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Field Theory: Equal Sets

Date: 06/12/2002 at 23:31:39
From: Lucia
Subject: Field theory-Abstract algebra

Hello, Dr. Math.

Thank you for the help you have given me in the past. Can you please 
help with the following:

  Show that Q(sqrt(i)) is isomorphic to Q(sqrt(2), i).

Thank you very much.

Date: 06/13/2002 at 11:58:06
From: Doctor Paul
Subject: Re: Field theory-Abstract algebra

Notice that 

  sqrt(i) = zeta_8

where zeta_8 is a primitive 8th root of unity.  A little bit of
geometry will help you see that

  zeta_8 = [sqrt(2) + i*sqrt(2)]/2

Write back if this isn't clear and I'll explain it in further detail. 
But for now I'll leave it as a statement for you to verify.


  sqrt(i) = [sqrt(2) + i*sqrt(2)]/2

We want to show that 

  Q(sqrt(i)) = Q(sqrt(2), i)

Using the fact that 

  sqrt(i) = [sqrt(2) + i*sqrt(2)]/2

we want to show that

  Q([sqrt(2) + i*sqrt(2)]/2) = Q(sqrt(2), i)

To do this we show inclusion in both directions.

One direction is obvious:

  Q([sqrt(2) + i*sqrt(2)]/2) 

is clearly contained inside 

  Q(sqrt(2), i) 

since a field that contains sqrt(2) and i certainly contains their 
product.  Thus it contains half of their product.  It also contains 
half of sqrt(2).  And we know that fields are closed under addition.  
So it will contain the sum of sqrt(2)/2 and i*sqrt(2)/2.  Thus it 
contains [sqrt(2) + i*sqrt(2)]/2.

To see that 

  Q(sqrt(2), i) 

is contained in 

  Q([sqrt(2) + i*sqrt(2)]/2)

we want to write sqrt(2) and i in terms of [sqrt(2) + i*sqrt(2)]/2.

For simplicity in notation, let 

  alpha = [sqrt(2) + i*sqrt(2)]/2


  2*alpha = sqrt(2) + i*sqrt(2) 

          = sqrt(2) * (1+i)

Thus we have:

  sqrt(2) = 2*alpha/(1+i)

We're almost there.  We wanted to write sqrt(2) in terms of alpha.  
We need to get rid of the i in the denominator.  Recall that 

  alpha = sqrt(i)

Thus alpha^2 = i.  So we can write:

  sqrt(2) = (2*alpha)/(1+alpha^2)

Thus sqrt(2) is an element of 

  Q([sqrt(2) + i*sqrt(2)]/2).

Now we want to show that i is an element of 

  Q([sqrt(2) + i*sqrt(2)]/2).

This is equivalent to showing that i is an element of Q(sqrt(i)).  
And showing this is easy: if a field contains an element x, then 
x^2 is also in the field since fields are closed under 
multiplication.  Thus 

  [sqrt(i)]^2 = i 

is in 

  Q(sqrt(i)) = Q([sqrt(2) + i*sqrt(2)]/2)

This shows that 

  Q(sqrt(2), i) 

is contained in 

  Q([sqrt(2) + i*sqrt(2)]/2).

We now have inclusion in both directions so the sets must in fact be 
equal.  Certainly two sets which are equal are isomorphic to each 

I hope this helps. Please write back if you'd like to talk about 
this some more.

- Doctor Paul, The Math Forum

Date: 06/13/2002 at 16:51:21
From: Lucia
Subject: Thank you (Field theory-Abstract algebra)

Dear Dr. Paul,

Thank you for your complete and easy-to-understend answer.  It was
very helpful.

Associated Topics:
College Modern Algebra

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