Normat 59:3-4, 131–143 (2011) 131
The inverse Problem of Galois theory
Christian U. Jensen
København Universitets matematiske Institut
Universitetsparken 5
DK- 2100 København
cujensenmath.ku.dk
The essence of Galois Theory: The systematically developed connection between two
seemingly unrelated subjects, the theory of fields and the theory of groups.
More specifically, but in the same line, is the idea of studying a mathematical
object by its group of automorphisms, an idea emphasized in Klein’s Erlanger Pro-
gram, which has been accepted as a powerful tool in a great variety of mathematical
disciplines.
The Galois Theory of field extensions combines the esthetic appeal of a theory
of nearly perfect beauty with the technical development and difficulty that reveal the
depth of the theory and that make possible its great usefulness primarily in algebraic
number theory and related parts of algebraic geometry.
(From Roger Lyndon in: Encyclopedia of Mathematics and Its Applications)
1 Basic concepts
For the convenience of the reader we start by briefly recalling the basic concepts.
All fields in this article are supposed to have characteristic 0, so there will be
no problems of separability. (In other words: every finite field extension will be
separable.)
If L/K is a field extension Aut(L/K) is defined as the group of all field automor-
phisms of L that are the identity on K.IfK is the fixed field of Aut(L/K),(i.e.no
element in L \ K is left invariant by all automorphisms of Aut(L/K))thenL/K
is called a Galois extension (or normal extension). If L/K is Galois Aut(L/K) is
usually denoted Gal(L/K).
Theorem 1.1. For a finite extension L/K the following conditions are equivalent:
i) L/K is a Galois extension.
ii) L is the splitting field over K for a polynomial f(x) in K[x], (i.e., L is a smallest
extension of K, in which f(x) splits completely in linear factors.)
iii) Every irreducible polynomial f(x) in K[x] having some root in L splits in linear
factors in L.
132 Christian U. Jensen Normat 3-4/2011
For the sake of brevity a Galois extension of a field K with Galois group G is called
a G-extension of K.Iff(x) is a polynomial in K[x] the Galois group of the splitting
field of f(x) over K is called the Galois group of f(x). As for the explicit form of
the Galois group the following is quite useful.
Theorem 1.2. If L is the splitting field over a field K of an irreducible polynomial
f(x) in K[x] of degree n, then the action of the Galois group Gal(L/K) on the roots
of f(x) gives rise to a permutation of these n roots. This yields an isomorphism of
Gal(L/K) onto a transitive subgroup of the symmetric group S
n
,calledtheGalois
permutation group of f(x).
Another useful tool in determining Galois groups is the following.
Theorem 1.3. Let M and L be extension fields of a field K, both of them being
contained in a common larger field. If M/K is a finite Galois extension then the
compositum ML is a Galois extension of L and Gal(ML/L) ƒ Gal(M/M flL).If
M and L are finite Galois extensions of K and M fl L = K then the compositum
ML is a finite Galois extension of K and Gal(ML/K) is isomorphic to the direct
product Gal(M/K) ◊ Gal(L/K).
The inverse problem of Galois theory asks whether every finite group can be re-
alized as the Galois group for some Galois extension of the rational number field
Q. This problem can als o be expressed in terms of permutation groups: Given a
transitive subgroup G of the symmetric group S
n
, the question is whether there
exists an irreducible polynomial p(x) in Q[x] of degre e n, such that G is the Galois
permutation group described in Theorem 1.2 with K = Q. Although many families
of finite groups can be realized as Galois groups over Q it is still an open problem
whether every finite group is realizable as Galois group over Q. In this article we
shall give a survey of some important results on this question. The account we give
is by no means complete. It is just our hope to give an impression of the type of
results and the type of methods that are used.
The author is grateful to Juliusz Brzezinski and Anders Thorup, who have read
the manuscript and made several valuable comments.
2 Some ”relatively easy” results
We start by a classical result which basically is just a study of cyclotomic fields,
i.e. fields generated by roots of unity.
Theorem 2. 1. Every finite abelian group A appears as a Galois group over Q.
Proof. (Sketch.) A is a direct product of cyclic groups. We restrict ourselves to
prove the theorem for a cyclic group. The proof for general A is slightly m ore
involved but depends on the same ideas.
