I'm struggling to understand this. Let $\mathbb{Q}(\xi)$ be the splitting field for $x^n - 1$, so here $\xi$ is the primitive n-th root of unity. If I consider the possible automorphisms of this field, I will get:

$$\iota \textrm{ (the identity)}, \qquad \textrm{and} \qquad \sigma_k: \xi \mapsto \xi^k, \quad2 \leq k< n.$$

I will have $n-1$ elements, same as $Z_{n-1}$. How can it be shown that this is actually $Z_n^{\times}$ instead? What am I missing?

  • 3
    $\begingroup$ Why do you have $n-1$ elements? Can $\xi$ be mapped to any $\xi^k$ or just those $k$'s relatively prime to $n$? $\endgroup$ Commented May 10, 2016 at 15:29
  • 2
    $\begingroup$ Write down the composition explicitly. For an example of what can go wrong, take $n = 4$. $\endgroup$
    – anomaly
    Commented May 10, 2016 at 15:51
  • $\begingroup$ Every $\mathbb Q$-automorphism of $\mathbb Q(\zeta)$ is a permutation of the set $\{1,\zeta,\zeta^2,\zeta^3,\dots\}$ that fixes 1 and preserves the multiplicative structure of $\langle \zeta \rangle \cong \mathbb Z / n \mathbb Z$, as a multiplicative subgroup of $\mathbb C$. So $\mathrm{Gal}(\mathbb Q(\zeta) / \mathbb Q)$ is contained in $\mathrm{Aut}(\mathbb Z / n \mathbb Z) \cong (\mathbb Z / n \mathbb Z)^\times$. Order consideration will get you equality. $\endgroup$
    – amcerbu
    Commented May 10, 2016 at 16:13

1 Answer 1


Simply, not all the maps $\sigma_k : \xi \mapsto \xi^k$ are field automorphisms. For example, consider $n = 4$; in this case, we usually denote the primitive root by $i$. Then, $\sigma_2$ maps both $\pm i$ to $-1$, so $\sigma_k$ is not an automorphism of $\Bbb Q[i]$ fixing $\Bbb Q$.

On the other hand, any automorphism must permute the roots of unity, so the map $\xi \mapsto \xi^k$ must be invertible if it is to be an automorphism, but this only holds if $k, n$ are coprime, that is, if $k \in \Bbb Z_n^{\times}$. Note that (for $n > 1$) $\Bbb Z_n^{\times} \cong \Bbb Z_{n - 1}$ iff $k, n$ are coprime for all $1 \leq k < n$, that is, iff $n$ is prime.

Again for $n = 4$ we have $\Bbb Z_4^{\times} = \{1, 3\}$, so the only two automorphisms are characterized by $i \mapsto i$ (the identity) and $i \mapsto i^3 = -i$ (complex conjugation).

  • 3
    $\begingroup$ Another reason is that the degrees are wrong. The degree of the extension gotten by adjoining an $n$-th root of unity $\zeta$ is equal to the degree of the minimal polynomial for $\zeta$, and this degree is not $n-1$, but $\varphi(n)$, the Euler number of integers $\le n$ that are relatively prime to $n$. $\endgroup$
    – Lubin
    Commented Jul 6, 2021 at 1:32

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .