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Finding Galois Groups and Subgroups of a Splitting Field of Degree 8

Date: 03/29/2010 at 18:02:45
From: Greeg
Subject: A Question in Galois theory

I have a problem I have been working on for a bit. The problem is to find
the Galois group and all intermediate fields of the splitting field of the
polynomial over Q with one root being sqrt(2 + sqrt(6)).

I know that -- because this polynomial is irreducible -- the Galois group
has to be a subgroup of S_4, the symmetric group on 4 variables.

The first thing that confuses me is that the roots of this polynomial are

   sqrt(2 - sqrt(6))
   -sqrt(2 - sqrt(6))
   sqrt(2 + sqrt(6))      and 
   -sqrt(2 + sqrt(6))

   a = sqrt(2 + sqrt(6))  and
   b = sqrt(2 - sqrt (6)), 
we have that the splitting field is Q(a,b).

But unless I'm mistaken, this extension is not equivalent to Q(a,i), which
means it has degree 16 -- contradicting the fact that

   gal(Q(a,b)/Q) <=S_4

Secondly, I'm somewhat confused as to how you would go about calculating
the actual Galois group in this example.

The more I think about it, the more confused I get.

Date: 03/30/2010 at 02:30:16
From: Doctor Jacques
Subject: Re: A Question in Galois theory

Hi Greeg,

Q(a) and Q(b) are both quadratic extensions of Q(sqrt(6)). This means that
the splitting field Q(a,b) is an extension of degree 4 if Q(a) = Q(b), and
of degree 8 otherwise.

You will find a discussion of this problem in the following article:

Splitting Fields of Quartic Polynomials 

Please feel free to write back if you require further assistance.

- Doctor Jacques, The Math Forum 

Date: 03/30/2010 at 23:57:02
From: Greeg
Subject: A Question in Galois theory


In this case, the splitting field is going to be of degree 8. How would
you go about finding these Galois groups?

In the link you forwarded, the examples deal with Galois groups of
splitting fields of polynomials of degree 4, not 8.

After doing some further reading, the way I tried to go about it  was to
call the 8 generators of the Galois group s_1,...,s_8; then note that

   s_i(Q(a,b)) = Q(s_i(a),s_i(b)) = Q(a,b);
and in that way, find the permutations of the roots in S_4 that each of
the s_i represent.

But I still get confused about some of the group elements.

Is there an easier way to go about getting the group and the intermediate
fields each subgroup relates to?

Date: 03/31/2010 at 04:57:28
From: Doctor Jacques
Subject: Re: A Question in Galois theory

Hi Greeg,

In this case, the minimal polynomial has degree 4. One way to compute it
is to let y = x^2. The minimal polynomial in y is a quadratic with roots:

   a^2 = 2 + sqrt(6)
   b^2 = 2 - sqrt(6)

And this polynomial is:

   y^2 - (a^2 + b^2)y + a^2b^2


   y^2 - 4y - 2               [1]

In terms of x, this gives the minimal polynomial:

   x^4 - 4x^2 - 2             [2]

To find the splitting field of [2], we use the usual technique: we first
find the roots of [1], giving a quadratic extension Q(sqrt(6)).

Each of the roots (+/-)a, (+/-)b of [2] is the solution of a quadratic
equation (x^2 = y) over Q(sqrt(6)). This shows that we have a tower of
extensions ...

   Q < Q(sqrt(6)) <= Q(a) <= Q(b)

... where each of the last two extensions has degree at most 2. Now,
Q(a)/Q has degree 4, since the polynomial [2] is irreducible; this shows
that Q(a)/Q(sqrt(6) has degree 2. The only remaining question is whether
Q(b)/Q(a) has degree 1 or 2. In this case, the answer is easy: as Q(a) is
real and Q(b) is not, we have a proper extension and

   [Q(b):Q(a)] = 2, 
which implies that 

   [Q(a,b):Q] = 8
The Galois group G has order 8.

G is a subgroup of S_4, because [2] is irreducible. Now, all the subgroups
of order 8 in S_4 are isomorphic to D_4 (the symmetry group of a square).
You can check this directly, or notice that -- as a subgroup of order 8 is
a Sylow 2-subgroup of S_4 -- all these subgroups are conjugate to each
other, and therefore isomorphic.

One of the three subgroups of order 8 in S_4 is the subgroup generated by
the permutations (1,2,3,4) and (1,2) (the other subgroups correspond to a
different numbering of the elements).

Let us now find the subgroups of G and the corresponding subfields. I'm
afraid there is no really short and easy way to do this; it requires some
intuition and some amount of work. (On the other hand, once you know the
answer, it is relatively easy to prove that it is correct.)

To find the subgroups of G, it may be convenient to use the geometric
representation of G as the symmetry group of a square whose vertices are
the roots of [2].

The first problem is to find in what order we should place the roots on
the vertices of the square. In D_4, we have two transpositions,
corresponding to reflections across the diagonals of the square. In our
equation, we have two automorphisms that permute two roots: the
transpositions (a, -a) and (b, -b). This shows that we should arrange our
roots as follows:

  a      b
  |      |
  |      |
 -b      -a

G has the following subgroups:

(1) H_0 = {()}, the trivial subgroup.