Normat 3-4/2011 Christian U. Jensen 133
Consider the cyclic group C
n
of order n. By Dirichlet´s theorem concerning prime
numbers in arithmetic progressions there exists a prime number p which is © 1 mod-
ulo n. The Galois group over Q of the p
th
cyclotomic field Q(’
p
), ’
p
being a primitive
p
th
root of unity, is cyclic of order p ≠ 1.Sincen divides p ≠ 1 the Galois group
Gal(Q(’
p
)/Q) contains a subgroup H such that the quotient Gal(Q(’
p
)/Q)/H is
cyclic of order n. By the main theorem of Galois theory, the fixed field of H will
then be a C
n
-extension of Q.
Remark. The realizability of two finite groups G and H as Galois groups over Q
does not automatically imply the realizability of the direct product G ◊ H as a
Galois group over Q. This, of course, holds if the orders |G| and |H| are relatively
prime, since in that case the com positum of a G-extension and an H-extension will
be a (G ◊ H)-extension. But in general one should be careful: There exist fields
K admitting a G-extension and an H-extension, but not a (G ◊ H)-extension.
However, from the ab ove theorem 2.1, theorem 1.3 and proposition 2.2 below (for
A being the family of finite abelian groups) it follows that if G is realizable as a
Galois group over Q,thensoisG ◊ A for every finite abelian group A.
Proposition 2.2. Let A be family of finite groups which is closed under formation
of finite direct p roducts. Let K be a field such that any group in A is a Galois group
over K.IfL is an arbitrary finite extension of K, then there exists for any A œA
an A-extension of K whose intersection with L is K.
Proof. Since all fields in this article have characteristic 0, there is a primitive el-
ement for the extension L/K, and hence by a theore m of Artin there are only
finitely many fields between L and K. Let t be the number of these fields. Let A
be a group in A.ThedirectproductA
t+1
of t +1 copies of A is by assumption
realizable as the Galois group of an extension M of K. We can w rite M as the
compositum of (t + 1) A-extensions M
i
, 1 5 i 5 t +1,whereM
i
fl M
j
= K for
i ”= j. Consider the fields M
i
flL, 1 5 i 5 t +1. Since there are only t fields between
K and L there must be two indices i and j such i ”= j and M
i
fl L = M
j
fl L,
whence M
i
fl L = M
j
fl L = M
i
fl L flM
j
fl L = M
i
fl M
j
fl L = K fl L = K.
The next two results may at first glance look like being ”almost” a solution of the
inverse problem of Galois theory.
Theorem 2.3. For any finite group G there exists an algebraic number field K
(i.e. a finite extension of the rational number field Q)thatadmitsaG-extension.
Proof. (Sketch.) For any odd prime p the polynomial
f(x)=(x
2
+ 4)(x ≠2)(x ≠ 4) ···(x ≠2(p ≠ 2)) ≠ 2
has degree p and is irreducible in Q[x] by Eisenstein’s criterion applied to the prime
number 2. Elementary calculus shows that f(x) has exactly (p ≠ 2) real roots. It
is a classical result that the splitting field over Q of an irreducible polynomial
in Q of prime degree p with e xactly p ≠ 2 real roots is Galois over Q with the
134 Christian U. Jensen Normat 3-4/2011
symmetric group S
p
as Galois group. Hence the splitting field M of f(x) over Q is
an S
p
-extension of Q.
Now for a suitable large prime p the group G sits as a subgroup of S
p
. Hence the
Galois group of M/Q c ontains a copy of G.IfK is the corresponding fixed field for
G the main theorem of Galois theory implies that M/K is Galois with G as Galois
group.
Remark. The sad thing about the above result is that we have no way of controlling
the ground field K, in particular of going down to Q. In some se nse the theorem
shows that the inverse problem of Galois theory is “just” a problem of descent.
The next result was first proved by E. Fried and J .Kollar [FK], but an error was
found in the proof by M. Fried. A relatively elementary proof was given by W.-D.
Geyer[Ge]. It will, however, take several pages, so we refrain from giving it here.
Theorem 2.4. Every finite group is the automorphism group of a finite extension
field L of the rational number field Q.
Remark. The sad thing here is that the field L constructed in this theorem will
usually not be a Galois extension of Q.
