Show that $\displaystyle f(x)=x^4+bx^2+d$ is irreducible over

$\displaystyle \mathbb Q(\sqrt{d(b^2-4d)})[x]$

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- Apr 13th 2009, 03:52 AMZetaXIrreduclible polynomials
Show that $\displaystyle f(x)=x^4+bx^2+d$ is irreducible over

$\displaystyle \mathbb Q(\sqrt{d(b^2-4d)})[x]$ - Apr 13th 2009, 12:04 PMNonCommAlg
- Apr 13th 2009, 02:32 PMZetaX
No this is correct, it is irreducible, that is what I am suppose to show.

- Apr 13th 2009, 03:08 PMNonCommAlg
the question doesn't make sense! for example, choose d = 1 and b = 3. then $\displaystyle \sqrt{d(b^2 - 4d)}=\sqrt{5}.$ now in $\displaystyle \mathbb{Q}(\sqrt{5})[x]$ we have:

$\displaystyle x^4+3x^2 + 1 = \left(x^2 + \frac{3 + \sqrt{5}}{2} \right) \left(x^2 + \frac{3 - \sqrt{5}}{2} \right).$ - Apr 14th 2009, 05:47 AMZetaX
I have that

$\displaystyle

f(x)=x^4+ax^2+b

$

is irreducible over rationals, and its galois group is

$\displaystyle

D_4

$

And M is the splitting field of $\displaystyle f(x)$

and I have that $\displaystyle

\mathbb Q(\sqrt{d(b^2-4d)})

$

is a quadratic subfield of M. - Apr 14th 2009, 11:07 AMThePerfectHacker
I think that

**ZetaX**is referring to this thread. In that thread he asks that given that $\displaystyle x^4 + bx^2 + d$ is irreducible over $\displaystyle \mathbb{Q}$ with Galois group $\displaystyle D_4$ then show that the three quadradic extensions (we know there has to be three since $\displaystyle D_4$ has three subgroups of index $\displaystyle 2$) are: $\displaystyle \mathbb{Q}(\sqrt{b^2-4d}),\mathbb{Q}(\sqrt{d}),\mathbb{Q}(\sqrt{d(b^2-4d)})$. He is asking to complete that solution. - Apr 14th 2009, 11:23 AMZetaX
I am not asking to complete this but the question I am asking is continuation of this.