubic
fields.
118 Juliusz BrzeziÒski and Ulf Persson. Normat 3-4/2011
2 Cyclic cubic fields
AcycliccubicfieldK can be generated over Q by any non-rational number x œ K.
In fact, the field Q(x) is then bigger than Q and must be e qual to K,sincethere
are no subfields between Q and K. The minimal polynomial of x has de gree 3. As
usual, we can choose a generator x such that its minimal polynomial has the form
Ï(X)=X
3
≠ pX + q,wherep, q are integers. Denoting by x
1
,x
2
,x
3
the zeros of
this polynomial, the discriminant of Ï(X) is the number (Ï)=[(x
1
≠ x
2
)(x
2
≠
x
3
)(x
3
≠x
1
)]
2
=4p
3
≠27 q
2
. Computing the discriminant, one can decide whether
the splitting field of Ï(X) is cyclic or not (see the article on cubic and quartic
equations in this volume):
Proposition 2.1. The splitting field of an irreducible polynomial Ï(X)=X
3
≠
pX + q over Q is cyclic if and only if its discriminant (Ï)=4p
3
≠ 27 q
2
is a
rational square.
Assuming that p, q are integers, the discriminant of an irreducible polynomial whose
splitting field is cyclic must be a square of an integer. For such a polynomial the
coefficient p must be positive. One of the the zeros of any cubic polynomial over
Q must be real. Therefore, all zeros are real, since the splitting field is generated
by the real zero. We shall denote by ”(Ï) the s quare root of (Ï) and call it the
reduced discriminant.
Example 1. It is easy to check that the polynomial Ï(X)=X
3
≠ 3X +1 is
irreducible over Q.Wehave(Ï)=9
2
, so the splitting field of Ï(X) is cyclic. Let
x be any zero. Then K = Q(x) is the splitting field. It is not difficult to check that
the remaining two zeros of Ï(X) are x
2
≠ 2 and 2 ≠ x ≠ x
2
.
Unfortunately, it is impossible to give a description of the splitting fields of cyclic
cubics using only the four arithmetical operations and extractions of roots if we
only use the real numbers. This is a consequence of a rather unexpected property
of cubic equations with 3 real zeros – in order to get a formulae for the zeros, it is
necessary to use complex numbers. This phenomenon, known as Casus Irreducibilis,
created many problems for mathematicians in the old ages. However, it suggests
that looking for a classification of cubic cyclic extensions, it may be convenient
to use complex numbers. In fact, it appears that the description of cyclic cubic
extensions s implifies dramatically if we only allow the 3rd root of 1 and extend
the cyclic cubic fields over Q to the cyclic cubic fields over the field of Eisenstein
numbers Q(Á),whereÁ is the primitive 3rd root of 1. In the next section, we
discuss the Eisenstein intege rs in order to apply them in classification of the cubic
cyclic extensions of rational numbers.
3 The Eisenstein integers
The Eisenstein integers, also known as Eulerian integers, are the numbers a + bÁ,
where a, b are integers and Á =
≠1+
Ô
≠3
2
is the primitive 3rd root of 1. These num-
bers form the ring of integers Z[Á ] in the quadratic field E = Q(
Ô
≠3). In general,
Normat 3-4/2011 Juliusz BrzeziÒski and Ulf Persson. 119
the algebraic integers in any algebraic number field are the zeros of polynomials
with integer coefficients and the highest coefficient equal 1. All integers in any al-
gebraic number field form a ring. Probably the best known such ring of integers
bigger than the rational integers Z is the ring of Gaussian integers Z[i] in the fi-
eld Q(i). The Eisenstein integers, like the Gaussian, have unique factorization into
Eisenstein primes. The Eisenstein primes fi are the numbers fi = p ,wherep is a
prime number such that p = 2 (mod 3) or the numbers fi = a + bÁ such that
Nr(a + bÁ)=(a + bÁ)(a + b¯Á)=a
2
≠ ab + b
2
is a prime number (in Z). For example, the numbers 2, 5, 2+Á, 1+3Á are Eisenstein
primes. The primes of the first type, that is, the “old” prime numbers which also
are Eisenstein primes, are called inert.Thenewprimesfi such that Nr(fi)=p =1
(mod 3) will be called split Eisenstein prim es. They split the old primes into
a product of two different prime factors, for example, 7=(1+3Á)(1 + 3¯Á).The
only integer prime, which, up to a sign, is a square of an Eisentstein integer is the
prime 3. We have, 3=≠(
Ô
≠3)
2
= ≠(1 + 2Á)
2
. The units in the ring of Eisenstein
integers, that is, the numbers ÷ œ Z[Á] such that ÷
≠1
is also an Eisenstein integer
are exactly ±1, ±Á, ±Á
2
(Á
2
= ≠1 ≠Á). Notice that only ±1 are cubes of units. The
unusually high number of units in the rings of Gauss and Eisenstein integers makes
these rings very special among the rings of integers in the quadratic number fields
whose integers normally have only two units: ±1. The Gauss integers have four:
±1, ±i, and the Eisenstein integers, as we have seen, six.
The following result, whose proof can be found in several textbooks, for example,
in [IR], p.13, gives the most important property of the Eisenstein integers. It is
usually proved using the fact that the Eisenstein integers form an Euclidean ring
with respect to the usual Euclidean norm (the square of the absolute value):
Theorem 3.1. The ring of Eisenstein integers is a unique factorization domain.
The theorem says that if – œ Z[Á] and – is neither 0 or a unit, then – = fi
1
···fi
k
,
where fi
i
are Eisenstein primes and the factorization is unique up to the order of
the factors and possible modifications of them by units. Recall that a modification
by a unit means that a factor fi is replaced by a factor ÷fi,where÷ is a unit.
We shall often work with cube roots of Eisenstein integers. For convenience of
references, we formulate the following simple fact:
Lemma 3.1. Every non-zero Eisenstein integer – can be represented in the form
– = µ‹
2
–
Õ
3
, where µ‹ is a unit or a product of different Eisenstein p rimes and –
Õ
is an Eisenstein integer.