3Hilbertenters
In some sense the inverse problem of Galois theory is as old as Galois theory itself;
however the first systematic attack was made by Hilbert in his famous paper [Hi]
in which the principal result was:
Theorem 3.1 (Hilbert’s Irreducibility Theorem). Let f(x
1
,...,x
m
,t
1
,...,t
n
) be
an irreducible polynomial in Q[x
1
,...,x
m
,t
1
,...,t
n
]. Then there exist infinitely
many n-tuples (q
1
,...,q
n
) of rational numbers such that the specialized polyno-
mial f(x
1
,...,x
m
,q
1
,...,q
n
) is either a constant in Q or irreducible viewed as a
polynomial in Q[x
1
,...,x
m
].
The proof is by no means easy. A relatively acce ssible proof can be found in Had-
lock[Ha]. Hilbert’s Irreducibility Theorem has a very important application in in-
verse Galois theory:
Theorem 3.2. Let Q(t
1
,...,t
n
) be a purely transcendental extension of Q, i.e. the
field of rational functions in n independent variables t
1
,...,t
n
.Iff(x, t
1
,...,t
n
)
is a polynomial in x over the field Q(t
1
,...,t
n
) with Galois gro up G, then there
exist infinitely many n-tuples (q
1
,...,q
n
) of rational numbers such that G is the
Galois group over Q of the specialized polynomial f(x, q
1
,...,q
n
). In particular, if
a finite group G can be realized as a Galois group over Q(t
1
,...t
n
) then G can also
be realized as a Galois group over Q.
Normat 3-4/2011 Christian U. Jensen 135
We omit the proof which may take a few pages. The usefulness of the above theorem
lies in the fact that it is much e asie r to realize a group as a Galois group over the
function field Q(t
1
,...,t
n
) than over Q: there is much more space and freedom to
operate in the function field. Here are two examples:
By the general polynomial of degree n we mean the polynomial
f(x, t
1
,...,t
n
)=x
n
+ t
1
x
n≠1
+ ···+ t
n≠1
x + t
n
viewed as a polynomial in x with coefficients in the rational function field Q(t
1
,...,t
n
)
in n independent variables t
1
,...t
n
. It c an shown (e.g. using the theory of sym-
metric polynomials) that the splitting field of f(x, t
1
,...,t
n
) over Q(t
1
,...,t
n
) is
a Galois extension of Q(t
1
,...,t
n
) with the symmetric group S
n
as Galois group.
Theorem 3.2 thus implies that S
n
can also be realized as a Galois group over Q.
In the same paper Hilbert showed that also the alternating group A
n
can be realized
as a Galois group over Q(t
1
,...,t
n
).(This is much harder than in the S
n
-case.)
Thereby one obtains a realization of A
n
over Q.
However, there is one defect in the above strategy: Both theorem 3.1. and theorem
3.2. are merely existence theorems and do not yield explicit Galois realizations of
the desired groups. One can find explicit polynomials over the function field whose
splitting field has the prescribed group as Galois group, but usually not an explicit
specialization to Q with the same Galois group.
For instance, let P
n
(t, x) be the following polynomial:
P
n
(t, x)=
I
x
n
+((≠1)
(n≠1)/2
nt
2
≠ 1)(nx + n ≠ 1) for odd n
x
n
+ nx
n≠1
+(≠1)
n/2
t
2
+(n ≠ 1)
n≠1
for even n.
The splitting field of P
n
(t, x) viewed as a polynomial in x over Q(t) is an A
n
-
extension of Q(t). So there exist infinitely many rational numbers q such that the
splitting field of P (q, x) over Q is an A
n
-extension. But the theory does not tell us,
how to find such q’s explicitly. The above results hold for a larger class of fields:
Definition. AfieldK is called Hilbertian if the assertion of Theorem 3.1 remains
true when Q is replaced by K.
Thus the symmetric groups S
n
and the alternating groups A
n
are realizable as
Galois groups over any Hilbertian field.
Examples. Every algebraic number field of finite degree is Hilbertian. For every field
K the field of all rational functions over K is Hilbertian. An algebraically closed
field (for instance the field C of complex numbers) is not Hilbertian. Similarly, the
field R of real numbers and the p≠adic number fields are not Hilbertian.
