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So I was playing around with trying to prove the regular 7-gon is not constructable under qualifier-exam conditions, so I didn't have a book open. I got it down to having (If I didn't make any basic math errors, I was doing this on pen and paper while my class was taking a calculus test), that $\sin (\frac {2\pi} 7)$ would by a root of $f(x)=64x^6 -80x^4+24x^2 -1$ I then noticed that by setting $c=(2x)^2 $, this gave me a polynomial $g(c)=c^3 -5c^2 +6c-1$, which is irreducable over the rationals by the rational root test and being a cubic.

So, the question is...does this make $f$ irreducible, so my algebraic degree is 6? Or does since $g$ is irreducible and I'd have to extract a square root to get back to $x$, does that then give me a field extension of degree 6? In any case, obviously my goal is to prove it's algebraic of a degree that's not a power of 2.

Alan
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  • Unfortunately, $f$ is not irreducible. It factors into a product of irreducible cubics (which is not a problem for the proof itself; it's just hard to show). – Kaj Hansen Jul 14 '15 at 03:47
  • Check, thanks! Been a while since I've broken out the field extension work, cramming on my Galois and constructible stuff. The hazards of taking a year off from algebra. – Alan Jul 14 '15 at 04:09
  • I made some changes to my answer that make it a lot more succinct, but the argument is the same. – Kaj Hansen Jul 14 '15 at 04:30
  • $\sin (\frac {2\pi} 7)$ is not a zero of your polynomial $f$. – jbuddenh Jul 14 '15 at 04:48
  • Decent chance I made a basic algebra error then in my scribbling. Still useful as an answer to what to do if something like this comes up. (I was playing around with Demoivre's formula with 7th powers and 1-cos^2 transforms and binomial formulas on scrap paper :) – Alan Jul 14 '15 at 05:00

3 Answers3

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As a general rule, $f(x)$ being irreducible does not mean $f(x^2)$ is irreducible.

For a simple example, $f(x)=x-1$.

Another example: $f(x)=x^2+x+1$ then $f(x^2)=f(x)f(-x)$.

Viktor Vaughn
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Thomas Andrews
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$\sin \left( \frac{2 \pi}{7}\right)$ is a zero of $f(x)=64x^6-112x^5+56x^3-7$. By Eisenstein's criterion with $p=7$, it is clear that this polynomial is irreducible.

Race Bannon
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Assuming everything you've done up to this point is correct, then based on your work, we know the roots of $f$ will be of the form $\displaystyle \frac{ \pm \sqrt{\alpha}}{2}$, where $\alpha$ is any of the three roots of that cubic.

Now notice that $\displaystyle \frac{\pm \sqrt{\alpha}}{2} \in \mathbb{Q}[\sqrt{\alpha}]$, and we have a tower of fields $\mathbb{Q} \subset \mathbb{Q}[\alpha] \subset \mathbb{Q}[\sqrt{\alpha}]$. Now consider the degrees of these extensions to finish up.

Given the tower of fields, we know the degree of $\mathbb{Q}[\sqrt{\alpha}]$ over $\mathbb{Q}$ is equal to the product of $[\mathbb{Q}[\alpha]:\mathbb{Q}]=3$ and $[\mathbb{Q}[\sqrt{\alpha}]:\mathbb{Q}[\alpha]]$. It follows that the degree of $\mathbb{Q}[\sqrt{\alpha}]$ over $\mathbb{Q}$ is divisible by $3$ and thus is not a power of $2$.

Kaj Hansen
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  • Yes, very nice...have to remember I don't need the exact degree, just to show I have a factor that's not 2. Useful for my toolkit for the upcoming test! – Alan Jul 14 '15 at 04:31
  • Glad I could help! Personally, I find that creating towers of fields like this comes in handy all the time in various situations. Good luck on your test :) – Kaj Hansen Jul 14 '15 at 04:36