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**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 (x

^{2}-a-b)

^{2}- 4ab. for

**Q**(sqrt a + sqrt b), i found the minimal polynomial to be x

^{4}- 2ax

^{2}- 2bx

^{2}+ a

^{2}+ b

^{2}- 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.

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