4 Emmy Noether and others enter
Let G be a subgroup of the symmetric group S
n
. Viewed as a group of permutations
of n independent variables t
1
,...,t
n
it gives rise to a group of automorphisms of
136 Christian U. Jensen Normat 3-4/2011
the rational function field Q(t
1
,...,t
n
). More precisely, if ‡ is a permutation in G
the corresponding automorphism ¯‡ of Q(t
1
,...,t
n
) is defined by
¯‡(f(t
1
,...,t
n
)=f(t
‡(1)
,...,t
‡(n)
)
where f(t
1
,...,t
n
) is a rational function in Q(t
1
,...,t
n
).
If F(G) is the fixed field of G then Q(t
1
,...,t
n
) is a Galois extension of F(G) with
Galois group G. Emmy Noether raised the following question:
(NP)”Noether´s Problem”: Is F(G) isomorphic to the field of rational functions
over Q in n independent variables?
If the answer is “ yes”, then Theorem 3.2 implies that G is realizable as a Galois
group over Q. It will have many other consequences which we shall mention later.
In the case, where G = S
n
the answer to (NP) is clearly “yes”. Indeed, the fixed
field F(S
n
) is just the field of the rational functions of the n elementary symmet-
ric polynomials in t
1
,...,t
n
. For the alternating groups A
n
the question is much
harder: it is known that the answer to (NP) is affirmative, if n 5 5, but the answer
is not known for any n>5. In the 1920’s and 1930’s (NP) was answered in the
affirmative for several groups of “small” orders. However, already in the case where
G is the cyclic group order 3 acting on Q(t
1
,t
2
,t
3
) by cyclic permutation of the
variables it is not trivial to prove that the fixed field is generated by 3 rational
functions in Q(t
1
,t
2
,t
3
).ThefixedfieldisQ(T
1
,T
2
,T
3
) where
T
1
= t
1
+ t
2
+ t
3
,T
2
= A(t
1
,t
2
,t
3
)/C(t
1
,t
2
,t
3
),T
3
= B(t
1
,t
2
,t
3
/C(t
1
,t
2
,t
3
),
and
A(t
1
,t
2
,t
3
)=3t
2
1
t
2
+3t
2
2
t
3
+3t
2
3
t
1
≠ t
3
1
≠ t
3
2
≠ t
3
3
≠ 6t
1
t
2
t
3
,
B(t
1
,t
2
,t
3
)=3t
1
t
2
2
+3t
2
t
2
3
+3t
3
t
2
1
≠ t
3
1
≠ t
3
2
≠ t
3
3
≠ 6t
1
t
2
t
3
,
C(t
1
,t
2
,t
3
)=t
2
1
+ t
2
2
+ t
2
3
≠ t
1
t
2
≠ t
2
t
3
≠ t
3
t
1
.
In 1925 Ph. Furtwängler [Fu] proved that (NP) could be answered in the affirmative
for all solvable transitive subgroups of S
p
, p being a prime 5 11. He noticed that
his method did not work for p = 47. For some time a (positive) solution of (NP)
was considered to be the ”right” way to solve the inverse problem of Galois theory.
Therefore, it came as a surprise that in 1969 R.Swan [Sw] and independently V.E.
Voskresenskii [Vo] proved that for G equal to the cyclic group C
47
of order 47 the
answer to (NP) is “no”. However, the fact that C
47
can be realized as a Galois group
over Q is, of course, a consequence of theorem 2.1. But the example showed that
Noether’s approach could not lead to a solution of the inverse problem of Galois
theory. For a full list of all references in this area the paper by H. Lenstra [Le] is
recommended. Here is also a complete classification of the abelian groups for which
(NP) has a negative answer and it is shown that the cyclic group of order 8 is the
smallest one.
Normat 3-4/2011 Christian U. Jensen 137
5 Solvable groups and embedding problems
For a long time it was an open problem whether every finite solvable group could
be realized as a Galois group over Q. Since every solvable group has a normal series
with abelian quotients and (theorem 2.1) every abelian group is Galois group over
Q, it may seem surprising that the realizability of solvable groups could be so hard.
