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I am trying to find the minimal polynomials of $\zeta_7+\zeta_7^{-1}$ and $\zeta_7+\zeta_7^2+\zeta_7^4$ over $\Bbb Q$.

For the first one, I found the following solution on the site (Minimal polynomial of $\omega:=\zeta_7+\overline{\zeta_7}$).

Here is the solution from Ewan Delanoy:

$$ \omega=\zeta+\zeta^6, \ \ \ \omega^2=\zeta^2+\zeta^5+2, \ \ \ \omega^3=\zeta^3+\zeta^4+3\omega $$

Adding all those three up, you obtain

$$ \omega^3+\omega^2+\omega =\sum_{k=1}^{6} \zeta^k+(3\omega+2)= -1+(3\omega+2)=3\omega+1 $$

So

$$ \omega^3+\omega^2-2\omega-1=0 $$

I can verify that this is indeed the the minimal polynomial of $\zeta_7+\zeta_7^{-1}$ over $\Bbb Q$.But I don't really understand the intermediate steps involved in finding that polynomial.I suspect that there might be some simplification rules for the root of unity which I don't know. So I am asking whether there are some properties of the root of unity used in the above calculation?

Also, how would I go about to find the minimal polynomial of $\zeta_7+\zeta_7^2+\zeta_7^4$ over $\Bbb Q$?

Could someone give me a hand? Thanks so much.

Community
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    The only properties used in Ewan's solution seems to be the definition of a root of unit, namely that $\zeta_7^7 = 1$, together with the fact that $\zeta_7 \neq 1$, which implies that $\sum_{k = 0}^6 \zeta_7^k = 0$. (Strictly speaking, by the way, one should show that $\omega$ does not satisfy a lower-order polynomial over $\Bbb Q$, that is, that the given polynomial doesn't factor, but this follows immediately from the Rational Root Test.) – Travis Willse Mar 20 '17 at 15:03
  • Did you mean that $\omega^3+\omega^2-3\omega-1=0$? – ancient mathematician Mar 20 '17 at 15:59

2 Answers2

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Surely all there is to know (algebraically) about $\zeta=\zeta_7$ is that it is a root of the irreducible polynomial $X^6+X^5+X^4+X^3+X^2+X+1$ ?

The other roots are the powers of $\zeta$ (except $1$), and algebraically they are indistinguishable.

So what we can say algebraically about $\zeta+\zeta^{-1}$ we can say about $\zeta^2+\zeta^{-2}$ and $\zeta^4+\zeta^{-4}$. [Any choice of $\zeta^i\not=1$ leads to one of these.]

So a bold guess for the minimal polynomial of $\zeta+\zeta^{-1}$ is $$ (X-(\zeta+\zeta^{-1}))(X-(\zeta^2+\zeta^{-2}))(X-(\zeta^4+\zeta^{-4})) $$ which, using $\zeta^6+\zeta^5+\zeta^4+\zeta^3+\zeta^2+\zeta+1=0$, is $$ X^3+X^2-3X-1,$$ which is irreducible and so the answer.

Similarly, $\alpha=\zeta+\zeta^2+\zeta^4$ and $\beta=\zeta^{-1}+\zeta^{-2}+\zeta^{-4}$ are the roots of $$ (X-\alpha)(X-\beta)=\dots=X^2+X+2, $$ again using only $\zeta^6+\zeta^5+\zeta^4+\zeta^3+\zeta^2+\zeta+1=0$.

ancient mathematician
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For $\zeta=\zeta_7$, here’s how I find the minimal polynomial of $\xi=\zeta+1/\zeta$: \begin{align} \zeta^3+\zeta^2+\zeta+1&+\zeta^{-1}+\zeta^{-2}+\zeta^{-3}=0\\ \xi^3=\zeta^3+3\zeta&+3\zeta^{-1}+\zeta^{-3}\\ =-\zeta^2+2\zeta-1&+2\zeta^{-1}-\zeta^{-2}\\ \xi^3+\xi^2=2\zeta+1&+2\zeta^{-1}\,, \end{align} and thus $\xi^3+\xi^2=2\xi+1$.

You can use the same method for any $\zeta_n+\zeta_n^{-1}$.

Lubin
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  • I think your method only works when $n$ is prime? For example, how should I use the method if $n=9$? – Y.X. Apr 13 '17 at 13:15
  • If you want the equation for $\zeta_9+1/\zeta_9$, @PropositionX, you must start with the minimal polynomial for $\zeta_9$, which is $X^6+X^3+1$. Proceed as before. (I’ve done it, it works.) – Lubin Apr 13 '17 at 13:19
  • I would really appreciate it if you could add this example to your answer, if I may ask. – Y.X. Apr 13 '17 at 13:21
  • But it’s exactly the same procedure, @PropositionX. You can do it for yourself. – Lubin Apr 13 '17 at 13:23