Results 1 to 11 of 11

Math Help - [SOLVED] Algebra-How do i show this is irreducible?

  1. #1
    Newbie
    Joined
    Oct 2008
    Posts
    23

    [SOLVED] Algebra-How do i show this is irreducible?

    My question asks me to find the minimum polynomial of .... i + root5

    I get x^4 - 8x^2 + 36 = 0 as the minimum polynomial

    How can i show this is irreducible.
    I've tried the rational root test but that would take too long.
    There must be another shorter way to do it.
    Anyone help please?
    Thanks.
    Last edited by CaptainBlack; May 24th 2009 at 11:58 PM. Reason: restore deleted question
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Newbie
    Joined
    Oct 2008
    Posts
    23
    Anyone please?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Math Engineering Student
    Krizalid's Avatar
    Joined
    Mar 2007
    From
    Santiago, Chile
    Posts
    3,654
    Thanks
    14
    To show irreducibility, observe that x^{4}-8x^{2}+36=\left( x^{2}-4 \right)^2+20>0.

    And do not bump.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Oct 2008
    Posts
    23
    I dont understand lol how can i show irreducibility from that? Thanks.

    And where does that > 0 inequality come from?
    Last edited by CaptainBlack; May 24th 2009 at 11:59 PM. Reason: restore deleted post
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Oct 2008
    Posts
    23
    If i let y=x^2 so that the polynomial becomes y^2-8y+36=0, then take the discriminant which gives a negative discriminant, hence the quadratic is irreducible, would that also imply that the quartic is irreducible?
    Follow Math Help Forum on Facebook and Google+

  6. #6
    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
    Quote Originally Posted by thegarden View Post
    If i let y=x^2 so that the polynomial becomes y^2-8y+36=0, then take the discriminant which gives a negative discriminant, hence the quadratic is irreducible, would that also imply that the quartic is irreducible?
    If you're all working over the real numbers, then you're correct.

    ~~~~~~~~~~~~~~~~~~~~~~~~~~

    As for (x^2-4)^2+20, we know that a square number is always positive.

    hence (x^2-4)^2+20 \geq 0+20 > 0

    therefore it is never equal to 0 (for real values) and thus it is not reducible.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Oct 2008
    Posts
    23
    Is this the same for rational numbers?
    I want to show it is irreducible in Q
    Last edited by CaptainBlack; May 25th 2009 at 12:01 AM. Reason: restore deleted post
    Follow Math Help Forum on Facebook and Google+

  8. #8
    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
    Quote Originally Posted by thegarden View Post
    Is this the same for rational numbers?
    rational numbers are real numbers.

    the thing that would change is if you're working over complex numbers (in the form a+ib)
    Follow Math Help Forum on Facebook and Google+

  9. #9
    Newbie
    Joined
    Oct 2008
    Posts
    23
    Quote Originally Posted by Moo View Post
    rational numbers are real numbers.

    the thing that would change is if you're working over complex numbers (in the form a+ib)
    Of course lol cheers
    Last edited by thegarden; December 2nd 2008 at 09:21 AM.
    Follow Math Help Forum on Facebook and Google+

  10. #10
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    Quote Originally Posted by Krizalid View Post
    To show irreducibility, observe that x^{4}-8x^{2}+36=\left( x^{2}-4 \right)^2+20>0.

    And do not bump.
    Quote Originally Posted by Moo View Post
    If you're all working over the real numbers, then you're correct.

    ~~~~~~~~~~~~~~~~~~~~~~~~~~

    As for (x^2-4)^2+20, we know that a square number is always positive.

    hence (x^2-4)^2+20 \geq 0+20 > 0

    therefore it is never equal to 0 (for real values) and thus it is not reducible.
    Irreduciblity does not mean not having any zeros.
    This does not fully complete the proof.
    Follow Math Help Forum on Facebook and Google+

  11. #11
    Flow Master
    mr fantastic's Avatar
    Joined
    Dec 2007
    From
    Zeitgeist
    Posts
    16,948
    Thanks
    5
    Quote Originally Posted by thegarden View Post
    My question asks me to find the minimum polynomial of .... i + root5

    I get x^4 - 8x^2 + 36 = 0 as the minimum polynomial

    How can i show this is irreducible.
    I've tried the rational root test but that would take too long.
    There must be another shorter way to do it.
    Anyone help please?
    Thanks.
    Thread closed because OP deleted question after getting reply.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 4
    Last Post: March 4th 2012, 05:04 PM
  2. Replies: 2
    Last Post: July 31st 2011, 08:54 PM
  3. Show that the Markov chain is irreducible.
    Posted in the Advanced Statistics Forum
    Replies: 2
    Last Post: October 7th 2009, 03:59 AM
  4. Show that x^3-2 is irreducible in Z/19Z [x]/<x^2-2>
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: November 11th 2008, 02:17 PM
  5. Show that a prime is not irreducible in Z[i]
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: January 21st 2008, 01:12 PM

Search Tags


/mathhelpforum @mathhelpforum