However, here it should be taken in consideration that there exist fields K such
that any finite abelian group appears as Galois group over K, but no non-abelian
group does. To be more specific, one has to study ”embedding problems”: Let L/K
be a finite Galois extension with Galois group G and let fi : H æ G be a surjective
homomorphism of the finite group H onto G. One then asks whether there e xists
a Galois extension M/K, M containing L and with Gal(M/K ) ƒ H such that the
restriction map Res : Gal(M/K) æ Gal(L/K)=G corresponds to fi,i.e.such
that Res = fi ¶ „ for some isomorphism „ : Gal(M/K) æ H. This is called the
embedding problem given by L/K and fi. In case of an affirmative answer we say
that M/K is a solution to the embedding problem. We illustrate the above by two
examples.
Examples. Let K = Q and L = Q(
Ô
2) and fi the canonical homorphism for the
cyclic group C
4
of order 4 onto the cyclic group C
2
of order 2. The correponding
embedding problem has a solution with M = Q(
2+
Ô
2).
However, with the same K but L = Q(
Ô
≠1) and the same fi it is just an exercise
in Galois theory to s ee that the corresponding embedding problem has no solution.
Some important embedding problems are solvable. Let G be the Galois group of a
finite Galois extension L/K and let A be a finite group, on which G acts, i.e. there
is given a homomorphism – : G æ Aut(A). One can then define the semi-direct
product A o G as the set of all pairs (a, g) with the group composition
(a
1
,g
1
)(a
2
,g
2
)=(a
1
–
g
1
(a
2
),g
1
g
2
).
This gives rise to a short exact sequence
1 æ A æ
i
æ A o G æ
fi
æ G æ 1,
where i is the injective map sending a to (a, 1), fi is the surjective mapping sending
(a, g) to g and the kernel of fi equals the image of i. One can then consider the
embedding problem with H = A o G and the fi as above.
The following result is basically due to M. Ikeda (Cf.[JLY] p. 108).
Theorem 5.1. Let K be a Hilbertian field (e.g. Q ) and let L/K be a Galois
extension with G as Galois group. Let A be an abelian group on which G acts.
Then the above embedding problem has a solution, that is, there is an (A o G)-
extension of K having L as its G-subextension.
A very special case is the following. Let K be Q, and L an extension of degree
2, i.e. L = Q(
Ô
q),whereq is a rational number which is not a square. Let A be
138 Christian U. Jensen Normat 3-4/2011
the cyclic group C
n
of order n and let the non-trivial element in Gal(Q(
Ô
q)/Q)
act on C
n
by sending each element in C
n
to its inverse. Then the corresp onding
semi-direct product is the dihedral group D
n
of order 2n. The theorem then says
that L = Q(
Ô
q) can be embedded in a D
n
extension. In particular, every dihedral
group is realizable as a Galois group over Q.
As for the realizability of solvable groups the first general step was taken in 1937
by Reichardt[Re] and Scholz[Sco], who considered the case of p-groups, i.e. groups
whose order is a power of a prime p. By solving successive embedding problems
controlling ramifications such that obstructions were eliminated they managed to
prove that for any odd prime p every p-group appears a Galois group over Q.
Shafarevich succeeded in extending this result to the case p =2and finally in
1954 [Sha] obtained the celebrated theorem that every finite solvable group can
be realized as Galois group over Q. The proof was extremely complicated and had
several inaccuracies. The first complete proof, correct in all details, was given in
the book [NSW]
6 Generic polynomials and regular extensions
Instead of just knowing that a finite group G can be realize d as a Galois group
over Q it would be desirable to get some kind of “universal parametrization” of
all G-extensions of Q. To put the question more precisely the following concept is
introduced.
Definition. Let P (t
1
,...,t
n
)(x) be a monic polynomial in Q(t
1
,...,t
n
)[x],where
t
1
,...,t
n
and x are independent variables and let G be a finite group. P (t
1
,...,t
n
)(x)
is called a generic polynomial for the group G, if the following two conditions are
satisfied:
(i) for every field K the splitting field of P (t
1
,...,t
n
)(x) over K(t
1
,...,t
n
) is a
G-extension of K(t
1
,...,t
n
),
(ii) for every field K every G-extension of K is the splitting field over K for
P (a
1
,...,a
n
)(x) for a suitable n-tuple (a
1
,...,a
n
) œ K
n
.
