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November 17th, 2017, 08:13 AM  #1 
Senior Member Joined: Jan 2015 From: usa Posts: 103 Thanks: 0  Roots of a polynomial
Let $L=\mathbb{Q}(\sqrt{2+\sqrt{3}})$. $L$ is a normal extension of $\mathbb{Q}$. >Express the roots of $m(x)=(x^22)^23$ (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: 3,261 Thanks: 894 
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 "hardnosed".) 
November 24th, 2017, 03:45 PM  #3  
Senior Member Joined: Aug 2017 From: United Kingdom Posts: 266 Thanks: 80 Math Focus: Algebraic Number Theory, Arithmetic Geometry  Quote:
Quote:
________ 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.  

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