1. Note first that G is naturally a subset of the set of
all functions
from L to L. The set of all such functions forms a group, where the
multiplication is defined as composition of functions.
Hence it suffices to prove that G is closed under multiplication and
inverses, that is that the composition and inverse of an automorphism
that fixes K, is an automorphism that fixes K. But the composition
and inverse of a function that fixes K certainly fixes K and it is not
hard to check that composition and inverse of an automorphism is an
automorphism.
2. Clearly H is a subset of G. On the other hand H = Gal(L/M), so
that H is also a group. Thus H is a subgroup.
3. As G fixes K, clearly K 
M
L, and so it suffices to prove
that M is closed under addition, additive inverses , multiplication and
multiplicative inverses. We check closure under addition; the other
cases are just as straightforward. Suppose that m and n ∈ M. We
have to check that m + n is fixed by every element of H. Pick
∈ H.
Then (m) = m and
(n) = n.
As
was arbitrary, m + n ∈ M, as required.
4. Suppose that m ∈ LK. Then
(m) = m, for every
∈ K. As
H K, it fol lows that
(m) = m, for every
∈ H. Thus m ∈ LH and
so 
.
Suppose that
∈ Gal(L/N). Then
(n) = n, for every n ∈ N. As
M N, it follows that
(m) = m, for every m ∈
M. Thus
∈
Gal(L/M) and so Gal(L/N) Gal(L/M).
5. Pick
∈ H. Then
(m) = m, for every m ∈ LH = M. Thus
∈ Gal(L/M) = K. Hence H K.
6. Pick m ∈M. Then
(m) = m, for all
∈ Gal(L/M). Thus m ∈ N
and M N.
7. We have already seen that the degree of L/Q is eight. Thus the
degree of an in termediary field M is two or four . Note that
are all intermediary fields of degree two, as each field
is generated by
an element α, whose square is rational. Conversely, we have seen in
class that if M/Q has degree two, then there is an α ∈ M such that
M = Q( α), where α2 ∈ Q. Now, by repeated application of the tower
law , a basis for L/Q is given as
Thus a general element α of L is given by
for some rational numbers a, b, c, d, e, f, g and h. It is easy to see that
the only way the square is rational is if there are no cross-terms, so
that all but one coefficient is in fact zero. Thus the list above certainly
exhausts all possible quadratic intermediary fields.
Now consider classifying all quartic extensions. Again it is easy to write
down quite a few examples of quartic extensions
Suppose that M is a quartic extension that contains
. Then M/Q()
is quadratic. Arguing as before, it follows that M is generated over
Q() be an element α whose square lies in
Q(). Now a basis for
L/Q( ) is given by 1,
and 
.
Thus
As before it is easy to see that the condition that α2 lies in Q()
implies that all cross-terms vanish, so that all but one coefficient is
zero. Arguing similarly for the other quadratic extensions, it follows
that the list given above exhausts all quartic extensions provided we
can show that a quartic extension must contain a quadratic one.
Now we turn to the problem of computing the Galois group and all of
its subgroups. Let
be an automorphism of L. Then
is determined
by its action on the generators of L. Now 
must a root of x 2 −2,
so that
and 
In particular G has at most eight elements. Now we turn to the problem
of showing that there is an automorphism 
which switches the sign
of and fixes the other two signs. Note first
that we may find
such that 
Indeed
and − are
both roots
of the same ir reducible polynomial . Now the extension L/Q(
is
a splitting field for (x2 − 3)(x2 − 5). It follows that we may find an
extension of σ that
fixes 
and 
.
Similarly we may find 
and
. Let H be the group they generate
inside G. Note that each of these elements has degree two and that
they commute with each other. It follows that H is an abelian group,
isomorphic to Z2 × Z2× Z2. As H and G both have eight elements, in
fact H = G.
Thus G has seven subgroups of order two , given by the seven non-zero
elements of G, each of which has order two. To find all subgroups of
order four is a little more involved. The trick is to identify G with
and to realise that a subgroup of order four corresponds to a linear
subspace of dimension two. But a plane in a three dimensional space
is the same as a line in the dual space, and so there are seven planes
as well, that is seven subgroups of order four.
Let M be an intermediary field of order 4. Let H = Gal(L/M). Then
H is a subgroup of G. Now L/M is quadratic and so there is a β∈ L
such that L = M( β ), where β2 ∈M. Thus the order of H is two (one
can either change the sign of β or not). It follows that M
LH. By
inspection, LH is always a quartic extension of K, and so M = LH.
But then there are at most seven quartic extensions.
We note that the lattice of subgroups and intermediary fields are the
same (but with an upside-down identification).
8.
t3 − 1 = (t − 1)(t2 + t + 1),
and t2 +t+1 is irreducible. Thus it suffices to find a splitting field for
t2 + t + 1. Let ω be a root of t2 + t + 1. Then ω3 = 1 and so ω2 is the
other root,
.
Thus Q(ω) is in fact a splitting field for t3 − 1. The degree of the
extension is two. Of course we may always re present ω as exp(2π /3).
Consider t4 + 5t2 + 6. This factors as
(t2 + 2)(t2 + 3),
so that 
. Just as the case with
, the degree of
this field extension is 4.
Finally consider t6 − 8. Suppose that ω is a primitive sixth root of
unity, and let α be a positive sixth root of 8, that is a positive square
root of 2. Then the roots of t6 − 8 are ωiα, where i
ranges from 0 up
to 5. Thus a splitting field is given as Q( α, ω). Now ω is a cube root
of −1. Put differently ,
x6 − 1 = (x3 − 1)(x3 + 1),
and the roots of the first factor are cube roots of 1, so that ω is a root
of the second factor. It is easy to check that x3 + 1 is irreducible over
Q. Thus ω has degree three over Q. On the other hand
has degree
two over Q. As 2 and 3 are coprime the degree of the splitting field is
6.
9. Note first that any two splitting fields define isomorphic extensions
of K, so that we are free to go through the construction given in class
and check the result for the splitting field so constructed.
We proceed by induction on n. If n = 1, then f is linear, f splits in L
and so L = K. Now suppose that n > 1. Suppose that f is reducible,
so that f = gh, where g and h have degree a and b respectively. Then
n = a + b. Let M/K be a splitting field for g over K and let L/M
be a splitting field for h over M. Then L/K is a splitting field for the
product f. By induction
[M : K]|a! and [L : M]|b!.
Thus
[L : K] = [L : M][M : K]|a!b!
and as a!b! divides (a + b)! = n!, we are done in this case.
Finally suppose that f is irreducible. We first adjoin a root α of f.
Thus we get a field extension K( α )/K, which has degree n, as f is
irreducible. Now
f(x) = (x −α )g(x),
where g(x) ∈ K(α )[x]. As the degree of g(x) is n − 1, by induction
[L : K(α )]|(n − 1)!.
Arguing as before, the result follows.
