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Question is to find Minimal Polynomial of $\zeta+\zeta^{-1}\in \mathbb{Q}(\zeta)$ over $\mathbb{Q}$ Where $\zeta$ is primitive $13^{th}$ root of Unity. What all i know is that Minimal polynomial of Degree 6.

Usually, given an element say $\sqrt{2}+\sqrt{3}$ to find minmal polynomial, we take $\alpha=\sqrt{2}+\sqrt{3}$ and the square it and then do further simplifications to get a linear Combination of powers of $\alpha$ (polynomial in $\alpha$) equal to zero and if that polynomial is irreducible, we say it as minimal polynomial for given element over given field.

Where as for this element $\zeta+\zeta^{-1}$ It is getting too complicated as i go in that way.

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Please suggest me some other procedure (if any) or a hint to simplify the calculation.

Thank You ðŸ™‚

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Galois theory provides some machinery for this:

Suppose $\rm L/K$ is Galois with $\rm G=Gal(L/K)$ and $\rm m(x):=minpoly_{\alpha,K}(x)$. Then

$\quad \rm m(\sigma(\alpha))=\sigma(m(\alpha))=\sigma(0)=0$ implies $\rm (x-\sigma\alpha)\mid m$ in $\rm L[x]$ for all $\rm \sigma\in G$,

$\quad \rm(x-\sigma\alpha)$ all coprime, $\rm \sigma\in G/S$, $\rm S=Stab_G(\alpha)$, implies $\rm f(x):=\prod\limits_{\sigma\in G/S}(x-\sigma\alpha)\mid m$ in $\rm L[x]$,

$\quad \rm \sigma f(x)=f(x)$ for all $\sigma\in G$ implies $\rm f(x)\in K[x]$; $\rm f(\alpha)=0$ implies $\rm m(x)\mid f(x)$ in $\rm K[x]$,

$\quad \rm f(x)\mid m(x)$ and $\rm m(x)\mid f(x)$ and both $\rm f,m$ monic implies $\rm f(x)=m(x)$.

Therefore the zeros of $\rm\alpha$’s minimal polynomial over $\rm K$ are precisely its $\rm Gal(L/K)$-conjugates.

As ${\rm Gal}\big({\bf Q}(\zeta_n)/{\bf Q}\big)=({\bf Z}/n{\bf Z})^\times$ it suffices to consider $\sigma:\zeta\mapsto\zeta^k$ for $k=1,\cdots,12$.

By symmetry we need only consider $1,\cdots,6$ for $\alpha=\zeta+\zeta^{-1}$. Thus

$$\begin{array}{ll} {\rm minpoly}_{\zeta+\zeta^{-1},\bf Q}(x) = & ~~~~\left(x-(\zeta+\zeta^{-1})\right)\left(x-(\zeta^2+\zeta^{-2})\right)\left(x-(\zeta^3+\zeta^{-3})\right) \\ & \times\left(x-(\zeta^4+\zeta^{-4})\right)\left(x-(\zeta^5+\zeta^{-5})\right)\left(x-(\zeta^6+\zeta^{-6})\right). \end{array}$$

Simplify the resulting expansion via the rules $\zeta^n=\zeta^{n\,\bmod\,13}$ and $\sum\limits_{k=0}^{12}\zeta^k=0$.

Zero-theory computational approach:

Since $\zeta^{13}-1=0$, $\sum\limits_{k=0}^{12}\zeta^k=0$ hence $1+\sum\limits_{k=1}^6x_k=0$, where $x_k=\zeta^k+\zeta^{-k}$. Let $x=x_1$, then the system $x^k=\sum\limits_{2i\leqslant k}{k\choose i}x_{k-2i}$ for every $k\geqslant1$ is lower triangular with unknowns $(x_k)_{k\geqslant1}$ and unit diagonal hence each $x_k$ is a linear combination with integer coefficients of $x^i$ for $i\leqslant k$, for example $x_1=x$, $x_2=x^2-2$, $x_3=x^3-3x$, $x_4=x^4-4x^2+2$, and so on.

Thus, $\sum\limits_{k=1}^6x_k=P_6(x)$ where $P_6$ is a monic polynomial of degree $6$ with integer coefficients, and $P_6(x)+1=0$.

