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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.

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