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### Splitting Fields of Quartic Polynomials

```Date: 05/14/2004 at 14:54:50
From: Nina
Subject: Splitting Fields

I found your web really useful and interesting for getting ideas and
wondering if I've gotten it right or not.  I picked the irreducible
polynomial x^4 - 8x^2 + 8 and I tried to find its splitting field (E)
and G(E/Q) set, but I got stuck after finding the roots.  Could you

```

```
Date: 05/16/2004 at 08:58:31
From: Doctor Jacques
Subject: Re: Splitting Fields

Hi Nina,

The computation of Galois groups in general is a very complex
question; special, ad hoc, techniques are available for specific
cases, and the bi-quadratic equation is one of them.

In general, the splitting field of an irreducible polynomial of degree
n can have degree up to n!; the Galois group is a subgroup of S_n.
For a general quartic, the degree of the splitting field is therefore
a divisor of 24.

However, we can reduce this in the case of a bi-quadratic equation.

f(x) = x^4 - 8*x^2 + 8

The obvious way to solve this is to let y = x^2, solve for y, and
compute x = (+/-) sqrt(y_i) for each root y_i.  Let us look closely at
what happens.

We start by solving y^2 - 8*y + 8 = 0. This gives us two roots: y1 =
4 + sqrt(2) and y2 = 4 - sqrt(2).  We have created an extension of
degree 2, namely Q(sqrt(2)).

We then proceed to solve the equations x^2 = y1 and x^2 = y2.  Each of
these equations can produce an extension of degree 2, although these
extensions could be the same.  This means that we end up with an
extension of degree 4 or 8 (the degree cannot be less than 4, since
the polynomial is irreducible; the first root generates an extension
of degree 4).

Now, up to isomorphism, S4 contains two subgroups of degree 4 (C4 and
C2 x C2) and one subgroup of degree 8 (D4).  There are therefore three
possible Galois groups.

In our example, the four roots of f(x) are:

a = sqrt(4 + 2*sqrt(2))
b = sqrt(4 - 2*sqrt(2))
c = -sqrt(4 + 2*sqrt(2)) = -a
d = -sqrt(4 - 2*sqrt(2)) = -b

Of course, a and c are in the same extension, and the same is true for
b and d.  The first question to ask is: does b belong to Q(a)?  If it
does, Q(a) contains all the roots and is the splitting field (of
degree 4); otherwise, the splitting field is Q(a,b) and has degree 8.

b belongs to Q(a) if and only if ab does.  We compute:

ab = sqrt((4 + 2*sqrt(2))(4 - 2*sqrt(2)))
= 2*sqrt(2)

and this is in Q(a), since a^2 = 4 + 2*sqrt(2). Specifically, we have:

b = a - 4/a = s(a)             [1]

We know therefore that the extension is of degree 4, generated by a
(or any root), and the Galois group can be C4 or C2 x C2.

One way to decide this is to look at the automorphism [1].  If we
apply it again, with a little calculation, you will find that:

(s o s)(a) = s(b) = c

and this shows that the order of s is not 2, which rules out C2 x C2;
the Galois group is the cyclic group C4 in this case.

There is another way to decide this; it may look a little more
complicated, but it makes the generalization easier.  The idea is to
look at the action of C2 x C2 on the following expressions:

u = (a + b)(c + d)
v = (a + c)(b + d)
w = (a + d)(b + c)

If the Galois group is C2 x C2, it contains the permutations (a,b)
(c,d), (a,c)(b,d), and (a,d)(b,c) (the action must be transitive,
since f is irreducible).

It is easy to see that u, v, and w are fixed by those permutations.
If this is indeed the Galois group, then, by definition, u, v, and w
must be rational.

We compute:

u = -(8 + 4*sqrt(2))
v = 0
w = -(8 - 4*sqrt(2))

As u and w are irrational, this confirms that the Galois group is not
C2 x C2.

For examples of the other cases, let us look first at:

f(x) = x^4 - 10*x^2 + 1

The roots are, in this case:

a = sqrt(2) + sqrt(3)
b = sqrt(2) - sqrt(3)
c = -sqrt(2) - sqrt(3) = -a
d = -sqrt(2) + sqrt(3) = -b

We have ab = 1, and this is certainly in Q(a), so the extension is of
degree 4.  We compute then:

u = (a + b)(c + d) = -8
v = (a + c)(b + d) = 0
w = (a + d)(b + c) = -12

and, as u, v, and w are rational, we conclude that the Galois group
is C2 x C2.  This is not surprising; intuitively, you can interchange
sqrt(2) and -sqrt(2) on the one hand, and sqrt(3) and -sqrt(3) on the
other hand, independently of each other.

For a final example, let us look at:

f(x) = x^4 - 6*x^2 + 2

The roots are, in this case:

a = sqrt(3 + sqrt(7));
b = sqrt(3 - sqrt(7));
c = -a
d = -b

We compute:

ab = sqrt(9 - 7) = sqrt(2)

and this is not in Q(a) (you cannot produce a sqrt(2) using powers of
a).  This shows that the extension is of degree 8 in this case, and
the Galois group is D4, the only type of subgroup of order 8 in S4.
In this case, you cannot generate the extension using only a single
root of f(x), although you can generate it using a single element
which is a polynomial in a and b--this element will have a minimal
polynomial of degree 8.  For example, the element:

y = a + ab = a + sqrt(2)

has the minimal polynomial:

y^8 - 20*y^6 + 88*y^4 - 104*y^2 + 36

(you need special software to produce this...)

If you are really interested in this, you may try to address the
general case of the polynomial:

f(x) = x^4 - 2*p*x^2 + q

You should try to prove that, assuming f is irreducible,

* If q is the square of a rational number, the group is C2 x C2

* Else, if (p^2/q - 1) is the square of a rational number, the group
is C4

* Else, the group is D4 and the extension is of degree 8.

more, or if you have any other questions.

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