Proof. If – is a unit, we can take µ = –, ‹ =1,–
Õ
=1.If– is a nonunit, then
we represent it as a product of Eisenstein primes. We group these primes in such a
way that µ is the product of different Eisenstein primes which in – have exponents
giving residue 1 modulo 3, and ‹ of those whose exponents give residue 2 modulo
3. The remaining factors of – give a third power of an integer –
Õ
,sowehave
– = µ‹
2
–
Õ
3
. 2
120 Juliusz BrzeziÒski and Ulf Persson. Normat 3-4/2011
In the representation – = µ‹
2
–
Õ
3
,theproductµ‹
2
is called the cube-free part
of – (it is unique up to a sign).
Now we use the Eisenste in integers in order to describe the cubic extensions of E.
In the next se ction, we will “go down” and use the Eisenstein integers in order to
describe the cubic cyclic extensions of the rational numbers.
Proposition 3.1. (a) For every cubic Galois extension L of E there is – œ E such
that L = E(
3
Ô
–).
(b) Two cubic extensions E(
3
Ô
–) and E(
3
Ô
—) of E are isomorphic if and only if
(the images of ) – and — generate the same subgroup of E
ú
/E
ú3
,thatis,–— or –
2
—
is a cube in E.
(c) Every cubic extension L of E is of the form L = E(
3
µ‹
2
), where µ, ‹ are
Eisenstein integers and µ‹
2
”=1is a unit or µ‹ is a product of different E isenstein
primes.
Proof. (a) This is a special case of a much more general result concerning arbitrary
cyclic Galois extensions of any degree n over fields containing n different nth roots
of 1 (see see e.g. [L], Theorem 6.2).
In order to prove this denote by ‡ a nontrivial automorphism of L over E and take
any x œ L \ E such that the se cond coefficient (by X
2
) of the minimal polynomial
Ï(X) of x over E is 0. The zeros of this polynomial are x, ‡(x),‡
2
(x) and, of
course, L = E(x). Moreover, we have Tr(x)=x + ‡(x)+‡
2
(x)=0. Take y =
x + Á‡(x)+Á
2
‡
2
(x). It is easy to check that ‡(y)=Á
2
y,soy/œ E if only y ”=0.
Then we have ‡(y
3
)=y
3
, that is, y
3
= – œ E.Thus,wehaveL = E(
3
Ô
–).If
y =0, then the equations y =0and Tr(x)=0easily imply (subtract the equation
and divide by 1 ≠ Á), that ‡(x)=Áx, that is, ‡(x
3
)=x
3
and already x
3
= – œ E.
(b) The abelian Kummer theory (see [L], Chap. VI, §8) says that cubic extensions
of E are in a one-to-one correspondence with the cyclic subgroups of E
ú
/E
ú3
,when
to the extension E(
3
Ô
–) corresponds the subgroup ge nerated by the image of – in
E
ú
/E
ú3
. Now – and — define the same subgroup if and only if —E
ú3
= –E
ú3
or
—E
ú3
= –
2
E
ú3
, which is equivalent to –
2
— œ E
ú3
or –— œ E
ú3
.
For convenience of the Reader, we give a direct proof. If –— or –
2
— is a cube in
E, then one checks immediately that E(
3
Ô
–)=E(
3
Ô
—). Let E(
3
Ô
–) and E(
3
Ô
—)
be isomorphic. Then the field E(
3
Ô
–) contains elements x, y such that x
3
= –
and y
3
= —.If‡ is a nontrivial automorphism of L over E,then‡(x)=Á
i
x
and ‡(y)=Á
j
y for i, j =1, 2. Assume that ‡(x)=Á
i
x and ‡(y)=Á
i
y.Then
‡(x/y)=x/y, which means that x/y = fl œ E. Hence – = fl
3
— or, equivalently,
–—
2
is a cube in E. Assume now that ‡(x)=Á
i
x and ‡(y)=Á
j
y,wherei ”= j (that
is, one of the exponents equals 1, the other equals 2). Then ‡(x/y
2
)=x/y
2
,which
means that x/y
2
= fl œ E. Hence – = fl
3
—
2
or, equivalently, –— is a cube in E.
(c) If L = E(
3
Ô
–), then we can always assume that – ”=0is an Eisenstein integer,
since E is the quotient field of Z[Á]. With the notations of Lemma 3.1, we have
L = E(
3
Ô
–)=E(
3
µ‹
2
). 2
Normat 3-4/2011 Juliusz BrzeziÒski and Ulf Persson. 121
4 Going up and down
In this section, we establish a correspondence between the cubic Galois extensions
of the rational numbers and the cubic Galois extensions of the Eisenstein numbers.
It appears that it is not difficult to read o ff the properties of the cubic Galois
extensions of the rational numbers by “moving” them up to the Eisenstein numbers
and then down to the rational numbers.
Proposition 4.1. There is a one-to-one correspondence between cyclic cubic ex-
tensions K ∏ Q and the cyclic cubic extensions L = EK of E such t hat L ∏ Q
is Galois with cyclic Galois group Z
6
. Moreover, two extensions K
1
and K
2
are
isomorphic if and only if the extensions L
1
= EK
1
and L
2
= EK
2
are isomorphic
over E.
Proof. It is clear that if K ∏ Q is a cyclic cubic extension, then EK is a cyclic
Galois extensions of E and Galois extensions of Q with Galois Group Z
2
◊Z
3
= Z
6
.
Moreover, if K
1
and K
2
are isomorphic, then EK
1
and EK
2
are isomorphic over
E.
