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Math Help - Degree/Basis

  1. #1
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    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
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  2. #2
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    For 1.

    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: 1, \sqrt{3}

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

    so entire thing has degree 2.
    this is wrong.

    a basis for the extension \mathbb Q (\sqrt{3}, \sqrt{7})/ \mathbb Q is: 1, \sqrt{3}, \sqrt{7}, \sqrt{21} so the degree is 4.
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  3. #3
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    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?
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  4. #4
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    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
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  5. #5
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    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?
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  6. #6
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    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.
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  7. #7
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    √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
    ???
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  8. #8
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    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).
    Last edited by Deveno; April 11th 2011 at 07:42 AM.
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  9. #9
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    ok that makes sense
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  10. #10
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    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?
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  11. #11
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    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....
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  12. #12
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    degree 6
    we need 5/6
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  13. #13
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    Quote Originally Posted by Deveno View Post
    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
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  14. #14
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    my bad, that was a typo, it should be a+b+2xy
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