Degree of $\mathbb{Q}(\xi_{p^{2}})$ over $\mathbb{Q}$. What is the degree of the extension $\mathbb{Q}(\xi_{p^{2}})$ over $\mathbb{Q}$ where $p$ is a prime and $\xi_{p^{2}}$ is a primitive $p^{th}$ root of unity?
 A: The degree of $\zeta_n$ over $\mathbb{Q}$ is $\phi(n)$ (Euler-phi function), for $n=p^2$ that is $p(p-1)$.
A: Let me expand Keith’s (KCd) comment to make the result clear, not only for $p^2$, but for $p^n$, $n\ge2$. All polynomials in this discussion will be in $\Bbb Z[x]$.
The primitive $p^n$-th roots of unity are the $p^n$-th roots that aren’t $p^{n-1}$-th roots, and so they’re roots of $f(x)=(x^{p^n}-1)\big/(x^{p^{n-1}}-1)$. As Keith says, we need to show that $f(x    +1)$ is an Eisenstein polynomial.
To make things easier, let me point out that if $g=p+\cdots+x^m$ is a monic Eisenstein polynomial of degree $m$, and $h$ is any monic polynomial of degree $r$ without constant term and congruent modulo $p$ to $x^r$, then $g(h(x))$ is a monic Eisenstein polynomial of degree $mr$.
Next, note that
$$
g(x)=\frac{(x+1)^p-1}x=x^{p-1}+px^{p-2}+\cdots+p
$$
is a monic Eisenstein polynomial of degree $p-1$, no matter what the prime $p$. Now, substitute for $x$ in $g$ the polynomial $(x+1)^{p^{n-1}}-1$. It’s monic of degree $p^{n-1}$, has no constant term, and is congruent to $x^{p^{n-1}}$ mod $p$. Do the substitution and lo and behold! the result is my $f(x+1)$ of the second paragraph, Eisenstein of degree $p^n-p^{n-1}$ in accordance with my third paragraph.