The variables t
1
,...,t
n
are called the parameters.
If there is a G-generic polynomial of the finite group G (with unspecified n), then
theorem 3.2 implies that G is realizable as a Galois group over Q and actually
over any Hilbertian field. It can be proved that if (NP) has a positive solution for
a group G then there exists a G-generic polynomial, but the conve rse is false. As
mentioned earlier (NP) has a negative answer if G is cyclic of order 47, but by the
following result of Lenstra there exists generic polynomial for the cyclic group of
order 47.
Theorem 6.1. There exist G-generic polynomials for a finite abelian group G if
and only if G has no element of order 8.
The theorem shows, in particular, that already for the cyclic group of order 8, the
answer to (NP) is negative. As for dihedral groups less information is available.
Normat 3-4/2011 Christian U. Jensen 139
For the dihedral group D
q
of order 2q, q being an odd number, there exist generic
polynomials, but if q is even there are - so far - only fragmentary results. For a
group which admits generic polynomials it is often quite cumbersome to find them
explicitly. Here are some examples.
The construction in the proof that a positive answer to (NP) for a group G implies
the existence of a G-generic polynomial usually leads to complicated polynomials
with more parameters than necessary. For instance, for S
3
and the cyclic group of
order 3 one would get 3 parameters. By alternative methods one gets much simpler
generic polynomials with just one parameter. Indeed, x
3
+ tx + t is generic for S
3
and x
3
≠tx
2
+(t≠3)x+1 is generic for the cyclic group of order 3. (See e.g. [Se] pp.
1-2 or [JLY], p. 30.) For the cyclic group of order 5 there exists a generic polynomial
with 2 parameters and it can be shown that there is no generic polynomial with
only one parameter.
Both the condition that (NP) has a positive solution for G and the condition that
there exists a G-generic polynomial are quite strong. There is a weaker condition
that is almost as good and for which no group is known for which it does not hold.
Definition. Let t
1
,...,t
n
be indeterminates and G a finite group. A G-extension M
of Q(t
1
,...,t
n
) is called regular if Q is relatively algebraically closed in M,i.e.ifno
element in M \Q is algebraic over Q. We often say that M is a regular G-extension
over Q.
The existence of a G-generic polynomial implies the existence of a regular G-
extension:
Theorem 6.2. Let f (t
1
,...,t
n
,x) œ Q(t
1
,...,t
n
)[x] be a G-generic polynomial.
Then the splitting field of f(t
1
,...,t
n
)(x) over Q(t
1
,...,t
n
) is a regular G-extension
over Q.
We omit the proof. Hence for a finite group G we have the following implications:
(NP) has a positive answer for G ∆ there exists a G-generic polynomial ∆ there
exists a G-regular extension over Q ∆ there exists a G-extension of Q.
Here the first two implications are strict: there exists a generic polynomial for the
cyclic group C
47
of order 47, but Noethers problem has a negative answer for C
47
.
Furthermore for the cyclic group C
8
of order 8, there exists a regular C
8
-extension
of Q, but (by theorem 6.1) no generic C
8
-polynomial. It is an open question whether
the third implication can be reversed: There is a conjecture that this implication
actually is an equivalence. The following two theorems are stability results for
regular extensions.
Theorem 6.3. If G is a finite group for w hich th ere is a regular G-extension over
Q, then for an arbitrary field K there exists a regular G-extension over K, i.e.
there exists a G-extension M of a rational function field K(t
1
,...,t
n
) such that K
is relatively algebraically closed in M .
Theorem 6.4. If G and H are finite groups such that there exists a G-regular
extension and an H-regular extension of Q, then there also exists a (G◊H)-regular
extension of Q.
140 Christian U. Jensen Normat 3-4/2011
Remark. Theorem 5.1 and theorem 3.2 (with Q replaced by a Hilbertian field) show
that if a group G can be realized as Galois group for a regular extension over Q
then G can be realized as a Galois group over any Hilbertian field.
