How can $p^f-p^{f-1}$ divide 2?

Question

How can $p^f-p^{f-1}$ divide 2 ?

If $p^f$ divides $\left\lvert\mu\left(\mathbb Q(\sqrt d)\right)\right\rvert$, the number of roots of unity in $Q(\sqrt d)$ with $d$ a squarefree integer, then how can we conclude that $(p^f-p^{f-1})\mid2$

Answer 1

We know that $[\mathbb{Q}(\zeta_n):\mathbb{Q}] = \varphi(n)$ so in order for $\zeta_n \in \mathbb{Q}(\sqrt{d})$, we need $\varphi(n) = 1$ or $2$ since we have a tower of fields $\mathbb{Q}(\sqrt{d}) \supseteq \mathbb{Q}(\zeta_n) \supseteq \mathbb{Q}$.

If $p$ divides $n$ then $p-1$ divides $\varphi(n)$ so only $p = 2$ and $3$ are possible.

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First case: $n$ divisible by $3$.

If $n = 3^ak$ with $a \geq 1$ and $\gcd(k, 3) = 1$ then $\varphi(n) = 3^{a-1}\cdot2\varphi(k)$ so $a = 1$ and $k = 1$ or $2$, giving $n = 3$ or $6$.

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Second case: $n$ divisible by $2$ and not by $3$ (so $n$ is a power of $2$).

If $n = 2^a$ with $a \geq 1$ then $\varphi(n) = 2^{a-1}$ so $a = 1$ or $2$, giving $n = 2$ or $4$.

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Third case: $n$ is not divisible by $2$ nor by $3$. Then $n = 1$.

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Finally we see that $n = 1, 2, 3, 4$ or $6$.

If $n = 4$ then $\sqrt{-1} \in \mathbb{Q}(\sqrt{d})$ so $d = -1$ and $\mu(\mathbb{Q}(\sqrt{d})) = \{1, -1, i, -i\}$.

Otherwise, all roots of unity are powers of $\zeta_6$ so $\mu(\mathbb{Q}(\sqrt{d}))$ is a subgroup of the group of $6$-th roots of unity, which means its order divides $6$.

By the way $\zeta_6 = -\zeta_3^{-1}$ so $\zeta_3 \in \mathbb{Q}(\sqrt{d})$ if and only if $\zeta_6 \in \mathbb{Q}(\sqrt{d})$ and this occurs for $d = -3$. So the subgroup cannot be of order $3$. Also $-1$ is always in $\mathbb{Q}(\sqrt{d})$ so the order cannot be $1$.

So we finally get that $\left|\mu(\mathbb{Q}(\sqrt{d}))\right| = 2, 4$ or $6$.

One can now check that if $p^f$ divides $\left|\mu(\mathbb{Q}(\sqrt{d}))\right|$ then $\varphi(p^f)$ divides $2$ because the only possibilities are $p = 3$ and $f = 1$ or $p = 2$ and $f = 1, 2$.