Sum of roots of unity while computing irreducible polynomial

Question

I'm trying to compute the minimal polynomial of a root of $\xi+\xi^{-1}$ where $\xi$ is a fourteenth root of unit, that is, $x^{14} = 1$.

Using Galois theory I was able to determine that the minimal polynomial is $(X-(\xi+\xi^{-1}))(X-(\xi^4+\xi^{-4}))(X-(\xi^2+\xi^{-2}))$ my problem now is that it seems not evident how to expand this product.

How does one compute for instance, $\xi^4+\xi^{-4}+\xi^2+\xi^{-2}+\xi+\xi^{-1}$ as a rational number which appears for instance in the $X^2$ of the expansion of this polynomial?

Answer 1

Well, one way to do this is to first find the minimal polynomial of $\xi$ itself. You can find that by dividing $x^{14}-1$ by factors which have as roots the non-primitive 14th roots of unity. In this case, that's pretty easy: we can first factor $x^{14}-1$ as $(x^7-1)(x^7+1)$, where the first factor has the 7th roots of unity. The roots of the second factor $x^7+1$ are the primitive 14th roots of unity together with $-1$, so to get the minimal polynomial of the primitive roots, we just divide by $x+1$ to get $$x^6-x^5+x^4-x^3+x^2-x+1.$$ So, that's the minimal polynomial of $\xi$. Now if we want to evaluate $$\xi^4+\xi^{-4}+\xi^2+\xi^{-2}+\xi+\xi^{-1}=\xi^4+\xi^{10}+\xi^2+\xi^{12}+\xi+\xi^{13}$$ we can just take the remainder when we divide the polynomial $$x^{13}+x^{12}+x^{10}+x^4+x^2+x$$ by the minimal polynomial $x^6-x^5+x^4-x^3+x^2-x+1.$

Answer 2

I'll give three different (not quite) methods:

1) The most direct approach, as in @Eric Wofsey 's answer, is to determine first the minimal polynomial of $\zeta = e^{i\pi/7}$, which is $X^6 - X^5 + X^4 -X^3 + X^2 - X + 1$. This is a "palindromic" polynomial. Putting $T=X+1/X$ and computing $T^2, T^3$, one gets the minimal polynomial of $ cos (\pi/7)$, which is $T^3-T^2-2T+1$. NB: this polynomial is not solvable by ruler and compass since $7$ is not a Fermat prime.

2) The second approach is to use a bit of Galois theory to reduce to the minimal polynomial of $cos (2\pi/7)$. Let $\eta=\zeta^2=e^{2i\pi/7}$. Because $(\mathbf Z/14)^*=(\mathbf Z/7)^*\cong C_6$, the two extensions $\mathbf Q(\zeta)/\mathbf Q$ and $\mathbf Q(\eta)/\mathbf Q$ coincide, and their Galois group $C_6$ is generated by the complex conjugation $\tau : \zeta \to \zeta^{-1}$ and the 3-cycle $\sigma: \zeta \to \zeta^3$. Since $\eta \in\mathbf Q(\zeta)$, one has $\mathbf Q(\zeta+\zeta^{-1})=\mathbf Q(\eta+\eta^{-1})$, and the two minimal polynomials of $cos(\pi/7)$ and $cos(2\pi/7)$, resp. $f=(X-(\zeta+\zeta^{-1}))(X-(\zeta^{2}+\zeta^{-2}))(X-(\zeta^{4}+\zeta^{-4}))$ (see your post) and $g=(X-(\eta+\eta^{-1}))(X-(\eta^{2}+\eta^{-2}))(X-(\eta^{4}+\eta^{-4}))$ are obtained one from another by applying $(\tau \sigma^2)^{\pm 1}$ to the respective three factors. I thought that a priori $g$ would be easier to obtain by the following method : since $\eta^{6}+\eta^{5}+ ...+1=0$, taking the real part will give $cos(6\theta)+cos(5\theta)+...+1=0$, with $\theta = 2\pi/7 $. Using the relations $cos(6\theta)=cos(\theta), cos(5\theta)=cos(2\theta), cos(4\theta)=cos(3\theta), cos(3\theta)=4cos(\theta)^3-3cos(\theta)cos(2\theta)=2cos(\theta)^2 -1$, and computing, one gets $g=X^3+X^2-2X-1$. But finally, the same argument as in 1) gives the same result.

3) The third, more general approach is to use the Tchebychev polynomials $T_n$ , characterized by $T_n (cos\theta)=cos(n\theta)$, to determine the minimal polynomials of $cos(2\pi/n)$ in a kind of recursive manner, like for the determination of the cyclotomic polynomials (but with less efficiency). See e.g. W. Watkins & J. Zeitlin, "The minimal polynomials of $cos(2\pi/n)"$, Amer. Math. Monthly, 1983, 471-474 .

EDIT. As noticed by some people, in the definition of $\sigma$, the exponent 2 must be replaced by 3. I made the necessary changes.