Irreduclible polynomials

• Apr 13th 2009, 03:52 AM
ZetaX
Irreduclible 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 PM
NonCommAlg
Quote:

Originally Posted by ZetaX

Show that $\displaystyle f(x)=x^4+bx^2+d$ is irreducible over $\displaystyle \mathbb Q(\sqrt{d(b^2-4d)})[x]$

this can't be right! did you mean reducible?
• Apr 13th 2009, 02:32 PM
ZetaX
No this is correct, it is irreducible, that is what I am suppose to show.
• Apr 13th 2009, 03:08 PM
NonCommAlg
Quote:

Originally Posted by ZetaX

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 $\displaystyle \sqrt{d(b^2 - 4d)}=\sqrt{5}.$ 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 AM
ZetaX
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 AM
ThePerfectHacker
Quote:

Originally Posted by NonCommAlg
this can't be right! did you mean reducible?

Quote:

Originally Posted by ZetaX
No this is correct, it is irreducible, that is what I am suppose to show.

Quote:

Originally Posted by NonCommAlg
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:

$\displaystyle x^4+3x^2 + 1 = \left(x^2 + \frac{3 + \sqrt{5}}{2} \right) \left(x^2 + \frac{3 - \sqrt{5}}{2} \right).$

Quote:

Originally Posted by ZetaX
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.

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 AM
ZetaX
I am not asking to complete this but the question I am asking is continuation of this.