Results 1 to 5 of 5

Math Help - Normal Closure....

  1. #1
    AAM
    AAM is offline
    Junior Member
    Joined
    Mar 2009
    Posts
    36

    Normal Closure....

    Hi guys,

    How do you find the normal closure Q(Sqrt(1+Sqrt(5))) over Q?

    I have absolutely no idea how to do this! :-s I think it has something to do with chains of subfields & conjugate, but if some could please explain the general method & how to apply it to this question I would be extremely greatful! :-)

    Many thanks in advance! x
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by AAM View Post
    Hi guys,

    How do you find the normal closure Q(Sqrt(1+Sqrt(5))) over Q?

    I have absolutely no idea how to do this! :-s I think it has something to do with chains of subfields & conjugate, but if some could please explain the general method & how to apply it to this question I would be extremely greatful! :-)

    Many thanks in advance! x
    The normal closure of \mathbb{Q}(\sqrt{1+\sqrt{5}}) is the splitting field of \text{min}(\sqrt{1+\sqrt{5}},\mathbb{Q}) over \mathbb{Q}?

    Is that enough or you have any other questions on why that is so?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    AAM
    AAM is offline
    Junior Member
    Joined
    Mar 2009
    Posts
    36
    Ok, so does that means the normal closure is:

    Q(Sqrt1+Sqrt(5)),Sqrt(1-Sqrt(5))) then?

    & could you please exaplin why this is the case? The definition for normal closure that I have is that it is the smallest extension of the current extension which is normal.

    Many thanks. :-)
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by AAM View Post
    Ok, so does that means the normal closure is:

    Q(Sqrt1+Sqrt(5)),Sqrt(1-Sqrt(5))) then?

    & could you please exaplin why this is the case? The definition for normal closure that I have is that it is the smallest extension of the current extension which is normal.

    Many thanks. :-)
    Let x=\sqrt{1+\sqrt{5}}\implies x^2 - 1 = \sqrt{5} \implies x^4 - 2x^2  - 4 = 0. This polynomial is also irreducible.

    The zeros of this polynomial are: x = \pm \sqrt{\sqrt{5}+1},\pm i\sqrt{\sqrt{5}-1}.
    Therefore, the normal closure is: N = \mathbb{Q}\left( \sqrt{\sqrt{5}+1}, i\sqrt{\sqrt{5}-1} \right)

    Notice that \left( \sqrt{\sqrt{5} + 1} \right) \left( i \sqrt{\sqrt{5}-1} \right) = i\sqrt{4} = 2i.

    Thus, it is also okay to write: N = \mathbb{Q}\left( \sqrt{\sqrt{5}+1}, i , \sqrt{\sqrt{5}-1}\right).
    -----

    Here is an explanation of why we take the normal closure to be splitting field. First we define what we mean by "normal closure". Let K/F be an algebraic field extension, the normal closure of K over F is the splitting field of \{ \min(a,F) | a\in K \} over F. The idea is that the normal closure is the "smallest" normal extension over F containing K. By "smallest" we mean that if E is a normal extension over F which satisfies K\subseteq E\subseteq N then E=N. To prove this let a\in K and consider \min (a,K), since E/K is normal it immediately means that \min (a,K) splits over E and so E contains all the roots of \min (a,K). But N is the field generated by the roots of this polynomials, hence, N\subseteq E which forces N=E. As a consequence we can prove that the normal closure of K=F(a_1,...,a_n) is the splitting field over \min(a_j,F) for each j=1,2,...,n. Let E be this splitting field, then obviously K\subseteq E\subseteq N with E/K normal, therefore it forces E=N. Your problem is a special case when K = F(a), so the normal closure is the splitting field of \min (a,F) over K.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    AAM
    AAM is offline
    Junior Member
    Joined
    Mar 2009
    Posts
    36
    Fantastic! :-D

    Thank you SO much theperfecthacker! :-)

    In an question online I've seen a chain of fields like:

    Q SUBFIELD OF Q(Sqrt(3)) SUBFIELD OF Q(Sqrt(1+2Sqrt(3)),Sqrt(1-2Sqrt(3))).

    Is this true in general?

    ie -

    Q SUBFIELD OF Q(Sqrt(A)) SUBFIELD OF Q(Sqrt(a+bSqrt(A)), Sqrt(a-bSqrt(A))) ?

    If so, this means that

    Q(Sqrt(A))(Sqrt(a+bSqrt(A)),Sqrt(a-bSqrt(A)
    = Q(Sqrt(a+bSqrt(A)),Sqrt(a-bSqrt(A)) , yes?

    Furthermore, is Q(Sqrt(A)) the ONLY quadratic subfield of Q(Sqrt(a+bSqrt(A)),Sqrt(a-bSqrt(A))) ? If not, what are the others? Indeed, what are all the subfields of Q(Sqrt(a+bSqrt(A)),Sqrt(a-bSqrt(A))) ?

    Many thanks. :-)
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Relation between topological closure and algebraic closure
    Posted in the Differential Geometry Forum
    Replies: 3
    Last Post: November 4th 2010, 03:45 PM
  2. closure of A
    Posted in the Differential Geometry Forum
    Replies: 1
    Last Post: October 19th 2009, 10:22 AM
  3. Closure
    Posted in the Advanced Algebra Forum
    Replies: 4
    Last Post: October 19th 2009, 02:59 AM
  4. Replies: 6
    Last Post: February 11th 2009, 01:56 PM
  5. Closure and MIV
    Posted in the Advanced Algebra Forum
    Replies: 3
    Last Post: April 27th 2008, 12:10 AM

Search Tags


/mathhelpforum @mathhelpforum