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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.

help with the following:

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

Thank you very much.
Lucia

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

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.

Thus

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

Then

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
other.

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

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

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

Dear Dr. Paul,

Lucia
Associated Topics:
College Modern Algebra

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