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(Credits: This que is taken from NBHM 2022, which is written in terms of equivalent condition of this using topology and closed set)

I tried $z^{15}-1$ is a trivial polynomial having one of its roots. And suspected that the least degree polynomial divides this.

So I tried to factorize this. $z^{15} -1=(z-1)(1+z+ ...z^{14})$

Then since clearly $z \ne 1 $, I suspected the least degree polynomial divides $(1+z+ ...z^{14})$

Tried to factorize (like grouping 5 terms and taking common term outside) and also(like grouping 3 terms).

But I don't clearly get the way to solve further claiming the root lies in which factor.

Eric Wofsey
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Muthu'Maths_student'
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    show that $z^{15}-1=(z-1)(z^2+z+1)(z^4+z^3+..1)Q(z)$ with $\deg Q=8=\phi(15)$ and then show that $Q$ is your required answer – Conrad Oct 14 '22 at 18:36
  • The minimal polynomial is $\Phi_{15}$, which has degree $\phi(15)=8$. – Dietrich Burde Oct 14 '22 at 18:43
  • This is just the 15th cyclotomic polynomial, which is the minimal polynomial of the 15th root of unity. You can find more about how to compute them on Wikipedia, for instance. – Mark Saving Oct 14 '22 at 18:44

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