Tutorial 7 : Galois Groups of Quartics and Solvability by
Radicals

1. Determine the Galois groups of the quartics:
and

(1) Put
. The roots of are
. Hence
splitting field of which has degree over
. Thus
.
(2) Put denote the resolvent cubic.
disc
and
. Thus
or. But splitting field is
which has degree over
.
Hence .
(3) Let
Then disc
and
is
irreducible over
. Hence .
(4) Let
Then disc
and
is
irreducible over
. Hence .
(5) Let
. A splitting field of is
. Hence
and
.
(6) Let
Then disc
. Moreover
is irreducible by
irreducibility
test. Hence is irreducible in
.Hence

2. Show that the resolvent cubic of
is
and

he resolvent cubic is and disc .Thus .

3. Show that the Galois group of an irreducible quartic
in
with exactly two real roots is either or

The degree of the splitting field of irreducible quartic over
is
atleast
Indeed, let be a real root and be a complex root of Then
Therefore we must rule out as the
Galois group. For this we show
that
disc
is not a square in
Let
be all the roots of . Note that

disc

Hence
disc Therefore
.

4. Let be a real root of an irreducible rational quartic
whose resolvent cubic is irreducible. Show that is not constructible
by ruler and compass. Can we construct the roots of the quartic by ruler and compass ?

Total charge is obtained by integrating the charge density over the volume, . The amount of charge enclosed inside a spherical surface of radius r is . Using Gauss’s law, the electric field inside the sphere is and outside the sphere it is .. Because of spherical symmetry, the gradient is simply a differentiation with respect to r. Inside the sphere, , where C is a constant Outside the sphere, the potential is , with C’ being constant. The potential being continuous at the surface r=R, the constants must be chosen so that Thus .

5. Show that
is irreducible over
and

Note that
is irreducible mod . Hence it is irreducible over
Its
resolvent cubic
is

The roots of are
Note that
the splitting field of Thus
has degree over
which implies that
.

6. Find sufficient conditions on the integers and so that
is a Galois extension of
with cyclic
Galois group of order

Let be nonzero integers and put

Then
Therefore
. Thus
is
splitting field
of a polynomial over
with roots
Check that
Consider
the automorphism of defined by
and
Then

Theferefore the Galois group is

7. Let
be an irreducible polynomial of prime degree Suppose that has exactly
two non-real roots. Show that is not solvable by radicals
over

Note that is irreducible mod . Hence it is irreducible over
Its
resolvent cubic
is

The roots of are
Note that
the splitting field of Thus
has degree over
which implies that
.

8. Let
be indeterminates and let
be the elementary symmetric polynomials of
Show that
is not a radical extension of
but
is a radical extension of

Let be the square root of the discriminant of
the polynomial
We know that is a splitting
field of over and
Moreover
Hence
is a simple radical extension. We show that is not a
simple radical extension. If it were, then is a splitting field of
an irreducible cubic
for some
But
is a Galois extension. Hence all the roots of are in If
are any two roots of then
Hence is a
primitive cube root over
This is a contradiction. Similar
argument shows that the other cubic extensions of in are not
simple radical extensions. Hence is not a radical extension.

9. Let be the Galois group of an irreducible quintic over
Show that or if has an element of order

Let be an irreducible quintic over
Then its Galois
group
is a transitive subgroup of Hence we see that the order of is a
multiple
of Thus
or We must show that it cannot have
order
or If the order is then it is cyclic. But there is no
permutation
of oder in
Suppose Then due to simplicity of is not a subgroup of
Hence
has an odd permutation say Then is either a product of
two disjoint
cycles of order and or a -cycle. In the latter case
which is
not possible. In the former case is a transposition. But such a group
is