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### Constructibility and Galois Groups

```Date: 04/30/2005 at 12:44:22
From: Erika
Subject: constructibility and Galois groups

Let a be a complex number and a root of an irreducible polynomial f
over the rationals.  Show that a is constructible if and only if the
Galois group of f is a 2-group.

I have proven that if the group is a 2-group, then a is constructible,
but I'm having trouble with the other direction.

So, if a is constructible, Q(a) is an extension of degree power of 2
(f has degree power of 2) but Q(a) is not necessarily a splitting
field.  I don't know what I can say about the Galois group.  Please
help!

```

```
Date: 05/01/2005 at 05:09:18
From: Doctor Jacques
Subject: Re: constructibility and Galois groups

Hi Erika,

First, note that we must understand "constructible" in the wide sense.
For example, Q(i) is a Galois extension of degree 2, but we cannot
really say that i is constructible with ruler and compass in the
geometric sense--we must understand that a number is constructible if
it can be computed only by taking square roots (it would be more

We must show that the degree of the splitting field of f is a power
of 2.

We can compute a root 'a' of f by constructing a chain of quadratic
extensions:

Q < Q(a1) < Q(a2) < ... Q(an)     [1]

where an = a.  We can compute any other root 'b' of f by choosing the
other sign in some of the square roots, giving another chain of

Q < Q(b1) < Q(b2) < ... Q(bn)

where bn = b; note that Q(b1) = Q(a1), but this will not necessarily
be true for the other extensions.  For example, if we consider the
polynomial:

f(x) = x^4 - 2*x^2 - 5

the four roots of f(x) are given by sqrt(1 + sqrt(6)), where we can
take both signs for the square roots.  In this case, the extensions
are:

Q < Q(sqrt(6)) < Q(sqrt(1+sqrt(6)))
Q < Q(sqrt(6)) < Q(sqrt(1-sqrt(6)))

and the latter two extensions are not the same, since the first one
is real and the second one is not (in this case, the Galois group is
D4, the dihedral group of order 8).

The trick is to build the splitting field by merging the chains like
[1] together.  More precisely, we build a chain:

Q < Q(a1) < Q(a1,a2) < Q(a1,a2,b2) < Q(a1,a2,b2,a3), ...   [2]

The chain [2] will terminate in the splitting field of f.

We know that Q(a1,b2) is of degree 2 over Q(a1).  This means that
Q(a1,a2,b2) is of degree 1 or 2 over Q(a1,a2), because b2 satisfies a
equation (not necessarily irreducible) over Q(a1,a2).

In general, depending on the degree of f, we can have more than two
chains like [1], and we can merge them together by including elements
from one level at a time--each new element is quadratic over a
previous extension, and therefore also at most quadratic over the
extension just before it in the merged chain.

This shows that all the extensions in the chain [2] are of degree 1 or
2, and therefore the degree of splitting field is a power of 2.

more, or if you have any other questions.

- Doctor Jacques, The Math Forum
http://mathforum.org/dr.math/

```

```
Date: 05/03/2005 at 02:10:32
From: Erika
Subject: constructibility and galois groups

Hi, thanks for your help.  There's little thing I'm still stuck on.  I
understand that we can build that chain of subfields for a, because a
is constructible, but why can we do it for the other roots?  Are the
other roots all constructible as well?  If so, why?  If not, why can
we do that?

Thanks!

Erika

```

```
Date: 05/03/2005 at 02:32:58
From: Doctor Jacques
Subject: Re: constructibility and galois groups

Hi again Erika,

All the roots of f are constructible, since the only definition of 'a'
is that it is a root of the irreducible polynomial f--loosely
speaking, we don't know in advance which root we will get.  Choosing
different signs for the square roots at each step will give you all
the roots of f.

Note that it is important that we understand "constructible" in the
algebraic sense, not the geometric sense.  For example, the
bi-quadratic equation I gave you has only two real roots.  If we
attempt to construct the roots geometrically, we will first intersect
a line and a circle, giving two choices for the first extension.

For one of these choices, we can find a first pair of roots by another
similar construction.  However, if we select the other intersection in
the first step, and attempt to find the other two roots, we will end
up trying to find the intersection of a line and a circle that do not
intersect each other.

For the purpose of computing Galois groups, however, this is
irrelevant--all operations take place in the splitting field of f, and
we are not restricted to the real part of it.

- Doctor Jacques, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Modern Algebra

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