Conversely, if L is a Galois extension of Q with Galois group Z
6
containing E,
then the fixed field of the only subgroup of Z
6
of order 2 (the identity and the
complex conjugation) is a cyclic cubic extension K of Q.Itisuniquelydefinedby
L.Thus,ifL
1
and L
2
are two isomorphic Galois extensions of Q with Galois group
Z
6
both containing E, then the corresponding cyclic cubic subfields K
1
and K
2
are
isomorphic. 2
Since the extension L = KE of the Eisenstein numbers obtained from a cubic
Galois extension K of the rational numbers is a cyclic Galois extension of Q,it
is important to find c onditions characterizing the typical cubic Galois extensions
L = E(
3
Ô
–) of E which are Galois over Q. Moreover, we want to characterize those,
which have a cyclic Galois group.
Proposition 4.2. If L = E(
3
Ô
–), where – œ E, is a cubic extension of E, then L
is a Galois extension of Q if and only if E(
3
Ô
–)=E(
3
Ô
¯–), which happens if and
only if Nr(–) œ Q
ú3
or – Nr(–) œ Q
ú3
. In the first case the Galois group G(L/Q )
is cyclic, and in the second, it is symmetric.
Proof. The minimal polynomial of
3
Ô
– over Q is Ï(X)=(X
3
≠ –)(X
3
≠ ¯–)=
X
6
≠ Tr(–)X
3
+ Nr(–), since it has rational coefficients and the degree 6 equal
to the degree [L : Q ].ThusifL over Q is Galois, we have
3
Ô
¯– œ L,sinceitisa
zero of the same irreducible polynomial which has
3
Ô
– as its zero. Hence, we have
L = E(
3
Ô
–)=E(
3
Ô
¯–). Conversely, if these equalities hold, then the field L is a
splitting field of the polynomial Ï(X), so it is a Galois extension of Q.
According to Proposition 3.1 (b), the equality E(
3
Ô
–)=E(
3
Ô
¯–) is equivalent to
–¯– = Nr(–) œ E
ú3
or –
2
¯– = – Nr(–) œ E
ú3
. By Proposition 3.1, we have – = µ‹
2
,
where µ‹ is a unit or a product of different Eisenstein primes. It is easy to see that
–¯– = µ‹
2
¯µ¯‹
2
is a cube if and only if ‹ = ÷¯µ, and –
2
¯– = µ
2
‹
4
¯µ¯‹
2
is a cube if and
only if ¯µ = ÷µ and ¯‹ = ±÷‹ (÷ a unit). We see that in the first case Nr(–) œ Q
ú3
,
and in the second, – Nr(–) œ Q
ú3
.
122 Juliusz BrzeziÒski and Ulf Persson. Normat 3-4/2011
Assume that the first case occurs. As a splitting field of Ï(X),thefieldL has 6
automorphisms mapping its zero
3
Ô
– (a fixed value) on one of the 6 others: Á
i
3
Ô
–,
Á
i
3
Ô
¯– for i =0, 1, 2. It is easy to check that the automorphism mapping
3
Ô
– onto
Á
3
Ô
¯– has order 6 (it follows from the ass umption that
3
Ô
–
3
Ô
¯– =
3
Nr(–) is a
rational number). Hence, the Galois group G(L/Q) is cyclic.
Conversely, assume that G(L/Q) is cyclic. We want to prove that Nr(–) is a rational
cube. In fact, we have
3
Ô
–
3
Ô
¯– =
3
Nr(–) œ L. Hence L contains a splitting field of
the polynomial X
3
≠Nr(–) with rational coefficients if this polynomial is irreducible.
But the splitting field of such an irreducible polynomial is of degree 6 over Q and
its Galois group is the symmetric group S
3
. This contradicts our assumption about
G(L/Q). Hence the polynomial X
3
≠ Nr(–) must be reducible, which shows that
Nr(–) is a rational cube.
Thus we have proved that the Galois group G(E(
3
Ô
–)/Q) is cyclic if and only if
Nr(–) is a rational cube. Consequently, it must be symmetric if and only if E(
3
Ô
–)
is Galois over Q and – Nr(–) is a cube in Q, since there are only two non-isomorphic
groups with 6 elements. 2
Now we are ready to prove that every cubic Galois extension of the rational numbers
can be uniquely defined by an Eisenstein integer, which we define below. Notice
that if z is a complex number such that Ÿz ·⁄z ”=0, then exactly one of the four
numbers ±z, ±¯z is in the first quadrant of the complex plane.
Theorem 4.1. For every cubic Galois field over Q there exists a unique Eisenstein
integer in the first quadrant of the complex plane called the invariant of t he field
such that two cubic Galois fields are isomorphic if and only if the corresponding
invariants are equal. More exactly:
(a) If K is a cubic Galois field and L = KE, then there exists a unique f in the
first quadrant of the complex plane such that L = E(
3
f
¯
f
2
) and f = ≠¯Á or f
is a product of different split Eisenstein primes. Two cubic Galois fields over the
rational numbers are isomorphic if a nd only if the corresponding numbers f are
equal.
(b) The field L = Q(
3
f
¯
f
2
) is the splitting field of the sextic polynomial
X
6
≠ Nr(f)Tr(f)X
3
+ Nr(f)
3
. (1)
The cubic subfield K of L is generated by x =
3
f
¯
f
2
+
3
¯
ff
2
, when the second cubic
root is chosen so that th e product of both equals Nr(f).Theminimalpolynomialof
x over Q is
X
3
≠ 3 Nr(f)X ≠ Tr(f ) Nr(f), (2)
and K is its splitting field.
Proof. (a) By Proposition 3.1 (c), we have L = E(
3
µ‹
2
),whereµ‹ is a unit or a
product of differe nt Eisenstein primes. In the first case, we note that L = E(
3
Ô
÷),
Normat 3-4/2011 Juliusz BrzeziÒski and Ulf Persson. 123
for any unit ÷ ”= ±1. The only unit in the first quadrant of the complex plane is
÷ = ≠¯Á, so taking f = ≠¯Á, we get L = E(
3
f
¯
f
2
).
