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I wanted to refer to this, but I can't remember if this a theorem, named or otherwise, and if it is, how to properly state it. The idea is if we have a solution in radicals to a polynomial with integer coefficients, then where we have a $n^{th}$ root expression within the solution we can multiply that expression by any $n^{th}$ root of unity and this will also be a solution.

i.e. if $p(2+\sqrt{3+\sqrt7})=0$ then

$p(2+\sqrt{3-\sqrt7})=p(2+\sqrt{3+\sqrt7})=p(2-\sqrt{3-\sqrt7})=0$

if $p(2+\sqrt{3+\sqrt[3]7})=0$ then

$p(2+\sqrt{3+\sqrt[3]7})=p(2-\sqrt{3+\sqrt[3]7})=p(2+\sqrt{3+\sigma_{1}\sqrt[3]7})=p(2-\sqrt{3+\sigma_{1}\sqrt[3]7})=p(2+\sqrt{3+\sigma_{2}\sqrt[3]7})=p(2-\sqrt{3+\sigma_{2}\sqrt[3]7})=0$

Where $\sigma_{1} = e^{\frac{2}{3} \pi i}$, $\sigma_{2} = e^{\frac{4}{3} \pi i}$

I think this relates to Field Theory where these other solutions are the 'conjugates' of polynomial used for extension of a field. But that's pretty fuzzy to me nowadays. I may just have guessed at this when I studied it a long time ago.

Also - this set of solutions is the set of solutions for an integer polynomial which will be a factor of $p$?

Also - possibly we not even had to suppose the original expression was a solution, this tells us how to construct the required polynomial?

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There are some limitations to the combinations of the roots of unity that can be used. Nested simplifiable square roots provide some. Also with cubics. Consider the polynomial $$ p(x)=x^3-6x-6. $$ Its zeros are $$ x_1=\root3\of2+\root3\of4,\quad x_2=\sigma_1\root3\of2+\sigma_2\root3\of4,\quad x_3=\sigma_2\root3\of2+\sigma_1\root3\of4. $$ But only these combinations of third roots of unity work. See a derivation of Cardano's formula for an explanation (or review Galois theory).

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    $\begingroup$ Other causes of failure of this (often true!) observation in the OP are the relations between roots of unity and other roots. Of these the better knowns are perhaps the relation between $\sqrt{-3}$ and the third roots of unity, and that between $\sqrt5$ and the fifth roots of unity (via $\cos(2\pi/5)$). $\endgroup$ – Jyrki Lahtonen Jan 24 '14 at 11:49
  • $\begingroup$ Thanks for all your observations. Do you know if the last question at least holds up? If I have an expression in radicals, will I construct an integer polynomial by taking all the combinations as described in the OP, and using them as the set of roots? $\endgroup$ – Neil W Jan 25 '14 at 12:40
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Counterexample: take $t = \sqrt{8 + 2 \sqrt{7}}$, $s = \sqrt{8 - 2 \sqrt{7}}$, $p(x) = x^2 - 2 x - 6$. Then $p(t) = 0$ but $p(s) = 4 - 4 \sqrt{7}\ne 0$. The reason is that these nested square roots simplify. (I am, of course, using the principal branch of the square root)

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