Results 1 to 14 of 14

Math Help - algebra, field

  1. #1
    Member
    Joined
    Jan 2008
    Posts
    114

    algebra, field

    Sorry for all the algebra questions everybody, I have my exam on Wednesday (which I'm going to fail) and seem to be struggling with everything as it's so abstract...

    Let g = X^3 +2X - 1
    Let F be the smallest subfield of R containing a root \alpha of g. What are the elements of F?

    Would this just be the identity element 1?

    What is the dimension of F considered as a vector space over the rationals?

    This I just do not understand at all....can anyone please explain?

    Thanks in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Hallo,

    (which I'm going to fail)
    Never say this before you write the exam

    Would this just be the identity element 1?
    1 is not a root, but -1 is. I don't know if it helps...
    I don't get the thing with the subspace F... sorry
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2008
    Posts
    114
    Ok cheers for the relpy Moo, think I'll just pray that those questions don't come up!

    I've got another question which seems to come up every year, seems a little easier but I still don't see what to do?!

    Write down the inverse of  \alpha^2 - 2\alpha in the form  \lambda \alpha^2 + \mu\alpha + v where  \lambda, \mu, v are all rational numbers.

    Can anyone please explain the method for this?

    Thanks again.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by hunkydory19 View Post
    Sorry for all the algebra questions everybody, I have my exam on Wednesday (which I'm going to fail) and seem to be struggling with everything as it's so abstract...

    Let g = X^3 +2X - 1
    Let F be the smallest subfield of R containing a root \alpha of g. What are the elements of F?

    Would this just be the identity element 1?
    Note that any field should contain 0 and 1. Here it should also contain \alpha according to the posed question. Now wonder what happens \alpha^2, \alpha^3,.... Since closure would force them into the field. The real thing you should observe here is that :

    g(\alpha) = 0 \Rightarrow \alpha^3 + 2\alpha - 1 = 0 \Rightarrow \alpha^3 = - 2\alpha + 1 -----------------------------------(*)
    This means every power of \alpha in this field can be brought down to the form \lambda \alpha^2 + \mu\alpha + v

    For example:
    Write down the inverse of in the form where are all rational numbers.

    Can anyone please explain the method for this?
    <br />
 (\alpha^2 - 2\alpha)^{-1}   = \lambda \alpha^2 + \mu\alpha + v \Rightarrow (\lambda \alpha^2 + \mu\alpha + v)\cdot(\alpha^2 - 2\alpha) = 1

    (\lambda \alpha^2 + \mu\alpha + v)\cdot(\alpha^2 - 2\alpha) = \lambda \alpha^4 - (2\lambda- \mu)\alpha^3 - (2\mu+v)\alpha^2 - 2v\alpha = 1

    Now use the property of \alpha we just proved in (*)

    \lambda (-2\alpha^2 + \alpha) - (2\lambda- \mu)(-2\alpha+1) - (2\mu+v)\alpha^2 - 2v\alpha - 1 = 0

    (-2\lambda-2\mu-v)\alpha^2 + (5\lambda- 2\mu -v)\alpha -2\lambda +\mu - 1= 0

    So solve this system of equations:

    -2\lambda-2\mu-v = 0

    5\lambda- 2\mu -v = 0

    -2\lambda +\mu-1 = 0

    Unless I have been careless in my arithmetic you should get the right answer
    Quote Originally Posted by hunkydory19 View Post
    What is the dimension of F considered as a vector space over the rationals?
    Clearly it is 3. Since \{1,\alpha,\alpha^2 \} is the basis.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Yo Iso,

    Why should F contain 0 and 1 ? Because they're null elements ?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by Moo View Post
    Yo Iso,

    Why should F contain 0 and 1 ? Because they're null elements ?
    Every Field should have additive and multiplicative identities since a field is both an additive and a multiplicative group.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Moo
    Moo is offline
    A Cute Angle Moo's Avatar
    Joined
    Mar 2008
    From
    P(I'm here)=1/3, P(I'm there)=t+1/3
    Posts
    5,618
    Thanks
    6
    Oh, I misunderstood "field" then -__-
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Member
    Joined
    Jan 2008
    Posts
    114
    Thank you so much Iso, you have literally just saved my life!
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Member
    Joined
    Jan 2008
    Posts
    114
    OK I've just worked through this example, and understood everything...EXCEPT where the  -2\alpha + 1 comes from? Obviously I see that it comes from rearranging, but what I mean is do you always have to know what g is before you can find this alpha equation?

