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Math Help - IS f(x)= x^4 + x^3 -x^2 - 2x -2 irreducable over Q??

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    IS f(x)= x^4 + x^3 -x^2 - 2x -2 irreducable over Q??

    IS f(x)= x^4 + x^3 -x^2 - 2x -2 irreducable over Q??

    Attempt:

    Ive shown that this has no linear factors as +-1 +-2 are not roots.

    Then I let F(x)=(x^2+ax+b)(x^2+cx+d)

    And I end up with
    a +c=1
    d+ac+b=-1
    ad+bc=-2
    bd=-2

    But i can seam to show why no a b c d exist (or Exist)

    Is there an easier approach?

    I have an exam in 12 hours so I hope soeone can help!

    Thanks in advance!

    Niall
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    Quote Originally Posted by Niall101 View Post
    bd=-2
    It is sufficient to assume irreduciblility over Z by Gauss's Lemma.
    Now this condition tells us that b=\pm 1 \text{ and }d=\mp 2 or b=\pm 2 \text{ and }d=\mp 1.
    Show that any one of these leads to a contradiction i.e. those equations are not solvable with integers.
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    Great thanks! Its so obvious now! I ment to say a b c d were in Z. Thats what a week of no sleep will do to you
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    Quote Originally Posted by Niall101 View Post
    IS f(x)= x^4 + x^3 -x^2 - 2x -2 irreducable over Q??

    Attempt:

    Ive shown that this has no linear factors as +-1 +-2 are not roots.

    Then I let F(x)=(x^2+ax+b)(x^2+cx+d)

    And I end up with
    a +c=1
    d+ac+b=-1
    ad+bc=-2
    bd=-2

    But i can seam to show why no a b c d exist (or Exist)

    Is there an easier approach?

    I have an exam in 12 hours so I hope soeone can help!

    Thanks in advance!

    Niall
    i will note this theorem.
    If f(x) is reducible over \mathbb{Q}, then f(x) is reducible over \mathbb{Z}.

    so, as a contrapositive of the statement,
    If f(x) is irreducible over \mathbb{Z}, then f(x) is irreducible over \mathbb{Q}.

    from this, a,b,c,d \in \mathbb{Z}

    bd = -2
    then WLOG, b = \pm 1 and d=\mp 2.

    try to solve the combinations..

    EDIT: aww, didn't notice that there were replies already..
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    with a=1 c=0 b=1 d=-2 I get (x^2+x+1)(x^2-2)=F(x) so its not irreducable! Thanks for all the replies! Hope I can do this in the exam later!
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    Sorry for another question. If i have F(x)=x^3-5 Is it sufficiant to show it have no linear factors by plugging in +-5 +-1 and showing not equal to zero? I was thinking because the only factors could be linear or of degree 3. Thanks again.
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    MHF Contributor kalagota's Avatar
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    yes, it is enough..
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    Hey thanks so much! I really need to get this exam to pass the year. I really appreciate your help and such quick response
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    If anyone has a chance to shed some light on this one it would be a great help

    Let A = cuberoot(5),
    and let
    F = a + bA + cA in R
    a; b; c in Q
    Find the multiplicative inverse of the element 2 +A +A^2 in the field F.

    For this I Know There is an element in the field which multiplied by this give the Multiplicitive identity. Is there a way of doing this by taking powers of A?
    Last edited by Niall101; August 17th 2008 at 08:42 PM.
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  10. #10
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Niall101 View Post
    If anyone has a chance to shed some light on this one it would be a great help

    Let A = cuberoot(5),
    and let
    F = a + bA + cA in R
    a; b; c in Q
    Find the multiplicative inverse of the element 2 +A +A^2 in the field F.

    For this I Know There is an element in the field which multiplied by this give the Multiplicitive identity. Is there a way of doing this by taking powers of A?
    should that red thing be squared?
    assuming yes,
    A = \sqrt[3]{5}

    F = a + bA + cA^2 \subseteq \mathbb{R} where a,b,c \in \mathbb{Q}<br />

    what we want is to find a + bA + cA^2 so that (2 +A +A^2)(a + bA + cA^2)=1

    if we try to multiply, we will get 2a + (2b+a)A + (a+2c+b)A^2 + (c+b)A^3 + (c)A^4 = 1

    however, A^4 = A^3A = 5A

    thus, we reduce it to 2a + (2b+a + 5c)A + (a+2c+b)A^2 + (c+b)A^3 = 1

    can you do it from here?
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    Yes I think I have it now. I actually had this a.A^3=5.A written down on one of my attempts. Thank you so much for all the help. Your a life saver!
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    Hey guys thanks again! The exam went great! only 2 more to go
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    MHF Contributor kalagota's Avatar
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    good for you!
    if you still have questions, don't hesitate to post it.. (but on another thread.. )
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    Great thanks! On my way out now to face Group theory!
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