# Find the minimal polynomial for a = π^2 over Q(π^3)

Printable View

• March 26th 2012, 10:11 AM
lteece89
Find the minimal polynomial for a = π^2 over Q(π^3)
If it were in Q then the polynomial would be p(x) = x - π^2, but I don't know if being over Q(π^3) changes that. I think that when you compare the fields you get degree 1, but again, I'm not completely sure.
• March 26th 2012, 10:33 AM
Sylvia104
Re: Find the minimal polynomial for a=π^2 over Q(π^3)
If the extension is $\mathbb Q(\pi)/\mathbb Q\left(\pi^3\right)$ the minimal polynomial is $x^3-\pi^6.$
• March 26th 2012, 11:54 AM
lteece89
Re: Find the minimal polynomial for a = π^2 over Q(π^3)
Is this because the minimal polynomial needs to be degree three?
• March 26th 2012, 01:40 PM
Sylvia104
Re: Find the minimal polynomial for a=π^2 over Q(π^3)
The minimal polynomial here is the monic polynomial $f(x)$ of minimal degree with coefficients in $\mathbb Q\left(\pi^3\right)$ such that $f\left(\pi^2\right)=0.$ If $f(x)$ has degree $1$ or $2,$ $f\left(\pi^2\right)$ is of the form $a_0+\pi^2$ or $a_0+a_1\pi^2+\pi^4,$ where $a_0,a_1\in\mathbb Q\left(\pi^3\right);$ clearly these cannot be $0.$
All times are GMT -8. The time now is 12:57 PM.
{"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": 13, "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": 0.9166180491447449, "perplexity_flag": "middle"}