Extension Fields

Date: 12/03/98 at 08:49:17
From: Andy Ullom
Subject: Modern Algebra Problems

I have looked at these problems and do not know where to start. Could 
you give me a little help?

We are working with Extension Fields.

1) Show that Q(sqrt(2), sqrt(3)) = Q(sqrt(2) + sqrt(3))

2) Find the splitting field of x^3 - 1 over Q. Express your answer in 
   the form of Q(a).

Thanks for any help that you can give me.

Sincerely,
Andy Ullom

Date: 12/03/98 at 11:43:40
From: Doctor Wilkinson
Subject: Re: Modern Algebra Problems

(1) The problem asks you to show that two sets are equal. The usual way 
of doing this is to show that everything in the first set is in the 
second set and that everything in the second set is in the first. 
(That's in fact the definition of set equality.)

In this case it is enough to show that sqrt(2) and sqrt(3) are both in
Q(sqrt(2) + sqrt(3)) and that sqrt(2) + sqrt(3) is in Q(sqrt(2), 
sqrt(3)). Now the second of these two statements is clearly true, so 
the only hard part is to show that sqrt(2) and sqrt(3) are in 
Q(sqrt(2) + sqrt(3)). That is, we want to be able to write sqrt(2) as 
some kind of expression involving sqrt(2) + sqrt(3) using just 
addition, subtraction, multiplication, and division, and numbers in Q.

A standard trick when you have a sum with square roots is to form the
corresponding difference and multiply, because then you can use the
"difference of two squares" factorization formula. So let's try looking 
at:

   (sqrt(2) + sqrt(3)) * (sqrt(2) - sqrt(3)) = sqrt(2)^2 - sqrt(3)^2
                                             = 2 - 1 = -1

Or:

   sqrt(2) - sqrt(3) = -1/(sqrt(2) + sqrt(3))

This shows that sqrt(2) - sqrt(3) belongs to Q(sqrt(2) + sqrt(3)).

But if you add sqrt(2) + sqrt(3) and sqrt(2) - sqrt(3), you get 
sqrt(2). This shows that sqrt(2) belongs to Q(sqrt(2) + sqrt(3)), and 
you should be able to finish the proof from here.

(2) This is really just a very fancy way of asking you to solve the
equation x^3 - 1 = 0.  Can you do that?  (Hint:  factor out x - 1).

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