In the second case, by Proposition 4.2, we have Nr(µ‹
2
)=µ‹
2
¯µ¯‹
2
is a cube and
µ‹ is divisible by at least one Eisenstein prime fi. It is easy to see that the product
contains a cube of fi if and only if ¯µ = ÷‹ ,where÷ is a unit. Hence, we have ¯‹ = ÷µ
and it follows that L = E(
3
µ¯µ
2
). Now notice that if an inert prime fi or the prime
fi =
Ô
≠3 divides µ,thenitscubedividesµ¯µ
2
, so it can be removed as a factor of µ.
Hence, we can assume that µ is a product of only split Eisenstein primes. Assume
that both the real and the imaginary parts of µ are non-zero. If the real part of µ
is not positive, we can replace it by ≠µ without changing the field L = E(
3
µ¯µ
2
).
If now µ is not in the upper half plane, we can replace it by ¯µ. Thus if finally µ is
in the first quadrant, we define f = µ and we have L = E(
3
f
¯
f
2
) as required. We
prove that Ÿf ·⁄f ”=0in (b).
Assume now that E(
3
f
¯
f
2
) and E(
3
Ò
f
Õ
¯
f
Õ
2
) are isomorphic. According to Proposi-
tion 3.1 (b), we get that either f
¯
f
2
f
Õ
¯
f
Õ
2
or f
¯
f
2
f
Õ
2
¯
f
Õ
4
is a cube. Similar argument
as above shows that in the first case , we have
¯
f
Õ
= ÷f and in the second, f
Õ
= ÷f
for a unit ÷. Taking this into account, we use again that the products are cubes and
get ÷ = ±1. However, the choice of f, f
Õ
in the first quadrant makes the equalities
¯
f
Õ
= ÷f and f
Õ
= ÷f impossible unless f
Õ
= f.
(b) As we noted in the proof of Prop os ition 4.2, the field L is a splitting field of the
polynomial (X
3
≠f
¯
f
2
)(X
3
≠
¯
ff
2
), which is just (1). The number x is fixed by the
automorphism of order 2 mapping
3
f
¯
f
2
on
3
¯
ff
2
and it is easy to find the cubic
polynomial whose zero is x.Sincex is not rational number, it generates the fixed
field K of the automorphism of order 2. The field K is a cubic Galois extension of
Q, which must be the splitting field of the minimal polynomial of x.
Finally notice that if ⁄f =0,thenL = Q(
3
f
¯
f
2
) is a real field, which is not
the case. If Ÿf =0,thenTr(f)=0and the polynomial (1) is reducible, which is
impossible. Thus, we have Ÿf ·⁄f ”=0. 2
The non-unit invariants of the cubic Galois fields are products of split Eisenstein
primes. If g is any complex number which is a product of such Eisenstein primes,
then exactly one of ±g, ±¯g is in the first quadrant. In this sense, we shall say that
each product of split Eisenstein primes defines uniquely a cubic Galois extension
of the rational numbers. According to Theorem 4.1 (b), it is easy to find a cubic
Galois field over Q when its invariant f is given – it is simply the splitting field of
the polynomial (2). We shall say that this polynomial is canonical for its splitting
field.
Now we can construct a list of cubic Galois extensions of Q using their invariants
f œ Z[Á]. We can order such polynomials with respect to the value of the norm
Nr(f) of s uch numbers f. First we note an observation concerning the number of
such polynomials:
Proposition 4.3. The number of non-isomorphic cubic Galois extensions of ra-
tional numbers with the invariant f such that Nr(f) is fixed equals 3 · 2
k≠1
, where
k is the number of different primes dividing Nr(f), when Nr(f) ”=1, and it is equal
124 Juliusz BrzeziÒski and Ulf Persson. Normat 3-4/2011
1, when Nr(f)=1. These cubic Galois fields are splitting fields of the polynomials
(2) with fixed p = 3 Nr(f).
Proof. According to Proposition 4.1 and Theorem 4.1, it suffices to count the
number of invariants f corresponding to fields with given Nr(f). For Nr(f )=1,
we get only one cubic Galois field over Q (see the first argument in the proof of
Theorem 4.1 (a)).
If Nr(f)=p
1
···p
k
is a product of prime numbers congruent to 1 modulo 3, and
we fix a prime fi
i
such that Nr(fi
i
)=p
i
for i =1,...,k, then each solution of
the equation Nr(f)=p
1
···p
k
has the form f = ÷–
1
···–
k
,where÷ is one of
±1, ±Á, ±Á
2
and –
i
= fi
i
or ¯fi
i
for i =1,...,k. Hence, we have 6 · 2
k
different
solutions. Now we need to count the number of different solutions f in the first
quadrant of the complex plane. Since the mappings f ‘æ ≠f and f ‘æ
¯
f do not
change the field E(
3
f
¯
f
2
) and exactly one of the numbers ±f,±
¯
f is in the first
quadrant, the number of such invariants f equals 3 · 2
k≠1
. 2
Example 2. It is easy to construct the cubic Galois field corresponding to a given
invariant f. In the next section, we list cubic Galois fields together with their
minimal canonical polynomials, which we define there. Already now, we c ould list
cubic Galois fields ordered by the increasing norm Nr(f) and defined by their
canonical polynomials. The first few values of Nr(f) are 1, 7, 13, 19, 31,....The
first composite number in the sequence is 91 = 7 · 13. For Nr(f)=1, we have only
one polynomial corresponding to the extension L = E(
3
Ô
Á) whose invariant is the
only 6th root of 1 in the first quadrant, that is, f = ≠¯Á =
1+
Ô
3
2
. In fact, it is easy
to check that L is the splitting field of X
6
≠X
3
+1with Galois group Z
6
and the
corresponding cyclic Galois field over Q is the splitting field of X
3
≠ 3X +1. For
Nr(f)=7, as for any prime number congruent to 1 modulo 3 in the sequence, we
have 3 polynomials. An easy computation shows that these are the splitting fields
of the trinomials X
3
≠21X ≠7, X
3
≠21X ≠28, X
3
≠21X ≠35 with corresponding
invariants f =2+3Á, f =3+2Á, f =3+Á.
