Field Theory: Equal SetsDate: 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. 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 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. 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, Thank you for your complete and easy-to-understend answer. It was very helpful. Lucia |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/