# Thread:

1. ## Irreduclible polynomials

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

2. 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?

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

4. 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:

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

5. 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.

6. 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
$<br /> f(x)=x^4+ax^2+b<br />$
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.

7. I am not asking to complete this but the question I am asking is continuation of this.
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 0, "mathjax_display_tex": 0, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 24, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9700907468795776, "perplexity_flag": "head"}