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Math Help - Irreduclible polynomials

  1. #1
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    Irreduclible polynomials

    Show that f(x)=x^4+bx^2+d is irreducible over
     \mathbb Q(\sqrt{d(b^2-4d)})[x]
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  2. #2
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    Quote Originally Posted by ZetaX View Post

    Show that f(x)=x^4+bx^2+d is irreducible over  \mathbb Q(\sqrt{d(b^2-4d)})[x]
    this can't be right! did you mean reducible?
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  3. #3
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    No this is correct, it is irreducible, that is what I am suppose to show.
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  4. #4
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    Quote Originally Posted by ZetaX View Post

    No this is correct, it is irreducible, that is what I am suppose to show.
    the question doesn't make sense! for example, choose d = 1 and b = 3. then \sqrt{d(b^2 - 4d)}=\sqrt{5}. now in \mathbb{Q}(\sqrt{5})[x] we have:

    x^4+3x^2 + 1 = \left(x^2 + \frac{3 + \sqrt{5}}{2} \right) \left(x^2 + \frac{3 - \sqrt{5}}{2} \right).
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  5. #5
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    I have that
    <br />
f(x)=x^4+ax^2+b<br />
    is irreducible over rationals, and its galois group is
    <br />
D_4<br />
    And M is the splitting field of  f(x)

    and I have that <br /> <br />
\mathbb Q(\sqrt{d(b^2-4d)})<br />
    is a quadratic subfield of M.
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  6. #6
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    Quote Originally Posted by NonCommAlg View Post
    this can't be right! did you mean reducible?
    Quote Originally Posted by ZetaX View Post
    No this is correct, it is irreducible, that is what I am suppose to show.
    Quote Originally Posted by NonCommAlg View Post
    the question doesn't make sense! for example, choose d = 1 and b = 3. then \sqrt{d(b^2 - 4d)}=\sqrt{5}. now in \mathbb{Q}(\sqrt{5})[x] we have:

    x^4+3x^2 + 1 = \left(x^2 + \frac{3 + \sqrt{5}}{2} \right) \left(x^2 + \frac{3 - \sqrt{5}}{2} \right).
    Quote Originally Posted by ZetaX View Post
    I have that
    <br />
f(x)=x^4+ax^2+b<br />
    is irreducible over rationals, and its galois group is
    <br />
D_4<br />
    And M is the splitting field of  f(x)

    and I have that <br /> <br />
\mathbb Q(\sqrt{d(b^2-4d)})<br />
    is a quadratic subfield of M.
    I think that ZetaX is referring to this thread. In that thread he asks that given that x^4 + bx^2 + d is irreducible over \mathbb{Q} with Galois group D_4 then show that the three quadradic extensions (we know there has to be three since D_4 has three subgroups of index 2) are: \mathbb{Q}(\sqrt{b^2-4d}),\mathbb{Q}(\sqrt{d}),\mathbb{Q}(\sqrt{d(b^2-4d)}). He is asking to complete that solution.
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  7. #7
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    I am not asking to complete this but the question I am asking is continuation of this.
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