```

```
Date: 05/17/2004 at 13:25:48
From: Nina
Subject: Splitting field

```

```
Date: 05/09/2007 at 18:14:37
From: Edie
Subject: Splitting Fields of Quartic Polynomials again

you could prove that last assertion about the general case of the
polynomial f(x) = x^4 - 2*p*x^2 + q?

Thanks so much!!

```

```
Date: 05/10/2007 at 03:47:53
From: Doctor Jacques
Subject: Re: Splitting Fields of Quartic Polynomials again

Hi Edie,

The last statements are just a generalization of the technique used
in the text.

The roots of the equation are:

a = sqrt ( p + sqrt (p^2 - q) )
b = sqrt ( p - sqrt (p^2 - q) )
c = -a
d = -b

Note that sqrt(p^2 - q) is irrational, because we assumed that the
polynomial is irreducible.

Q(a) and Q(b) are extensions of degree 4.  We must first decide
whether these extensions are the same: if this is the case, the
Galois group will have order 4, otherwise it will have order 8 (and
will be isomorphic to D4, since this is the only type of subgroup of
order 8 in S4).

As explained in the text, b belongs to Q(a) if and only if ab does.
Now, we have ab = sqrt(q).

We must therefore decide if sqrt(q) belongs to Q(a).  To do this, we
consider the two possible Galois groups of order 4 : C2 x C2 and C4.

Case 1 : C2 x C2
----------------

If the Galois group is C2, then, as explained in the text, the
expressions:

u = (a + b)(c + d)
v = (a + c)(b + d)
w = (a + d)(b + c)

must be rational.  We always have v = 0, and we have:

u = -2 (p + sqrt(q))
w = -2 (p - sqrt(q))

These expressions will be rational if sqrt(q) is rational.

Conversely, if sqrt(q) = ab is rational, then Q(a) = Q(b) and the
group is of order 4.  In addition, we can write:

a =  sqrt (M) + sqrt (N)
b =  sqrt (M) - sqrt (N)
c = -sqrt (M) + sqrt (N)
d = -sqrt (M) - sqrt (N)

where M = (p + sqrt(q))/2 and N = (p - sqrt(q))/2 are rational.  The
C2 x C2 structure is quite obvious in these expressions.

Case 2 : C4
-----------

If sqrt(q) is irrational, the group cannot be C2 x C2.  To check if
sqrt(q) belongs to Q(a), we look at the subfields of Q(a).  If the
group is C4, it has only one proper subgroup of order 2, and this
means that Q(a) contains only one quadratic extension.

sqrt(z), defined up to a rational factor.  As Q(a) contains

a^2 = p + sqrt(p^2 - q)

this quadratic extension is generated by sqrt(p^2 - q).  It will be
the same as Q(sqrt(q)) if and only if p^2 - q and q differ by the
square of a rational number; this is equivalent to the condition that

(p^2 - q)/q

be rational.

In the remaining case, ab (and therefore b) does not belong to Q(a),
and the extension is of order 8.

Does this help?  Please feel free to write back if you want to
discuss this further.

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

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Associated Topics:
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