Irreduclible polynomials

**ZetaX** · April 13th 2009, 03:52 AM

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

**NonCommAlg** · April 13th 2009, 12:04 PM

Originally Posted by ZetaX

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?

**ZetaX** · April 13th 2009, 02:32 PM

No this is correct, it is irreducible, that is what I am suppose to show.

**NonCommAlg** · April 13th 2009, 03:08 PM

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 $\sqrt{d(b^2 - 4d)}=\sqrt{5}.$ now in $\mathbb{Q}(\sqrt{5})[x]$ we have:

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

**ZetaX** · April 14th 2009, 05:47 AM

I have that
$ f(x)=x^4+ax^2+b $
is irreducible over rationals, and its galois group is
$ D_4 $
And M is the splitting field of $f(x)$

and I have that $ \mathbb Q(\sqrt{d(b^2-4d)}) $
is a quadratic subfield of M.

**ThePerfectHacker** · April 14th 2009, 11:07 AM

Originally Posted by NonCommAlg

this can't be right! did you mean reducible?

Originally Posted by ZetaX

No this is correct, it is irreducible, that is what I am suppose to show.

Originally Posted by NonCommAlg

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:

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

Originally Posted by ZetaX

I have that
$ f(x)=x^4+ax^2+b $
is irreducible over rationals, and its galois group is
$ D_4 $
And M is the splitting field of $f(x)$

and I have that $ \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 $x^4 + bx^2 + d$ 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.

**ZetaX** · April 14th 2009, 11:23 AM

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

Thread: Irreduclible polynomials

LinkBack

Thread Tools

Display

Irreduclible polynomials

Similar Math Help Forum Discussions

Roots of 3rd degree polynomials and Characteristic Polynomials

Polynomials written as a product of irreducible polynomials

Irreduclible polynomial

Irreduclible polynomials

Dividing Polynomials AND Factoring Polynomials

Search Tags