Question about the Cyclotomic polynomial and Cyclotomic field. I am studying for my Algebra PHD Qualifying and I can not solve the first 3 parts of this problem. I think I got the 4 correct. I have some ideas but I can't put them together to solve it. Anyone can help me ? Thanks.
Let $p$ be prime. Consider the number field $E = \mathbb{Q}(\zeta_p)$, with $\zeta_p = \exp\left(\frac{2\pi i}{p}\right)$. Suppose that $\gcd(k,p-1) = 1$. 


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*Find $T_{E|\mathbb{Q}}(\zeta_p^{k})$ and $N_{E|\mathbb{Q}}(\zeta_p^k)$

*Let $\mathcal{O}[\zeta_p]$ be the set of all integral elements of the cyclotomic extension E. Show that $p$ is no longer irreducible in $\mathcal{O}[\zeta_p]$.

*Find $T_{E|\mathbb{Q}}(1-\zeta_p^{k})$ and $N_{E|\mathbb{Q}}(1-\zeta_p^k)$

*Assuming that $\mathcal{O}[\zeta_p] = \mathbb{Z}[\zeta_p]$, find the field discriminant of the cyclotomic extension E. 
For 1) I coud write the matrix general matrix and try to find the trace and determinant but I think it will be messy. Also its not possible to calculate the characteristic polynomial to obtain the coefficients. So I have no idea.
I know that the cyclotomic polynomial is the minimal polynomial. 
For 2) I can not figure it out the factorization
For 3) I suspect that after we could solve 1, then 3 will be easy using the Einsestein criterion for with the cyclotomic polynomial evaluated at $x+1$.
My proof of 4)
Let ${\displaystyle \varphi (n)}$ be the Euler's totient function. The discriminant of the extension for $n \in \mathbb{N}$ in general is given by
$${\displaystyle (-1)^{\varphi (n)/2}{\frac {n^{\varphi (n)}}{\prod_{p|n}{\displaystyle p^{\varphi (n)/(p-1)}}}}}$$
Using that $n = p$ is prime and $\varphi(p) = p-1$ we obtain the formula for field extension discriminant
$$\displaystyle (-1)^{(p-1)/2}\frac {n^{(p-1)}}{p} \hspace{5pt} \square$$
 A: 1) Note that $\{\zeta_p^k \mid 1\le k \le p-1 \}$ are the roots of the $p-$th cyclotomic polynomial $\Phi_p$. For the primes the cyclotomic polynomial has the following form:
$$\Phi_p(x) =x^{p-1} + x^{p-2} + \cdots + x + 1$$
This will help you determine that $T(\zeta_p^k) = -1$ and $N(\zeta_p^k) = 1$
3) You can note that $1-\zeta_p^k$ is a root of the polynomial $\Phi_p(1-x)$, which is irreducible. Use this to find your answer. If you want to avoid the minus sign, just note that $N(1-\zeta_p^k)=N(\zeta_p^k-1)$ and $T(1-\zeta_p^k)=-T(\zeta_p^k-1)$ and use $\Phi_p(x+1)$ to find the trace and norm of $\zeta_p^k-1$
2) From above you would be able to conclude that $N(1-\zeta_p^k) =p$. So $1-\zeta_p^k \mid p$ As this is true for all $k$ we have that $p = \prod_{i=1}^{p-1}(1-\zeta_p^i)$. This is the factorization you want in $\mathcal{O}[\zeta_p]$
4) You can use the formula $\text{disc}(1,\alpha,\dots,\alpha^{n}) = (-1)^{\frac{n(n-1)}{2}}N(f'(\alpha))$, where $f$ is the minimal polynomial of $\alpha$ over $\mathbb{Q}$. Here $f=\Phi_p$ and we have the following:
$$x^p-1=(x-1)\Phi_p(x) \implies px^{p-1} = \Phi_p(x) + (x-1)\Phi_p'(x)$$
Evaluate the expression at $\zeta_p$. To get $\Phi_p'(\zeta_p) = \frac{p\zeta_p^{p-1}}{\zeta_p-1}$. Finally:
$$N(\Phi'(\zeta_p)) = \frac{N(p)N(\zeta_p)}{N(\zeta_p-1)}=\frac{p^{p-1}\cdot 1}{p} = p^{p-2}$$
