Consider the polynomial $x^n - 1$, then a splitting field for $x^n- 1$ over $\mathbb{Q}$ is $Q(\omega)$ where $\omega$ is the nth primitive root of unity. If we look at the values we can send $\omega$ to that keeps the criteria that whatever we send it to is also an nth root of unity, then it seems that we can send $$\omega \rightarrow \omega, \omega^2, \omega^3, ..., \omega^{n-1}$$ Because $(\omega^k)^n = 1$ where $k < n$. To me, it seems that this is perfectly valid and these are the only automorphisms for $\mathbb{Q}(\omega)$ that fix $\mathbb{Q}$, which implies that $|\text{Gal}(\mathbb{Q}(\omega)/\mathbb{Q})| = n - 1$, but I'm almost sure I'm wrong, because I know that $$\text{Gal}(\mathbb{Q}(\omega)/\mathbb{Q}) \approx U(n)$$ And it is not necessary for $|U(n)| = n - 1$, so where am I going wrong?
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1$\begingroup$ When $n=2$, $\omega=-1$ and $1$ is also a second root of unity; is there an automorphism mapping $-1$ to $1$? $\endgroup$– Angina SengDec 9, 2017 at 7:53
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1$\begingroup$ You've only listed $n-1$ maps though? $\omega$ cannot go to $1$, since the latter is rational. A very useful case to study is $n=4$, where $\omega=i$. $\endgroup$– Steve DDec 9, 2017 at 7:54
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$\begingroup$ My apologies, I made a counting mistake, I have edited the question $\endgroup$– q.ThenDec 9, 2017 at 7:56
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1$\begingroup$ When $n=4$, $\omega=i$, and $-1$ is also a fourth root of unity; is there an automorphism mapping $i$ to $-1$? $\endgroup$– Angina SengDec 9, 2017 at 8:04
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$\begingroup$ @LordSharktheUnknown, I've thought about it for a while, and don't understand why can't there be an automorphism mapping $i$ to $-1$? $\endgroup$– q.ThenDec 9, 2017 at 8:10
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You list the possible nontrivial group homomorphisms of $μ_n = \{ζ ∈ ℚ(ω); ζ^n = 1\}$. Of course, every field automorphism on $ℚ(ω)$ restricts to such a homomorphism, but that restriction is then an automorphism as well. So you cannot send $ω$ to each of $ω, …, ω^{n-1}$, but only to those of the same order.
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1$\begingroup$ it sounds like $\phi$ needs to send $\omega$ to $\omega^k$, and $(k, n) = 1$? $\endgroup$– q.ThenDec 9, 2017 at 8:13
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1$\begingroup$ @q.Then Yes, because $μ_n \cong ℤ/nℤ$ via $ω^k \leftrightarrow k$, and so the elements of order $n$ in $μ_n$ correspond to those $k=1, …, n$ with $(k,n) = 1$. $\endgroup$– k.stmDec 9, 2017 at 8:41