# Degree/Basis

• Apr 10th 2011, 11:29 AM
mathematic
Degree/Basis
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
• Apr 10th 2011, 12:13 PM
zoek
For 1.

Quote:

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
a basis is: $\displaystyle 1, \sqrt{3}$

Quote:

(7)^1/2 has minimal polynomial x^2-7 so degree 2 over Q
a basis is $\displaystyle 1, \sqrt{7}$

Quote:

so entire thing has degree 2.
this is wrong.

a basis for the extension $\displaystyle \mathbb Q (\sqrt{3}, \sqrt{7})/ \mathbb Q$ is:$\displaystyle 1, \sqrt{3}, \sqrt{7}, \sqrt{21}$ so the degree is 4.
• Apr 10th 2011, 12:26 PM
Deveno
1. are you sure about this? remember Q(√3) is a subfield of R. the factorization of x^2 - 7 in R is (x + √7)(x - √7). if Q(√3,√7) has degree 2 over Q, wouldn't that force Q(√3,√7) to have degree 1 over Q(√3)? isn't this the same as saying √7 = a+b√3 for some rational a,b?

i think you should look for a polynomial of degree 4 as the minimal polynomial. the basis elements should be obvious to write down.

it is my contention that Q(√3,√7) = [Q(√3)](√7).

2. think about the basis elements you (hopefully) found for part (1). can you write √a + √b in terms of those basis elements, adapted to a,b instead of 3,7?

if so, then Q(√a+√b) is contained in Q(√a,√b). look at your minimal polynomial. what can you say about the dimension over Q of Q(√a+√b)? can you use a theorem from linear algebra about subspaces here?
• Apr 10th 2011, 12:32 PM
mathematic
for 2, I guess I should show (a)^1/2 + (b)^1/2 is in Q((a)^1/2, (b)^1/2) and a^1/2, b^1/2 in Q((a)^1/2 + (b)^1/2)
would this involve degrees and basises?
a basis for Q((a)^1/2 , (b)^1/2) is 1, a^1/2, b^1/2, (ab)^1/2
a^1/2+ b^1/2 in Q((a)^1/2 , (b)^1/2)
Q((a)^1/2 + (b)^1/2) has basis 1, (a)^1/2 + (b)^1/2?
Not quite sure about finding this basis
• Apr 10th 2011, 12:44 PM
mathematic
1.basis elements are 1, root 3, root 7, and root(21)
2.root a + root b= root a+ root b since root a and root b would be basis elements
looking at the minimal polynomial, has dimension 4
Would I use this theorem: Let V ba an n dimensional vector space. If W is any subspace of V, then W=V iff W has dimension n?
• Apr 10th 2011, 01:51 PM
Deveno
that would be the one. you could show that you could express √a and √b in terms of powers of (√a+√b), but that seems like more work to me.
• Apr 10th 2011, 02:55 PM
mathematic
√a and √b in terms of powers of (√a+√b), but that seems like more work to me.
I want to look at both ways, so if I were to do it this way:
x=(√a+√b)
x^2=a+2√ab+b
???
• Apr 10th 2011, 04:15 PM
Deveno
well, the elements 1, √a+√b, (√a+√b)^2 = (a+b) + 2√a√b, (√a+√b)^3 = (2a+b)√a + (a+2b)√b all have to be linearly independent, since [Q(√a+√b):Q] = 4.

so if we set √a+√b = u, √a = (1/(a-b))(u^3 - (a+2b)u), and √b = (1/(b-a))(u^3 - (2a+b)u) (you might want to double-check my algebra).

this shows that Q(√a,√b) is contained in Q(√a+√b) (note as well, that this gives a different basis for Q(√a,√b):

(it is an interesting exercise to show that if we set √a = x, √b = y, that the set {1,x+y,a+b+2xy,(2a+b)x+(a+2b)y} is linearly independent if {1,x,y,xy} is).
• Apr 10th 2011, 04:24 PM
mathematic
ok that makes sense
• Apr 10th 2011, 04:57 PM
mathematic
Ok this question is related to 1.
What about degree and basis for Q(2^1/2,2^1/3) over Q
2^1/2 has minimal polynomial x^2-2
a basis would be 1,2^1/2
2^1/3 has minimal polynomial x^3-2
basis would be 1, 2^1/3,2^2/3
basis for Q(2^1/2,2^1/3) has degree 5 and basis 1, 2^1/3,2^2/3, 2^1/2,2^1/6?
• Apr 10th 2011, 05:34 PM
Deveno
no, 2*3 is not 5.

x^3 - 2 still doesn't have any roots in Q(√2). any root still lies in a cubic extension of Q(√2).

look at your set of "powers" of two: {0, 1/6, 1/3, 1/2, 2/3}. can you see what is missing from that list?

hint: 1/3 = 2/6, 1/2 = 3/6, 2/3 = 4/6....
• Apr 10th 2011, 06:17 PM
mathematic
degree 6
we need 5/6
• Apr 10th 2011, 06:37 PM
mathematic
Quote:

Originally Posted by Deveno
well, the elements 1, √a+√b, (√a+√b)^2 = (a+b) + 2√a√b, (√a+√b)^3 = (2a+b)√a + (a+2b)√b all have to be linearly independent, since [Q(√a+√b):Q] = 4.

so if we set √a+√b = u, √a = (1/(a-b))(u^3 - (a+2b)u), and √b = (1/(b-a))(u^3 - (2a+b)u) (you might want to double-check my algebra).

this shows that Q(√a,√b) is contained in Q(√a+√b) (note as well, that this gives a different basis for Q(√a,√b):

(it is an interesting exercise to show that if we set √a = x, √b = y, that the set {1,x+y,a+b+xy,(2a+b)x+(a+2b)y} is linearly independent if {1,x,y,xy} is).

I thing I understand the first part
Now for showing Q(√a+√b) is contained in Q(√a,√b)
Let √a = x, √b = y
I have x+y=√a+√b
(x+y)^2=a+b+2xy
• Apr 11th 2011, 07:43 AM
Deveno
my bad, that was a typo, it should be a+b+2xy