(2) H_1 = {(),(a,-a)} and H_2 = {(),(b,-b)}. Each of these subgroups of
order 2 is generated by a reflection across a diagonal.

(3) H_3 = {(),(a,b)(-a,-b)} and H_4 = {(),(a,-b)(-a,b)}. Each of these
subgroups of order 2 is generated by a reflection across a median.

(4) H_5 = {(),(a,-a)(b,-b)}. This subgroup of order 2 is generated by a
central symmetry (or a rotation by 180 degrees).

(5) H_6 = {(),(a,-a),(b,-b),(a,-a)(b,-b)}. This subgroup of order 4 is
generated by the reflections across the diagonals; i.e., it is generated
by H_1 and H_2.

(6) H_7 = {(),(a,b)(-a,-b),(a,-b)(-a,b),(a,-a)(b,-b)}. This subgroup of
order 4 is generated by the reflections across the medians, i.e., it is
generated by H_3 and H_4.

(7) H_8 = {(),(a,b,-a,-b),(a,-a)(b,-b),(a,-b,-a,b)}. This subgroup of
order 4 is generated by a rotation of 90 degrees.

(8) G itself.

To find the corresponding subfields, we look in each case for an
expression that is invariant under the permutations of the corresponding
subgroup (but not under any larger subgroup); we also use the inclusion
relations between the subgroups to infer relations between the subfields.
We already know that the fixed field of H_0 is the whole extension Q(a,b),
and the fixed field of G is Q. For each subgroup H, we will write Fix(H)
for the fixed field of H.

H_1 fixes b and -b. This implies that Fix(H_1) must contain Q(b). As the
subfield must have degree |G|/|H_1| = 4, and Q(b) has degree 4, the fixed
field is Q(b). In the same way, Fix(H_2) is Q(a).

As H_6 = <H_1,H_2>, Fix(H_6) is the intersection of the above two fields,
Q(a) and Q(b). Both Q(a) and Q(b) contain

   Q(a^2) = Q(b^2) = Q(sqrt(6)).
As we are looking for an extension of degree 2, Fix(H_6) is Q(sqrt(6)).

H_3 exchanges a and b, and therefore fixes a + b; Fix(H_3) (of order 4)
must contain Q(a + b). As Q(a + b) is of degree 4, we conclude that
Fix(H_3) is Q(a + b). (With some work, it is possible to compute the
minimal polynomial of (a + b). It is x^4 - 8x^2 + 24.)

H_4 exchanges a and -b, and therefore fixes a - b. Using the same argument
as for H_3, we find that Fix(H_4) is Q(a - b). Note that the minimal
polynomial of (a - b) is the same as that of (a + b), because the
subfields are isomorphic, although distinct.

As H_7 = <H_3,H_4> the fixed field of H_7 (of degree 2) is the
intersection of Fix(H_3) and Fix(H_4). We can check that both H_3 and H_4
leave (a + b)^2 invariant. We compute:

 (a + b)^2 = a^2 + b^2 + 2ab
           = 2 + sqrt(6) + 2 - sqrt(6) + 2sqrt(2^2 - 6)
           = 4 + 2sqrt(-2)

This shows that Fix(H_7) is Q(sqrt(-2)).

H_5 is the intersection of H_6 and H_7. The corresponding subfield
Fix(H_5) (of degree 4) is therefore the subfield generated by Fix(H_6) =
Q(sqrt(6)) and Fix(H_7) = Q(sqrt(-2)), i.e.,

   Fix(H_5) = Q(sqrt(6),sqrt(-2)).

Finally, we want to find the fixed field of H_8; this extension has degree
2. We try to find an expression invariant under H_8 (and no larger
subgroup). We notice that the cyclic permutation s = (a,b,-a,-b), the
generator of H_8, changes a + b into b - a, and changes a - b into a + b.
This shows that s changes the sign of (a^2 - b^2).

To get rid of the change of sign, we could try squaring, but

   (a^2 - b^2)^2 = 24
This is invariant under the whole group. Another possibility is to notice
that s also changes the sign of ab, and therefore the expression ab(a^2 -
b^2) is invariant under H_8, and is contained in Fix(H_8).

We compute:

  ab(a^2 - b^2) = sqrt(-2)*2sqrt(6)
                = 4sqrt(-3)

As this generates an extension of degree 2 (the index of H_8), we have

   Fix(H_8) = Q(sqrt(-3)

We could also have found Fix(H_8) using a little trick. As H_8 contains
H_5, Fix(H_8) is a subfield of

   Fix(H_5) = Q(sqrt(6),sqrt(-2)).

Now, this field contains Q(sqrt(-3)), and, as we only have one remaining
subgroup H_8, it must correspond to that subgroup.

Well, I understand that this looks rather complicated, but as I said
before, there is no general way to easily compute Galois groups and
subfields (except in some simple cases; this case still falls in the
"relatively easy" category!). There are (complicated) computer algorithms
for this, but in practice, they are limited to polynomials of degree about
15, which confirms that you should not expect an easy solution.

- Doctor Jacques, The Math Forum 

Date: 04/01/2010 at 16:29:47
From: Greeg
Subject: Thank you (A Question in Galois theory)

Thanks for the help. It gave me a lot of insight.
Associated Topics:
College Modern Algebra

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