Theorem 5.4 implies that if G and H are finite groups such that there exist regular
G- and regular H-extensions over Q, then there exists a regular (G ◊H)-extension
over Q, in particular G ◊ H is realizable as a Galois group over Q (Cf. also the
remark after theorem 1.1.) Combining the above results it follows that for a finite
group G the existence of a G-regular extension over Q implies the existence of
infinitely many distinct G-extensions of any Hilbertian field. Therefore, the exis-
tence of a regular G-extension of Q yields considerably more information than the
existence of just a G-exte nsion does. It is known that every finite abelian group is
realizable as Galois group for regular extension of Q. So are many dihedral groups,
for instance, the dihedral group D
q
of order 2q for every odd q. However, it is
still unknown whether every finite solvable group can appear as Galois group for a
regular extension over Q, in particular, it is unknown if every finite solvable group
is realizable as Galois group over every Hilbertian field. Clearly every symmetric
group S
n
is realizable as Galois group of regular extension over Q. So is every al-
ternating group A
n
.Indeed,thesplittingfieldoverQ(t) for the polynomial P
n
(t, x)
in section 3 is a regular A
n
-extension of Q(t).
In the last decades large classes of linear groups and non-abelian simple groups
have been realized as Galois groups of regular extensions over Q. We briefly recall
the definitions. For a commutative ring R the special linear group SL
n
(R) of degree
n is the group of all (n ◊ n) matrices with entries in R having determinant 1. The
projective special linear group PSL
n
(R) of degree n is the quotient of SL
n
(R)
with respect to its centre. If R is a field it is a classical result that PSL
n
(R) is a
non-abelian simple group except for the case where n =2and R is a field with
2 or 3 elements. By arithmetic-geometric methods, e.g. by use of elliptic curves
and modular forms, several of the above groups ( for n = 2) have been realized
as Galois groups of regular extensions of Q. To give just one example, Shih [Shi]
proved in 1974 that PSL
2
(F
p
)
1
is realizable as Galois group for a regular extension
of Q provided p is a prime number for which 2, 3 or 7 is a quadratic non-residue
modulo p.
It is well known that every finite group is realizable as a Galois group of the splitting
field over Q(t) for a polynomial with coefficients in Q(t) where Q is the field of all
algebraic numbers. The problem is to obtain conditions which ensure that the
polynomial can be defined over Q. Certain very technical “rigidity” conditions on
the group G imply that G can be realized as Galois group of a regular extensions of
Q(t). A large class of simple groups can in this way be realized as Galois groups of
a regular extension of Q(t), for instance all the sporadic simple groups except the
Mathieu group M
23
, (but including the monster group!) and many classical groups
of Lie type and some exce ptional groups of Lie type. For a full account of these
results we refer to the monograph [MM].
1
If p is a prime number F
p
n
denotes the field with p
n
elements
Normat 3-4/2011 Christian U. Jensen 141
7 Explicit examples
The previous sections mainly dealt with the mere existence of polynomials whose
splitting fields had some prescribed Galois groups. Hilbert proved that all symmet-
ric groups S
n
and all alternating groups A
n
can be realized as Galois groups over
Q. However, he gave no explicit numerical examples. Issai Schur (1930-31)[Scu]
gave some explicit examples. The Galois group of the splitting field over Q for the
n-th Laguerre polynomial
L
n
(x)=
e
x
n!
d
n
(x
n
e
≠x
)
dx
n
=1≠
3
n
1
4
x
1!
+
3
n
2
4
x
2
2!
≠···+(≠1)
n
x
n
n!
is the symmetric group S
n
. He also proved that the Galois group of the splitting
field over Q for the polynomial
E
n
(x)=1+
x
1!
+
x
2
2!
+ ···+
x
n
n!
is A
n
for n © 0 mod 4 and S
n
for n ”© 0 mod 4.
In some sense “most” polynomials of degree n in Q[x] have S
n
as Galois group. To
be more precise: let g(n, t) be the number of monic polynomials in Z[x] of degree
n such that the absolute value of each coefficient is 5 t. Let h(n, t) be the number
of the above polynomials whose Galois group over Q is S
n
. By a famous result of
van der Waerden[Wa] the quotient h(n, t)/g(n, t) converges to 1 as t æŒ.Soto
realize that the Galois group of some polynomial is a symmetric group, is from a
“statistical” point of view not surprising, but the actual verification may require
some effort. The simplest example of a polynomial of degree n such that the Galois
group over Q - for every n -isS
n
is x
n
≠x ≠1. The proof that this holds for all n
depends on non-trivial results from algebraic number theory.