We end this section showing how to find the invariant f when a cubic Galois field
K over the rational numbers is given as a splitting field of some cubic polynomial.
Proposition 4.4. Let K be the splitting field of the cyclic cubic Ï(X)=X
3
≠pX +
q, where (Ï)=4p
3
≠ 27 q
2
= d
2
(d = ”(Ï)). Then K is the splitting field of the
polynomial (2) with the invariant f defined in the following way: 4(d +3q
Ô
≠3) =
f
¯
f
2
h
3
, where h œ Z[Á], the factor f is in the first quadrant of the complex plane
and is a unit or a product of different split E isenstein primes.
Proof. The equality (Ï)=4p
3
≠27 q
2
= d
2
implies that p =
27
4
s
2
+
1
4
r
2
,where
d = rp and q = sp for some parameters r, s. Let x
1
be one of the zeros of Ï(X) in
K. It is not difficult to find the remaining two zeros x
2
,x
3
expressed by r, s.Then,
as in the proof of Proposition 3.1 (b), we have x = x
1
+ Áx
2
+ Á
2
x
3
, which satisfies
x
3
= – œ Q (Á) (we have x ”=0as p ”=0). A short computation shows that – equals
4(d +3q
Ô
≠3) up to a third power in Q(Á). Hence, by Theorem 4.1 (a), we ge t a
representation of 4(d +3q
Ô
≠3) in the desired form. 2
Normat 3-4/2011 Juliusz BrzeziÒski and Ulf Persson. 125
Example 3. Let K be the splitting field of the polynomial X
3
+3X
2
≠ 88 X ≠
25 over Q. First we transform the polynomial to a trinomial using the standard
transformation X ‘æ X ≠ 1. We get Ï(X)=X
3
≠ 91 X + 65. We check that
(Ï) = 1073
2
= d
2
,soK is a cubic Galois field over Q. What is its invariant and the
canonical polynomial? Using Proposition 4.4 and the equality
Ô
≠3=2Á+1,wefind
4(d+3q
Ô
≠3) = 4(1703+195
Ô
≠3) = 8·13(73+15Á)=≠8·(4+3Á)(1≠3Á)
2
(1≠2Á)
3
.
Since Nr(4 + 3Á) = 13, we get f =4+3Á. Thus the canonical polynomial (2) is
X
3
≠ 39X ≠ 65.
Corollary 4.1. If a cubic Galois field K with invariant f is a splitting field of
X
3
≠ pX + q, then Nr(f) | p.
Proof. By Proposition 4.4, we have 4(d +3q
Ô
≠3) = f
¯
f
2
h
3
,whereh is an integer
in K. Hence 64p
3
=4· 16(d
2
+ 27q
2
) = Nr(f)
3
Nr(h)
3
.Sincef is a unit or f is a
product of different split Eisenstein primes in Z[Á], the norm Nr(f) is either 1 or a
product of different prime numbers congruent to 1 modulo 3. Thus Nr(f) | p. 2
5 Minimal canonical polynomials
As we already know, the cubic Galois fields cannot be described by binomials,
so the next “simple” choice were trinomials X
3
≠ pX + q. Using the canonical
trinomials given by Theorem 4.1, we get a solution of our problem, but one can ask
whether there are polynomials with smaller integer coefficie nts, which have the same
splitting field. In the present section, we adress this problem and characterize the
minimal canonical polynomials having the smallest possible value of the coefficient
of X. It appears (see Theorem 5.1) that the polynomial (2) corresponding to f is
already minimal in this sense unless its reduced discriminant is divisible by 27. We
shall prove that every cyclic cubic extension K of Q is a splitting field of exactly
one cubic polynomial X
3
≠pX ≠q for which p, q > 0 and p is minimal. This cubic
polynomial will be called minimal canonical for K and it is given in the following
way:
Theorem 5.1. Let K be a cubic Galois extension of Q with the invariant f =
a + bÁ. Then K has the minimal canonical polynomial Ï(X) given by (2) if and
only if 27 - ”(Ï), and by X
3
≠ Nr(f)X ≠
1
27
”(Ï) when 27 | ”(Ï). Moreover, these
two cases are equivalent to 3 - b and 3 | b, respectively, and the minimal canonical
polynomial in the second case is equal X
3
≠ Nr(f)X ≠
1
3
Nr(f)b.
In order to prove this, we need some auxiliary notions and results. Let K be a cubic
Galois extension of Q with Galois group G and denote by I(K) the ring of integers
in K. We shall denote by I(K)
0
the integers whose trace equals to 0. It is clear
that I(K)
0
is an abelian group (a Z-module) of rank 2 as a lattice on the kernel of
the linear map Tr : K æ Q, which has dimens ion 2 over the rational numbers. If ‡
is an automorphism of K and x œ K is an integer with trace equal to 0, then also
‡(x) is an integer in K and has trace equal to 0. Thus, the abelian group I(K)
0
is
a G-module of rank 2 over Z . The ideals I in the ring Z[Á] can be also c onsidered
126 Juliusz BrzeziÒski and Ulf Persson. Normat 3-4/2011
as G-modules if ‡ · – = Á– for – œ I. It is well-known that I(K)
0
as a G-module
is isomorphic to an ideal in Z[Á] as a G-module (see [CR], (74.3), p.508).