This answer is in response to the comment by @anon.

It isn’t the case that Chebyshev polynomials are an alternative to Galois theory. In an approach based on Chebyshev polynomials, you still need to first determine the degree $[{\bf Q}(\zeta+\zeta^{-1}):{\bf Q}]$ via Galois theory. But once you know this degree, then you can write down the minimal polynomial by means of Chebyshev polynomials.

Let $\zeta$ be a primitive $n$-th root of unity, where $n$ is an arbitrary odd prime. Writing $\alpha:=\zeta+\zeta^{-1}$, we know via Galois theory that the minimal polynomial of $\alpha$ over ${\bf Q}$ has degree $(n-1)/2$. Thus, if we can write down a monic polynomial over ${\bf Q}$ of degree $(n-1)/2$ which has $\alpha$ as a root, then this polynomial will automatically be the minimal polynomial of $\alpha$ over ${\bf Q}$. To achieve this, recall that the Chebyshev polynomial $T_n(X)$ is a degree-$n$ polynomial in ${\bf Q}[X]$ which satisfies the functional equation

$$T_n(\cos\theta)=\cos(n\theta).$$

Write $z=e^{i\theta}=\cos\theta+i\sin\theta$, so that $z+z^{-1}=2\cos\theta$ and thus

$$T_n\left(\frac{z+z^{-1}}2\right)=\frac{z^n+z^{-n}}2.$$

It follows that

$$T_n\left(\frac{Z+Z^{-1}}2\right)=\frac{Z^n+Z^{-n}}2$$

is an equality of rational functions, since it is true for infinitely many values of $Z$. Note that

$$T_n\left(\frac{\alpha}2\right)=\frac{\zeta^n+\zeta^{-n}}2=1,$$

so $\alpha$ is a root of $T_n(X/2)-1$. Now write $X=Z+Z^{-1}$, so that

$$T_n\left(\frac{X}2\right)-1 = \frac{Z^n+Z^{-n}}2 – 1 = \frac12 \left(Z^{n/2}-Z^{-n/2}\right)^2.$$

One can show directly that

$$\frac{Z^{n/2}-Z^{-n/2}}{Z^{1/2}-Z^{-1/2}}$$

is a polynomial $h(X)$ in ${\bf Z}[X]$ (in fact it is $U_{(n-1)/2}(X/2)+U_{(n-3)/2}(X/2)$, where $U_m(X)$ is the degree-$m$ Chebyshev polynomial of the second kind; see my answer to this other question). Thus

$$T_n\left(\frac{X}2\right)-1 = \frac12 (Z^{1/2}-Z^{-1/2})^2 \cdot h(X)^2 =

\frac12(Z-2+Z^{-1})\cdot h(X)^2 = \frac{X-2}2\cdot h(X)^2.$$

Since $\alpha$ is a root of the left side, and $\alpha$ is not a root of $X-2$, it follows that $\alpha$ is a root of $h(X)$. But $h(X)$ has degree $(n-1)/2$, and hence is a constant times the minimal polynomial of $\alpha$ over ${\bf Q}$. In fact it’s easy to check that $h(X)$ is monic, so it actually equals the minimal polynomial of $\alpha$ over ${\bf Q}$. Hence the minimal polynomial of $\alpha$ over ${\bf Q}$ is $U_{(n-1)/2}(X/2) + U_{(n-3)/2}(X/2)$. Now, for instance, we can write out the coefficients of this minimal polynomial, since we know that

$$

U_m\left(\frac{X}2\right) = \sum_{k=0}^{\lfloor m/2\rfloor}\binom{m-k}k (-1)^k X^{m-2k}.$$

Strangely, this expression for the coefficients of $U_m$ doesn’t seem to be in the wikipedia article for Chebyshev polynomials. For a derivation of it from first principles, you can see for instance the beginning of my paper with Abhyankar and Cohen. In that paper we used Dickson polynomials rather than Chebyshev polynomials; the translations to the notation of this answer are $U_m(X/2)=E_m(X,1)$ and $2T_m(X/2)=D_m(X,1)$ (these translations follow from the relevant functional equations).

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