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 November 17th, 2017, 08:13 AM #1 Senior Member   Joined: Jan 2015 From: usa Posts: 104 Thanks: 1 Roots of a polynomial Let $L=\mathbb{Q}(\sqrt{2+\sqrt{3}})$. $L$ is a normal extension of $\mathbb{Q}$. >Express the roots of $m(x)=(x^2-2)^2-3$ (the minimal polynomial of $\alpha=\sqrt{2+\sqrt{3}}$ over $\mathbb{Q}$) in $\mathbb{Q}(\alpha)$ in terms of $\alpha$. Thanks. Last edited by skipjack; June 24th, 2018 at 07:23 AM. November 22nd, 2017, 06:03 AM #2 Math Team   Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 I am concerned that you show no attempt of your own to do these problems- and, the calculations, if you know the definitions are pretty straightforward. It should be obvious that if $\displaystyle (x- 2)^2- 3= 0$ then $\displaystyle (x- 2)^2= 3$ so that $\displaystyle x- 2= \pm\sqrt{3}$, $\displaystyle x= 2\pm \sqrt{3}$. With $\displaystyle \alpha= \sqrt{2+ \sqrt{3}}$ one of those roots, $\displaystyle 2+ \sqrt{3}$ is immediately $\displaystyle \alpha^2$. To find the other root, $\displaystyle 2- \sqrt{3}$, in terms of $\displaystyle \alpha$, note that $\displaystyle \alpha^2= 2+ \sqrt{3}$ so that $\displaystyle \sqrt{3}= \alpha^2- 2$ and then $\displaystyle 2- \sqrt{3}= 2- (\alpha^2- 2)= 4- \alpha^2$. (By the way- technically, equations have "roots", expressions have "zeros". Rather than "roots of $\displaystyle (x- 2)^2- 3$" you should have said either "the zeros of $\displaystyle (x- 2)^2- 3$" or "the roots of $\displaystyle (x- 2)^2- 3= 0$". I will confess that is being a little "hard-nosed".) Thanks from Benit13 November 24th, 2017, 03:45 PM   #3
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Quote:
 Originally Posted by Country Boy It should be obvious that if $\displaystyle (x- 2)^2- 3= 0$ then $\displaystyle (x- 2)^2= 3$ so that $\displaystyle x- 2= \pm\sqrt{3}$, $\displaystyle x= 2\pm \sqrt{3}$. With $\displaystyle \alpha= \sqrt{2+ \sqrt{3}}$ one of those roots, $\displaystyle 2+ \sqrt{3}$ is immediately $\displaystyle \alpha^2$. To find the other root, $\displaystyle 2- \sqrt{3}$, in terms of $\displaystyle \alpha$, note that $\displaystyle \alpha^2= 2+ \sqrt{3}$ so that $\displaystyle \sqrt{3}= \alpha^2- 2$ and then $\displaystyle 2- \sqrt{3}= 2- (\alpha^2- 2)= 4- \alpha^2$.
The polynomial is $(x^2-2)^2 - 3$, not $(x-2)^2 - 3$. The fact that your findings suggest $\alpha$ is not a root of its own minimal polynomial should be a giveaway that something went wrong! :P

Quote:
 Originally Posted by Country Boy (By the way- technically, equations have "roots", expressions have "zeros". Rather than "roots of $\displaystyle (x- 2)^2- 3$" you should have said either "the zeros of $\displaystyle (x- 2)^2- 3$" or "the roots of $\displaystyle (x- 2)^2- 3= 0$". I will confess that is being a little "hard-nosed".)
It's pretty standard to refer to the "zeros" of a polynomial as "roots". In fact, I challenge you to find a book/paper that doesn't call them "roots". I'm sure you'll find one eventually, but you may have to search for a bit!

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As for the problem, firstly note that $m(x) = m(-x)$, so the roots will be $\pm \alpha, \pm \beta$ for some $\beta$. It's easy enough to compute beta explicitly (e.g. you could just find all the roots explicitly, or note that the product of the roots is the constant coefficient of $m(x)$). Once you've got that, work out $\alpha \beta$ and that'll give an easy way to write $\beta$ in terms of $\alpha$.

Last edited by cjem; November 24th, 2017 at 04:02 PM. Tags polynomial, roots Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post annakar Calculus 8 December 16th, 2012 04:52 AM zaidalyafey Algebra 5 November 8th, 2012 10:48 AM parastvand Abstract Algebra 5 October 17th, 2011 11:41 AM greg1313 Algebra 4 May 9th, 2009 10:05 AM johnny Algebra 13 January 15th, 2008 05:19 AM

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