    Also there were a few signs wrong in your working, but got the answers to be  \lambda = -\frac{6}{11}, \mu = -\frac{1}{11} v = -\frac{14}{11}

    Is there a way to check these answers correct after doing this method?

    Thanks again
    Last edited by hunkydory19; May 3rd 2008 at 09:52 AM.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by hunkydory19 View Post
    OK I've just worked through this example, and understood everything...EXCEPT where the  -2\alpha + 1 comes from? Obviously I see that it comes from rearranging, but what I mean is do you always have to know what g is before you can find this alpha equation?
    Yes! The field generated is due to a particular polynomial. Its called the generating polynomial of the field.


    Quote Originally Posted by hunkydory19 View Post
    Also there were a few signs wrong in your working, but got the answers to be [tex] \lambda = -\frac{6}{11}, \mu = -\frac{1}{11} and v = -\frac{14}{11} [tex]
    Sorry for that... I did not work it out on paper

    Quote Originally Posted by hunkydory19 View Post
    Is there a way to check these answers correct after doing this method?

    Thanks again
    Yes multiply out the obtained answer with \alpha^2 - 2\alpha and check if you get 1
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Member
    Joined
    Jan 2008
    Posts
    114
    Iso do not dare apologise to me, haha! Would be failing this exam if it wasn't for you! Was just pointing it out incase you got to different answers so thought it would save time. Thank you SO much for all your help!
    Follow Math Help Forum on Facebook and Google+

  12. #12
    Member
    Joined
    Jan 2008
    Posts
    114
    I've just tried your tip for checking if (\alpha^2 - 2\alpha)(-\frac{6}{11}\alpha^2 -\frac{1}{11}\alpha -\frac{14}{11}) multiply out to give 1 and they don't, nowhere near since none of the alphas cancel! What am I doing wrong?
    Follow Math Help Forum on Facebook and Google+

  13. #13
    Lord of certain Rings
    Isomorphism's Avatar
    Joined
    Dec 2007
    From
    IISc, Bangalore
    Posts
    1,465
    Thanks
    6
    Quote Originally Posted by hunkydory19 View Post
    I've just tried your tip for checking if (\alpha^2 - 2\alpha)(-\frac{6}{11}\alpha^2 -\frac{1}{11}\alpha -\frac{14}{11}) multiply out to give 1 and they don't, nowhere near since none of the alphas cancel! What am I doing wrong?
    Well I get 1 perfectly...

    Let b =-\frac1{11}

    (\alpha^2 - 2\alpha)(6b\alpha^2 + b\alpha + 14b)

    = 6b\alpha^4+(b - 12b)\alpha^3 + (14b - 2b)\alpha^2 - 28b\alpha

    = 6b(-2\alpha^2 + \alpha) -11b(-2\alpha + 1) + 12b\alpha^2 - 28b\alpha

    = -12b \alpha^2 + 6b\alpha + 22b\alpha -11b + 12b\alpha^2 - 28b\alpha

    = (-12b \alpha^2 + 12b\alpha^2)+(6b\alpha + 22b\alpha - 28b\alpha) - 11b

    = 0+0 - 11b = -11b = 1

    Hence proved
    Follow Math Help Forum on Facebook and Google+

  14. #14
    Member
    Joined
    Jan 2008
    Posts
    114
    Thank you so much Iso!! Absolute legend.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Borel field Algebra
    Posted in the Advanced Algebra Forum
    Replies: 2
    Last Post: April 3rd 2011, 02:45 AM
  2. Borel field- sigma- Algebra
    Posted in the Differential Geometry Forum
    Replies: 0
    Last Post: March 28th 2011, 10:57 AM
  3. Algebra over a field
    Posted in the Advanced Algebra Forum
    Replies: 5
    Last Post: January 10th 2010, 12:29 PM
  4. Algebra over a field - zero element
    Posted in the Advanced Algebra Forum
    Replies: 6
    Last Post: September 9th 2009, 05:05 AM
  5. complex field division algebra?
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: September 14th 2008, 08:47 PM

Search Tags


/mathhelpforum @mathhelpforum