What is the minimal polynomial for Q(a1/2, b1/2)?

[kathrynag]

1.Find the degree and basis for Q(3^1/2,7^1/2) over Q.

2.For any positive integers a, b, show that Q(a^1/2+b^1/2)=Q(a^1/2,b^1/2)


Ideas:
1. Well I know if I looked at (3)^1/2 over Q
Then (3)^1/2 has minimal polynomial x^2-3, so degree 2 over Q
(7)^1/2 has minimal polynomial x^2-7 so degree 2 over Q
so entire thing has degree 2.
Really unsure about basis

2.I started by computing the minimal polynomial of a^1/2+b^1/2 over Q
x=a^1/2+b^1/2
x-a^1/2=b^1/2
x^2+2(a)^(1/2)x+a=3
x^2+a-3=2(a)^1/2
x^4+x^2a-3x^2+x^2a+a^2-3a-3x^2-3a+9=4a
x^4+2x^2a-6x^2+a^2-6a+9-4a=0
I don't know how that would hekp me, though

 

[Mark44]

1.Find the degree and basis for Q(3^1/2,7^1/2) over Q.

2.For any positive integers a, b, show that Q(a^1/2+b^1/2)=Q(a^1/2,b^1/2)


Ideas:
1. Well I know if I looked at (3)^1/2 over Q
Then (3)^1/2 has minimal polynomial x^2-3, so degree 2 over Q
(7)^1/2 has minimal polynomial x^2-7 so degree 2 over Q
so entire thing has degree 2.

It has been a very long time since I did any abstract algebra, so caveat emptor. It would seem to me that the minimal polynomial for Q(3^1/2, 7^1/2) would be (x² - 3)(x² - 7), which is a fourth degree polynomial.

Really unsure about basis

2.I started by computing the minimal polynomial of a^1/2+b^1/2 over Q
x=a^1/2+b^1/2
x-a^1/2=b^1/2
x^2+2(a)^(1/2)x+a=3

You have a sign error in the line above. Also when you square b^(1/2), shouldn't you get b, not 3?

x^2+a-3=2(a)^1/2

You seem to have lost a factor of x in the line above.

In any case, I think you're going in the wrong direction. What you need to do for this part is to show that Q(a^1/2 + b^1/2) = Q(a^1/2, b^1/2).

x^4+x^2a-3x^2+x^2a+a^2-3a-3x^2-3a+9=4a
x^4+2x^2a-6x^2+a^2-6a+9-4a=0
I don't know how that would hekp me, though