An element a + b‡ of Z[G] acts on – œ I by (a + b‡) · – =(a + bÁ)–. Since, as a
consequence of the unique factorization of the Eisenstein integers, every element
of I is a multiple (a + bÁ)–
0
,where–
0
is a generator of I as an ideal in Z[Á], an
isomorphism of I with I(K)
0
maps such a product onto (a+b‡)·x
0
= ax
0
+b‡(x
0
),
where x
0
is an image of –
0
. Hence I(K)
0
is generated by x
0
as a module over Z[Á]
when its structure as such a module is defined by (a + bÁ) · x = ax + b‡(x) for
x œI(K)
0
(observe that x
0
is also a generator of I(K)
0
as a G-module, since
‡
2
(x
0
)=≠x
0
≠ ‡(x
0
)). Notice that I(K)
0
, as every non-ze ro ideal in Z[Á], has 6
different generators.
Lemma 5.1. Let x
1
,x
2
,x
3
be the zeros of X
3
≠ pX ≠ q and let 4 p
3
≠ 27 q
2
=
d
2
. Then, with a suitable choice of the sign of d and arbitrary a, b, the numbers
ax
1
+ bx
2
,ax
2
+ bx
3
,ax
3
+ bx
1
satisfy the equation X
3
≠ P (a, b)X ≠ Q(a, b)=0,
where
P (a, b)=p(a
2
≠ ab + b
2
),
Q(a, b)=qa
3
+
d ≠ 3q
2
a
2
b ≠
d +3q
2
ab
2
+ qb
3
Proof. We have
±d =(x
1
≠x
2
)(x
2
≠x
3
)(x
3
≠x
1
)=(x
1
≠x
2
)(x
1
+2x
2
)(2x
1
+x
2
)=2x
3
1
+3x
2
1
x
2
≠3x
1
x
2
2
≠2x
3
2
,
and
q = x
1
x
2
x
3
= ≠x
1
x
2
(x
1
+ x
2
)=≠x
2
1
x
2
≠ x
1
x
2
2
.
The numbers ax
1
+bx
2
, ax
2
+bx
3
and ax
3
+bx
1
satisfy an equation X
3
≠P (a, b)X ≠
Q(a, b)=0,wherethecoefficientP (a, b) is the sum of 3 products of two of these
numbers. Taking into account that x
3
= ≠x
1
≠ x
2
, we easily get the equality
P (a, b)=p(a
2
≠ ab + b
2
).
The coefficient Q(a, b) is the product of these 3 numbers. The coefficients of a
3
and
b
3
are the products x
1
x
2
x
3
= q. The coefficient of a
2
b is x
2
1
x
2
+ x
2
2
x
3
+ x
2
3
x
1
=
x
3
1
+3x
2
1
x
2
≠ x
3
2
, and ab
2
is x
2
1
x
3
+ x
2
3
x
2
+ x
2
2
x
1
= x
3
1
+3x
1
x
2
2
≠ x
3
2
. Choosing
d =(x
1
≠x
2
)(x
2
≠x
3
)(x
3
≠x
1
), we get that the first is (d ≠3q)/2 and the second
≠(d +3q)/2. 2
Remark 5.1. Let X
3
≠ pX ≠ q be given, where 4p
3
≠ 27q
2
= d
2
and choose
a =1,b= ≠Á in Lemma 5.1. Then P (a, b)=0and Q(a, b)=
d+3q
2
+3qÁ =: – œ Z[Á].
If we now set ’ = x
1
≠ Áx
2
,then’
3
= – (cf. Proposition 4.4). Furthermore, given
’ =
3
Ô
– (a fixed value), we can recapture x
i
(i =1, 2, 3) and thus get the Cardano
formulas for the solutions of a cubic equation. In fact, since x
1
,x
2
are real, we have
¯
’ = x
1
+(1+Á)x
2
,so
¯
’ ≠’ =(1+2Á)x
2
, which gives x
2
=
¯
’≠’
Ô
≠3
and x
1
,x
3
can be
easily computed.
Normat 3-4/2011 Juliusz BrzeziÒski and Ulf Persson. 127
Remark 5.2. Let K be the cubic Galois field defined by a zero x of a polynomial
X
3
≠ pX ≠ q and consider I(K)
0
as a module over the Eisenstein integers. Then
for ⁄ = a + bÁ œ Z[Á],thenumber⁄ · x = ax + b‡(x) is a zero of the polynomial
X
3
≠ P (a, b)X ≠ Q(a, b),whereP (a, b) = Nr(⁄)p and Q(a, b)=2Ÿ(⁄
3
–
0
) and
–
0
=
9q≠d≠2dÁ
18
. This follows from –
0
= ≠
1+2Á
9
–,where– is defined in Remark 5.1
(cf. also Remark 5.3).
Proposition 5. 1. Let K be a cubic Galois field over Q. Then K has the minimal
canonical polynomial, which can be defined as the minimal polynomial of a suitable
generator of I(K)
0
as a Z[Á]-module. If X
3
≠ pX ≠ q is the minimal canonical
polynomial of K, then its zeros are 3 generators I(K)
0
, while the remaining 3
generators satisfy the equation X
3
≠ pX + q =0. Moreover, for any polynomial
X
3
≠ p
Õ
X + q
Õ
with p
Õ
,q
Õ
œ Z having K as its splitting field, we have p | p
Õ
.
Proof. Let x
0
be a generator of I(K)
0
as Z[Á]-module and let X
3
≠pX ≠ q be its
minimal polynomial. We have p>0,since4 p
3
≠27 q
2
= d
2
and we can assume that
q>0,since≠x
0
is also a generator and satisfies the equation X
3
≠pX +q =0. Now
let x
Õ
0
be another generator of the Z[Á]-module I(K)
0
.Thenx
Õ
0
= ax
0
+ b‡(x
0
),
where a, b œ Z. Let x
Õ
0
be a zero of the polynomial X
3
≠p
Õ
X ≠q
Õ
. As before, we may
assume that q
Õ
> 0. According to Lemma 5.1, we have p
Õ
=(a
2
≠ab + b
2
)p, that is,
p | p
Õ
. By symmetry, we have also p
Õ
| p,sop
Õ
= p. Moreover, we have a
2
≠ab+b
2
=1,
where a, b are integers. Thus, we get six possibilities corresponding to the six units
in the ring Z[Á]: (a, b)=(±1, 0), (0, ±1), (1, 1), (≠1, ≠1). Using again Lemma 5.1,
this time for the free coefficients of the cubic polynomials, we get q
Õ
= q.Thus
the generators of I(K)
0
satisfy one of the equations X
3
≠ pX ± q =0. Moreover,
the arguments above show that if X
3
≠ pX ≠ q
Õ
is any integer polynomial whose
splitting field is equal K and q
Õ
> 0,thenq
Õ
= q.
