Is it true that the splitting field for $x^n-1$ over $\mathbb{Q}$ is $\mathbb{Q}(\xi_n)$ where $\xi_n$ is a primitive n$^{th}$ root of unity, making it an extension of degree $\phi(n)$ (Euler phi function)? Every element of this extension should look like $\sum a_k\xi_n^k$?
[Math] splitting field of $x^n-1$ over $\mathbb{Q}$
abstract-algebraextension-fieldfield-theorygalois-theorysplitting-field
Related Solutions
First reduce the $\zeta$'s: you have $$\mathbb{Q}(\zeta_{77}, \zeta_{15}) = \mathbb{Q}(\zeta_{lcm(77,15)}) = \mathbb{Q}(\zeta_{1155}).$$ This is an extension of $\mathbb{Q}$ of degree $\varphi(1155) = \varphi(77)\cdot \varphi(15) = 480.$
What is the intersection $\mathbb{Q}(\zeta_{1155}) \cap \mathbb{Q}(\sqrt[15]{5})$?
Subfields of cyclotomic fields are abelian (the converse is also true), that is, they are Galois with abelian Galois groups. However, the nontrivial subfields of $\mathbb{Q}(\sqrt[15]{5})$ - $\mathbb{Q}(\sqrt[3]{5})$, $\mathbb{Q}(\sqrt[5]{5})$ and $\mathbb{Q}(\sqrt[15]{5})$ - are not Galois. So you have $\mathbb{Q}(\zeta_{1155}) \cap \mathbb{Q}(\sqrt[15]{5}) = \mathbb{Q}$ and $$[\mathbb{Q}(\zeta_{1155}, \sqrt[15]{5}) : \mathbb{Q}] = 480 \cdot 15 = 7200.$$
If by $\Bbb{Z}_{p}$ you mean the field $\frac{\mathbb{Z}}{p\mathbb{Z}}$ that is the field of integers modulo $p$. Then do as follows:-
for any non-zero $a\in\frac{\mathbb{Z}}{p\mathbb{Z}}$ you have $a^{p-1}=1$ by Fermat's Little Theorem.
But this polynomial itself has degree $p-1$ and hence can have atmost $p-1$ roots.
So all the $p-1$ elements of $\Big(\frac{\mathbb{Z}}{p\mathbb{Z}}\Big)^{\times}$ are itself the roots of this polynomial.
And since $\frac{\mathbb{Z}}{p\mathbb{Z}}$ does not have any proper subfield of order $p-1$. , it follows that it is the splitting field.
More generally a field of order $p^{n}$(denoted by $\mathbb{F}_{p^{n}}$) is precisely the splitting field of the polynomial $x^{p^{n}}-x$.
You can argue like above result to prove this. See that all elements of the field are roots of this polynomial and the smallest subfield containing $p^{n}$ elements of $\mathbb{F}_{p^{n}}$ is $\mathbb{F}_{p^{n}}$ hence it is the splitting field.
Consequently you can use the uniqueness of splitting field to conclude that any finite field of order $p^{n}$ is precisely the splitting field of the polynomial $x^{p^{n}}-x\in \frac{\mathbb{Z}}{\mathbb{pZ}}[x]$. (Note that any field of characteristic $p$ contains $\frac{\mathbb{Z}}{\mathbb{pZ}}$ as it's prime subfield.
Also a slight piece of advice regarding notation. Most people use $\mathbb{F}_{p}$ to denote the field of integers modulo $p$ in order to avoid confusion with the p-adic integers.
Best Answer
Yes, everything you said is true. I have only added that the $a_k$ should be elements in $\mathbb Q$. These fields are known as the Ciclotomic Fields.