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?

  • $\begingroup$ $\varphi(p^2)=p(p-1)$ unless $p=2$, and $2$ if $p=2$ $\endgroup$
    – ahulpke
    Mar 11, 2017 at 0:57
  • 6
    $\begingroup$ See almost any textbook treatment of Galois theory, and the case of cyclotomic extensions will be handled. Your question is about a special case. When $\xi$ is a root of unit of prime power order $p^r$, its minimal polynomial over $\mathbf Q$ is $f(X) = (X^{p^r}-1)/(X^{p^{r-1}}-1)$, which is irreducible over $\mathbf Q$ because $f(X+1)$ is Eisenstein with respect to the prime $p$. $\endgroup$
    – KCd
    Mar 11, 2017 at 1:31

2 Answers 2


The degree of $\zeta_n$ over $\mathbb{Q}$ is $\phi(n)$ (Euler-phi function), for $n=p^2$ that is $p(p-1)$.

  • $\begingroup$ Could you explain why? $\endgroup$
    – PercyF2519
    Mar 11, 2017 at 0:59
  • $\begingroup$ The proof when $n$ is a prime power is easy enough, but when more than one prime divides $n$, the question becomes rather harder. $\endgroup$
    – Lubin
    Mar 11, 2017 at 3:21
  • $\begingroup$ The minimal polynomial of $\zeta_n$ over $\mathbb{Q}$ is called the $n$'th cyclotomic polynomial $\Phi_n(x)$. There are two ways to define this polynomial: (1) $\Phi_n(x) = \prod_{{\rm gcd}(k,n)=1} (x-\zeta_n^k)$, or (2) $\Phi_n(x)$ is $x^n-1$ divided by the product of the $\Phi_d(n)$ for $d|n$ and $d<n$. Either way, the degree is $\phi(n)$. The proof that $\Phi_n(x)$ is irreducible is found in textbooks right after the section on separable polynomials, as that is a key ingredient needed in the proof. $\endgroup$
    – Mark
    Mar 11, 2017 at 19:41

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.

  • $\begingroup$ sorry, but my original posting omitted a necessary condition in the third paragraph. Now corrected. $\endgroup$
    – Lubin
    Mar 11, 2017 at 17:37

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