Let now X
3
≠ p
Õ
X + q
Õ
,wherep
Õ
,q
Õ
œ Z, be any polynomial whose splitting field
is equal K and x
Õ
some of its zeros. Then x
Õ
= ax
0
+ b‡(x
0
) for some integers a, b.
By Lemma 5.1, we have p | p
Õ
,sop is really the least value of the coefficients of
x in polynomials having K as its splitting field and it divides the corresponding
coefficient of X in any trinomial of this type which splits in K. 2
Proof of Theorem 5.1. Let X
3
≠ pX + q b e the minimal canonical polynomial
of a cubic Galois extension K of Q with the invariant f. According to Corollary
4.1, Prop os ition 5.1 and Theorem 4.1, we have Nr(f) | p and p | 3 Nr(f ). Hence
p = 3 Nr(f) or p = Nr(f),sinceNr(f) is a product of different prime numbers or
Nr(f)=1.Ifp = 3 Nr(f), then the polynomial (2) is up to the sign of the free
term, the minimal canonical polynomial of K.
Assume that p = Nr(f) and let x denote a zero of the polynomial (2). According to
Lemma 5.1, equation (2) implies 3 Nr(f)=(a
2
≠ ab + b
2
) Nr(f) for some integers
a, b such that x = ax
0
+ b‡(x
0
),wherex
0
is a zero of the minimal canonical
polynomial of K. Hence, we have a
2
≠ab+b
2
=3. The last equation has 6 solutions:
(a, b)=±(1, ≠1), ±(1, 2), ±(2, 1). Taking into account x = ax
0
+ b‡(x
0
),wehave
‡(x)=a‡(x
0
)+b‡
2
(x
0
)=a‡(x
0
) ≠ b(x
0
+ ‡(x
0
)) = (a ≠ b)‡(x
0
) ≠ bx
0
,
128 Juliusz BrzeziÒski and Ulf Persson. Normat 3-4/2011
and solving the system of these two equations, we get
x
0
=
(a ≠ b)x ≠ b‡(x)
a
2
≠ ab + b
2
=
1
3
((a ≠ b)x ≠ b‡(x)) .
An uncomplicated computation (most conveniently using, for example, Maple),
shows that s uch an x
0
is a zero of one of the equations X
3
≠ Nr(f)X ±
d
27
(the
choice of sign depends on the choice of a, b). This can happen if and only if 27 | d.
Since
d
2
=4p
3
≠27 q
2
=4·27 Nr(f)
3
≠27 Tr(f)
2
Nr(f)
2
= 27 Nr(f)
2
!
4 Nr(f) ≠ Tr(f)
2
"
,
we have 27 | d if and only if 27 divides the last factor. An easy computation gives
4 Nr(f) ≠ Tr(f)
2
=3b
2
, which is divisible by 27 if and only if 3 | b. 2
Remark 5.3. Let (p, q, d) denote the coefficients and the square root of the di-
scriminant of the polynomial X
3
≠ pX ≠ q. Choose a =2,b =1in Lemma 5.1.
Then P (a, b)=3p, Q(a, b )=d and (3p, d, 27q) is the triple corresponding to the
polynomial X
3
≠ 3pX ≠ 27q. Notice that if we have 4p
3
= 27q
2
+ d
2
and multiply
by 27, we get 4(3p)
3
= (27q)
2
+ 27d
2
,sod and q “change places”. This explains the
relation between these two polynomials (in fact, (2 + Á)
2
= ≠3 up to a unit, so if
we do the transformation twice, we get (9p, 27q, 27d)). Notice also that if the first
polynomial is minimal canonical, then the second is the one given by (2).
As we know, there are 3 ·2
k≠1
non-isomorphic cubic Galois fields over Q for which
the norm of the invariant is a product of k primes. For one third of the corresponding
canonical polynomials (2), the minimal canonical polynomial is different:
Proposition 5.2. There exist 2
k≠1
non-isomorphic cubic Galois fields for which
the norm of the invariant is a product of k fixed primes and the minimal canonical
polynomial is not equal to the canonical polynomial.
Proof. Let K be a cubic Galois field over Q and let the norm of its invariant f
be a product of k Eisenstein primes. The numbers Áf and Á
2
f have the same norm
and it is easy to check using Proposition 3.1 and Theorem 4.1 that they define non-
isomorphic cubic Galois fields, which both are non-isomorphic to the field defined
by f. Now we prove that for exactly one of the invariants f,Áf,Á
2
f,thereduced
discriminant is divisible by 27, that is, the minimal canonical and the canonical
polynomial are not equal. Of course, this will prove the Proposition.
Let f = a + bÁ.ThenÁf = ≠b +(a ≠ b)Á and Á
2
f =(b ≠ a) ≠ aÁ. Now Nr(f )=
a
2
≠ ab + b
2
= 1 (mod 3) says that e ither exactly one of a, b is divisible by 3
or a, b give the same residue modulo 3. In any of these cases, exactly one of the
coefficients a, b, a ≠ b is divisible by 3. By last part of Theorem 5.1 this proves our
claim concerning the numbers f,Áf,Á
2
f. 2
Normat 3-4/2011 Juliusz BrzeziÒski and Ulf Persson. 129
Example 4. We continue the previous Example 3. We obtained there the canonical
polynomial X
3
≠39X ≠65 corresponding to the polynomial X
3
≠91X + 65 .Since
f =4+3Á, we ge t according to Theorem 5.1 that the minimal canonical polynomial
is X
3
≠ 13X ≠ 13.
