Galois Traces and the Möbius Function (14.5.6,11)

Dummit and Foote Abstract Algebra, section 14.5, exercise 6:

(6) Show $$\text{Tr}_{\mathbb{Q}(\zeta_n)/\mathbb{Q}}(\zeta_n)=\mu(n)$$ (11) Show the primitive $n^{th}$ roots of unity form a basis of $\mathbb{Q}(\zeta_n)$ if and only if $n$ is squarefree.

Proof: (6) Let $n=p_1^{α_1}...p_k^{α_k}$. Since for any choice of primitives we have $\zeta_n=(\zeta_{p_1^{α_1}}...\zeta_{p_k^{α_k}})^z$ for some $z$, by distributing powers we may simply assume $\zeta_n=\zeta_{p_1^{α_1}}...\zeta_{p_k^{α_k}}$. Since $\mu(ab)=\mu(a)\mu(b)$ for $(a,b)=1$, it suffices to show $\text{Tr}(\zeta_a\zeta_b)=\text{Tr}(\zeta_a)\text{Tr}(\zeta_b)$ for $(a,b)=1$ and that the proposition holds for $n$ a prime power.

Note that for $1≤m≤n$ relatively prime to $n$, we have $(\zeta_{p_1^{α_1}}...\zeta_{p_k^{α_k}})^m=\zeta_{p_1^{α_1}}^{r_1}...\zeta_{p_k^{α_k}}^{r_k}$ for $0 ≤ r_i < p_i^{α_i}$ and $(r_i,p_i^{α_i})=1$, and as well for such a selection of $r_i$ we have $\zeta_{p_1^{α_1}}^{r_1}...\zeta_{p_k^{α_k}}^{r_k}=(\zeta_{p_1^{α_1}}...\zeta_{p_k^{α_k}})^m$ for some $(m,n)=1$ by the Chinese Remainder Theorem is the two-sided inverse of this operation. Since $$\text{Tr}(\zeta_n)=\sum_{(m,n)=1, 1≤m < n} \zeta_n^m$$ we therefore have $\text{Tr}(\zeta_{p_1^{α_1}}...\zeta_{p_k^{α_k}})=\text{Tr}(\zeta_{p_1^{α_1}})...\text{Tr}(\zeta_{p_k^{α_k}})$.

Now, we may observe $\text{Tr}(\zeta_p)=\zeta_p+\zeta_p^2+...+\zeta_p^{p-1}=-1$. As well, assume $q=p^a$ for $a≥2$. Then $$\text{Tr}(\zeta_q)=\zeta_q+\zeta_q^2+...+\zeta_q^{p-1}+\zeta_q^{p+1}+...+\zeta_q^{2p-1}+\zeta_q^{2p+1}+...+\zeta_q^{p^a-1}=$$ $$(\zeta_q+...+\zeta_q^{p-1})+\zeta_q^p(\zeta_q+...+\zeta_q^{p-1})+\zeta_q^{2p}(\zeta_q+...+\zeta_q^{p-1})+...+$$ $$\zeta_q^{p(p^{a-1}-1)}(\zeta_q+...+\zeta_q^{p-1})=$$ $$(1+\zeta_q^p+\zeta_q^{2p}+...+\zeta_q^{p(p^{a-1}-1)})(\zeta_q+\zeta_q^2+...+\zeta_q^{p-1})=f(\zeta_q^p)(\zeta_q+\zeta_q^2+...+\zeta_q^{p-1})$$ where $f(x)=\dfrac{x^{p^{a-1}}-1}{x-1}$ which is valid since $\zeta_q^p≠1$. Thus we note $f(\zeta_q^p)=0$ as $(\zeta_q^p)^{p^{a-1}}-1=\zeta_q^{p^a}-1=0$ and so $\text{Tr}(\zeta_q)=0$ and $\text{Tr}(\zeta_q)$ agrees with $\mu(q)$ on prime powers, and since both are relatively multiplicative the proposition is complete.

(11) ($⇒$) By the above, we see $\text{Tr}(\zeta_n)=0$ if $n$ is not squarefree. ($⇐$) Let $n=p_1p_2...p_k$. Then we see $\mathbb{Q}(\zeta_n)=\mathbb{Q}(\zeta_{p_1}\zeta_{p_2}...\zeta_{p_k})$ is the composite of $\mathbb{Q}(p_k)$ and $\mathbb{Q}(\zeta_{p_1...p_{k-1}})$. By induction on prime width, we see the latter field has for basis over $\mathbb{Q}$ the elements $\zeta_{p_1}^{α_1}...\zeta_{p_{k-1}}^{α_{k-1}}$ for $1≤α_i < p_i$. Since the fields have relatively prime degree, we see the basis $\zeta_{p_k}^{α_k}$ for $1≤α_k < p_k$ for $\mathbb{Q}(\zeta_{p_k})$ over $\mathbb{Q}$ remains linearly independent over $\mathbb{Q}(\zeta_{p_1...p_{k-1}})$ (Corollary 13.2.22). Hence a basis for $\mathbb{Q}(\zeta_n)$ over $\mathbb{Q}$ is $\zeta_{p_1}^{α_1}...\zeta_{p_k}^{α_k}$ for $1≤α_i < p_i$, which by the Chinese Remainder Theorem are all the primitive $n^{th}$ roots.$~\square$