As mentioned in section 6 Shih[Shi] proved in 1974 that PSL
2
(F
p
) is realizable as
the Galois group of a regular extension of Q if 2, 3 or 7 is a quadratic non-residue
modulo p. This, in particular, applies to PSL
2
(F
7
). This group has order 168
and is the smallest simple group which is neither c yclic nor an alternating group.
Until 1970 PSL
2
(F
7
) was the smallest simple group that has not been realized as
a Galois group over Q. In the search for a septimic polynomial with this group
as a Galois group mathematicians from Karlsruhe University applied a “trial and
error” method, i.e. thousands of septimic polynomials were computer tested and
finally the polynomial T (x)=x
7
≠ 3x +7was found to be a good candidate: it is
irreducible, has exactly 3 real roots and the discriminant is the square of rational
number. Hence from knowledge of the transitive subgroups of S
7
it follows that the
Galois group of T (x) is either the alternating group A
7
or PSL(2, F
7
). But it was
then a hard task to rule out A
7
, before PSL(2, F
7
) was proved to be the Galois
group of T (x) over Q. To-day computer programs have been developed which in a
few seconds can compute Galois groups of irreducible polynomials over Q of degree
5 11.
For an irreducible polynomial p(x) œ Q[ x] the Galois permutation group over Q
is a transitive subgroup of S
n
(cf. theorem 1.2). This group is determined up to
142 Christian U. Jensen Normat 3-4/2011
conjugacy. The transitive subgroups of S
n
, n 5 12, have been classified up to
conjugacy. All these groups appear as permutation Galois groups of explicitly known
polynomials in Q[x]. A complete list of these polynomials can be found in [MM].
At the moment it is difficult to say which is the smallest group that so far has
not be en realized as Galois group over Q.Butitseems that it might be a certain
extension of the group PSL
2
(F
16
) of degree 2, which has order 8160.
8 Concluding Remarks
If K is a field such that every finite group is re alizable as a Galois group over K
then it is easy to see that every finite group is realizable as a Galois group over
any finite field extension L of K. (Use e.g. theorem 1.3 and proposition 2.2.) In
particular, if the inverse problem of Galois theory has a positive solution, then
every finite group appears as a Galois group over every algebraic number field of
finite degree.
However, there exists a finite field extension L/K such that every finite group is
a Galois group over L but not every finite group is a Galois group over K. An
amusing example of this phenomenon is due to Florian Pop [Po]. Recall that an
algebraic number – is called totally real, if – and all its conjugates are real, in other
words, if all the roots of –´s minimal polynomial over Q are real. The totally real
algebraic numbers form a subfield Q
tr
of the field, Q of all algebraic numbers. Not
every finite group is realizable as a Galois group over Q
tr
, for instance no group
of odd order (> 1) is realizable; more precisely, the only groups that are realizable
as Galois groups over Q
tr
are the groups which can be generated by involutions
(i.e. the elements of order 2). However, over the field Q
tr
(
Ô
≠1), which has degree
2overQ
tr
, every finite group is realizable as a Galois group!
Taking into ac count that the finite simple groups are the building blocks for all
finite groups the natural strategy concerning the inverse problem of Galois theory
has been to start by realizing the simple groups as Galois groups and then gradually
building up bigger Galois extensions. So far many simple groups have been realize d
as Galois groups over Q. One might think that if a large family of simple groups
have been realized as Galois groups over Q then the remaining simple groups would
follow automatically. But this is not the case. Indeed, if we divide the family of
finite simple groups arbitrarily into disjoint classes A and B,thenthereexistsa
field K such that every group in A is realizable as a Galois group over K while
no group in B is. Similarly, even if one had realized all finite s imple groups as
Galois groups over Q the remaining finite groups would not follow automatically.
Actually, even “worse” things can happen. By the length of a finite group G we
mean the length of one (and then of any) composition series of G. Then for any
positive integer t there exists a field K
t
such that every group of length 5 t appears
as a Galois group over K
t
, but there exists a group of length t +1 that does not
appear a Galois group over K
t
. So roughly speaking there is no easy short cut to
the solution of the inverse problem of Galois theory.
Normat 3-4/2011 Christian U. Jensen 143
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