It is not difficult to construct a list of cubic Galois fields over the rational numbers
in terms of their invariants and the minimal canonical polynomials using Theorem s
4.1 and 5.1. On the inside of the back c over you can find a table of all cases with
Nr(f) Æ 100.
6 Sieving cyclic Galois fields
In this section, we show how to list non-isomorphic cubic Galois fields using a
sieving process similar to the sieve of Eratosthenes applied to the sequence of all
cubic traceless trinomials defining such fields. In order to introduce this method, let
us start with a similar task for (real) quadratic fields. Every such field is a splitting
field of a quadratic polynomial X
2
≠ q with q>0. Of course, the splitting field
of such a polynomial is quadratic if and only if q is not a square. Two quadratic
number fields (subfields of the complex numbers) are isomorphic if and only if they
are equal, so we want to know when X
2
≠q and X
2
≠q
Õ
define the same field. We
start listing the sequence of all non-square integers (binomials X
2
≠ q) up to any
given limit, starting with the least one, that is, the number 2. This number defines
our first quadratic field Q(
Ô
2). Now we mark all integers in the s equence which
define the same field, that is, the numbers 2a
2
for integers a>1 (or polynomials
X
2
≠ 2a
2
). The next unmarked number is 3.ThusthenextfieldisQ(
Ô
3) and we
mark all the numbers 3a
2
with a>1.Thenextnumberis5 and so on. We get
the sequence 2, 3, 5, 6, 7, 10, 11, 13, 14,.... The polynomials (or the integers), which
remain on our list unmarked give non-isomorphic quadratic fields and have the
least pos sible value of q among the binomials having the same splitting field as
X
2
≠ q. Of course, our sequence is the sequence of all square-free integers – those
which are marked are exactly multiples of the square-free integers by a square ”=1.
We describe this pro cess in detail in order to show that a similar sieving procedure
can be used in order to get all cubic Galois fields.
First we note that two cubic Galois fields over Q (contained in the complex num-
bers) are also isomorphic if and only if they are equal – this is a general property of
Galois fields. Hence we want to list all different cubic Galois fields. Every such field
is a splitting field of an irreducible cubic polynomial X
3
≠pX+q,wherep, q are posi-
tive integers and the discriminant is a square, that is, 4p
3
≠27q
2
= d
2
for an integer
d. We can write down the sequence of all such polynomials starting with the least
possible p (in fact, p =3) and then for each fixed p, we can list all irreducible polyno-
mials with q in increasing order – since 27q
2
< 4p
2
, the number of such polynomials
is finite. Identifying the polynomials with the pairs of their coefficients (p, q),the
beginning of the sequence is (3, 1), (7, 7), (9, 9), (12, 8), (13, 13), (19, 19), (21, 7),....
We start with the first polynomial, which defines the “least” cubic Galois field – the
splitting field of X
3
≠3X +1 given by the pair (3, 1). According to Lemma 5.1, we
130 Juliusz BrzeziÒski and Ulf Persson. Normat 3-4/2011
have to mark on our list all polynomials corresponding to (P (a, b),Q(a, b)),where
P (a, b) = 3(a
2
≠ab + b
2
), Q(a, b)=a
3
+3a
2
b ≠6ab
2
+ b
3
, a, b œ Z,sincetheydefine
the same field as (3, 1). The next pair is (7, 7) and in practise, we check whether it
is marked by solving P (a, b)=7,Q(a, b)=7. It is unmarked, so it defines a new
field – the splitting field of X
3
≠ 7X +7. Now we mark all the pairs, which give
polynomials defining the same field as this polynomial according to Lemma 5.1, but
in practise, we check whether a pair is marked when we know the status of all ear-
lier pairs. For example, the next pair is (9, 9) and we check that this pair is marked
because of the pair (3, 1) as the corresponding system P (a, b)=9, Q(a, b)=9has
a solution a = ≠1,b =1. Hence , we mark this polynomial. The same can be said
about (12, 8), but the next pair (13, 13) is not represented as P (a, b),Q(a, b) neither
for (3, 1) or (7, 7), so it gives a new field. We continue the algorithm in this way. The
polynomials, which are unmarked on our list give different cubic Galois fields and
have the least possible coefficients among the trinomials having the same splitting
field. Thus, they are the minimal canonical polynomials defining these fields.
The reason this works is, because as remarked before Lemma 5.1, the traceless
integral elements of a cubic Galois field form a rank one module over Z[Á],which
is generated (up to a unit) by an element x
0
. The polynomial corresponding to
x
0
is the original unmarked polynomial (the minimal canonical), and furthermore
all “multiples” of the minimal canonical polynomials form a partition of all cubic
(traceless) polynomials, i.e. they never meet unless they coincide. Finally the pro-
cess works even if we are not able to mark all the infinite number of “multiplies”.
Say that we are only interested in (p, q) such that p Æ P
0
with P
0
given in advan-
ce. We then look at the finite number of pairs (a, b) such that the corresponding
p(a, b) Æ P
0
and restrict our marking to those. What is left unmarked remains
unmarked no matter how many more (a, b ) we will consider. This process can be
easily implemented on a computer and on the inside of the back cover you can find
a table presenting the beginning of a run.
Referenser
[C] H. Cohen, A Course in Computational Algebraic Number Theory, GTM 138,
Springer-Verlag, New York, Heidelberg, Berlin, 1993.
[CR] C. W. Curtis, I. Reiner, Representation Theory of Finite Groups and Associative
Algebras, AMS, Chelsea Publishing, Providence, Rhode Island, 1962.
[IR] K. Ireland, M. Rosen, A Classical Introduction to Modern Number Theory, GTM
84, Springer-Verlag, New York, Heidelberg, Berlin, 1990.
[L] S. Lang, Algebra, GTM 211, Springer-Verlag, New York, Heidelberg, Berlin, 2002.
