Minimal polynomial for $\zeta+\zeta^5$ for a primitive seventh root of unity $\zeta$ Minimal polynomial for $\zeta+\zeta^5$ for a primitive seventh root of unity $\zeta$
I have asked a similar problem Minimal Polynomial of $\zeta+\zeta^{-1}$ and i tried to repeat similar idea
Consider $\alpha=\zeta+\zeta^5$ and consider $\sigma \in Gal(\mathbb{Q}(\zeta)/\mathbb{Q})$ defined as $\sigma(\zeta)=\zeta^2$
so, we consider conjugates of $\alpha$ under $\sigma$


*

*$\alpha=\zeta+\zeta^5$

*$\sigma(\alpha)=\sigma(\zeta+\zeta^5)=\zeta^2+\zeta^3$

*$\sigma^2(\alpha)=\sigma(\zeta^2+\zeta^3)=\zeta^4+\zeta^6$

*$\sigma^3(\alpha)=\alpha$


So i would consider $(x-\alpha)(x-\sigma(\alpha))(x-\sigma^2(\alpha))$
This is giving me $x^3+x^2+(2+\zeta+\zeta^2+\zeta^4)x-1$
I do not know how to make coefficient of $x$ as real..
Please help me to fill this gap..
 A: Because $7$ is prime, we know that, $Aut(\mathbb{Q}[\zeta_7]/\mathbb{Q}) \cong \mathbb{Z}_7^\times \cong \mathbb{Z}_6$.
From here, we wish to find a generator for that automorphism group.  Note that $\mathbb{Q}[\zeta_7]$ is the splitting field for the irreducible polynomial $f(x) = x^6 + x^5 + \cdots + x + 1$ whose roots are the $6$ nontrivial $7$th roots of unity.  Now, it is a theorem that the automorphism group acts transitively on the roots of the polynomial $\iff$ that polynomial is irreducible.  Therefore, we know there must exist an automorphism within this group such that $\zeta \mapsto \zeta^n$ for any $n \in \{1, 2, ..., 5\}$.  The easiest way I know how to do this is through experimentation.
Once you've found your generator, $\phi$, then your automorphism group is simply $\{id, \phi, \phi^2, ..., \phi^5\}$.  
The motivation behind all this work was that you need to let your entire automorphism group act on $\zeta + \zeta_5$ to generate the minimal polynomial.  If we denote $\{\alpha_1, \alpha_2, ..., \alpha_k \}$ as the orbit of $\zeta + \zeta^5$ under this action, then the minimal polynomial will be be $m(x) = \displaystyle \prod_{i=1}^k\Big(x-\alpha_i\Big)$.
The moral of the story is that you had the right idea, but the previous sentence is where you went astray.

Edit: It is not necessary to actually find the generator, but it is necessary to let the entire automorphism group act on $\zeta + \zeta^5$.  Note that there will be an automorphism that maps $\zeta \mapsto \zeta^k$ for all $k$ by the reasoning above, and each one must (obviously) be unique.  Thus, the $6$ of them are:


*

*$\zeta \mapsto \zeta$

*$\zeta \mapsto \zeta^2$

*$\zeta \mapsto \zeta^3$


etc.
To conclude, if possible, I would recommend you find a copy of Artin and read Chapter 14, Section 4, Proposition 4.4.  I think that would get to the crux of where you went wrong initially.  You can find a .pdf version of this text here:  http://www.drchristiansalas.org.uk/MathsandPhysics/AbstractAlgebra/ArtinAlgebra.pdf
A: The map $\tau: \zeta \rightarrow \zeta^3$ is a generator of the automorphism group of $Q(\zeta)$, $\zeta^7=1$, $\zeta\ne 1$. Your automorphism $\sigma: \zeta \rightarrow \zeta^2$ is $\tau^2$, so it represents an element of order 3 in the Galois group.  So its fixed field is a quadratic extension of $K$ of  $Q$. 
The coefficients of your cubic polynomial  $f$ lie in $K$ not in $Q$.  You have computed the minimal polynomial over $K$ instead of $Q$. The field $K$ has one non-trivial automorphism induced by complex conjugation so to obtain the irreducible polynomial over $Q$ we just compute $g=f\bar{f}$ (applying complex conjugation to the coefficients). 
Complex conjugation is also an automorphism of the field $Q(\zeta)$ also and it is given by $\zeta\rightarrow \zeta^{-1}=\zeta^6$ (it is $\tau^3$). Complex conjugation does not belong to the subgroup generated by your element $\sigma$. 
Your cubic is $f=x^3+x^2+\frac{3+\sqrt{-7}}{2}x-1$, $\bar{f}=x^3+x^2+\frac{3-\sqrt{-7}}{2}x-1$ and $g=f\bar{f}=x^6+2x^5+4x^4+x^3+2x^2-3x+1$. 
This is the exactly the same as multiplying all $\prod_{i=0}^5 (x-\tau^i(\zeta+\zeta^5))$, since $f=\prod_{i=0}^2 (x-\tau^{2i}(\zeta+\zeta^5))$ and $\bar{f}=\prod_{i=0}^2 (x-\tau^{2i+1}(\zeta+\zeta^5))$
A: Credits for this answer are for Jyrki Lahtonen and i.m.soloveichik
I have considered Only $(x-(\zeta+\zeta^5))(x-(\zeta^2+\zeta^3))(x-(\zeta^4+\zeta^6))$ and simplified this to $$x^3+x^2+(2+\zeta+\zeta^2+\zeta^4)x-1$$
Now i would consider other conjugates of $\zeta+\zeta^5$ namely $\zeta^6+\zeta^2;\zeta^5+\zeta^4;\zeta^3+\zeta$
With Jyrki Lahtonen's link Is $\sqrt 7$ the sum of roots of unity? we see that for $S=\zeta+\zeta^2+\zeta^4$ we have 
$$(2S+1)^2=-7\Rightarrow S=\dfrac{i\sqrt{7}-1}{2}\Rightarrow 2+\zeta+\zeta^2+\zeta^4=2+S=\dfrac{i\sqrt{7}+3}{2}$$
So, My polynomial is $$f(x)=(x^3+x^2+(\dfrac{i\sqrt{7}+3}{2})x+1)$$
As i was missing conjugates I would consider conjugate of the polynomial $f(x)$ and multiply with $f(x)$
Conjugate of $f(x)$ is 
$$f(x)=(x^3+x^2+(\dfrac{-i\sqrt{7}+3}{2})x+1)$$
Now i multiply (I did with pen and paper and do not want to write those calculations here) and realized that their product is 
$$g(x)=x^6+2x^5+4x^4+x^3+2x^2-3x+1$$
