Can Q(sqrt a + sqrt b) be equal to Q(sqrt a, sqrt b) in abstract algebra?

[crappyjones]

field extension question (abstract algebra)

Homework Statement

For any positive integers a, b, show that Q(sqrt a + sqrt b) = Q(sqrt a, sqrt b).

Homework Equations

The Attempt at a Solution

i proved that [Q(sqrt a):Q] for all n belonging to Z⁺ is 2 whenever a is not a perfect square and 1 when it is. also, for Q(sqrt a, sqrt b) i found the minimal polynomial to be (x²-a-b)² - 4ab. for Q(sqrt a + sqrt b), i found the minimal polynomial to be x⁴ - 2ax² - 2bx² + a² + b² - 2ab.

i can show that both extensions have degree 4 (for both a and b non perfect squares, 3 for when either a or b is a perfect square, 2 for when both a and b are perfect squares) and that the basis of Q(sqrt a, sqrt b) is {1, sqrt a, sqrt b, sqrt (ab)}.

any thoughts on how to solve this problem? am i on the right track? completely off? any comments on what i am doing would be appreciated as well.

thanks,

cj.

 

[Dick]

I can agree with some stuff you are saying. Not everything. I would say, for example, that the extension where a and b are perfect squares has degree 1. I.e. it's not really an extension at all. sqrt(a) and sqrt(b) are then rational. Q(sqrt(a),sqrt(b)) is just Q. But to go back to the original question, it should be pretty obvious that Q(sqrt(a)+sqrt(b)) is contained in Q(sqrt(a),sqrt(b)), right? Can you say why? The harder job is to show that Q(sqrt(a),sqrt(b)) is contained in Q(sqrt(a)+sqrt(b)). You need to show that sqrt(a), sqrt(b) and sqrt(ab) are contained in Q(sqrt(a)+sqrt(b)). Consider powers of (sqrt(a)+sqrt(b)).