Conductor of a full cyclotomic field

For every $N\ge 1$ the index $t(N)$ is reduced and $\mathbb{Q}({\zeta }_N)=\mathbb{Q}({\zeta }_{t(N)})$: this is immediate by definition when $N$ is odd or $4\mid N$, and if $N=2m$ with $m$ odd then $t(N)=m$ is odd and [F3] gives $\mathbb{Q}({\zeta }_N)=\mathbb{Q}({\zeta }_m)$.

If $\mathbb{Q}({\zeta }_r)\hookrightarrow E$ is a $\mathbb{Q}$-algebra embedding into a field $E$, then the image $\xi$ of ${\zeta }_r$ is a primitive $r$-th root of unity and the subfield $\mathbb{Q}(\xi )$ is a splitting field of $t^r-1$ inside $E$, hence a copy of $\mathbb{Q}({\zeta }_r)$; in particular an embedding $\mathbb{Q}({\zeta }_r)\hookrightarrow \mathbb{Q}({\zeta }_s)$ exhibits $\mathbb{Q}({\zeta }_r)$ as a subfield of $\mathbb{Q}({\zeta }_s)$.

By step 1.1, $\mathbb{Q}({\zeta }_n)=\mathbb{Q}({\zeta }_r)$ is a splitting field of $t^r-1$ over $\mathbb{Q}$ that contains ${\zeta }_r$, so the identity embedding shows that $r$ is admissible for $\mathbb{Q}({\zeta }_n)$; hence the conductor of $\mathbb{Q}({\zeta }_n)$ is at most $r$.

Let $g\ge 1$ be admissible for $\mathbb{Q}({\zeta }_n)$ and put $s:=t(g)$. By step 1.1, $s$ is reduced and $\mathbb{Q}({\zeta }_g)=\mathbb{Q}({\zeta }_s)$, so admissibility gives a $\mathbb{Q}$-algebra embedding $\mathbb{Q}({\zeta }_n)\hookrightarrow \mathbb{Q}({\zeta }_s)$, i.e., since $\mathbb{Q}({\zeta }_n)=\mathbb{Q}({\zeta }_r)$ by step 1.1, an embedding $\mathbb{Q}({\zeta }_r)\hookrightarrow \mathbb{Q}({\zeta }_s)$; step 1.2 then makes $\mathbb{Q}({\zeta }_r)$ a subfield of $\mathbb{Q}({\zeta }_s)$.

Let $p$ be a prime with $p\mid r$, and write $L=\mathbb{Q}({\zeta }_r)$ and $M=\mathbb{Q}({\zeta }_s)$, so step 2.2 gives $L\subseteq M$. By [F4], choose a prime $\mathfrak{Q}$ of ${\mathcal{O}}_M$ above $p$; define $\mathfrak{P}:=\mathfrak{Q}\cap {\mathcal{O}}_L$. The inclusion ${\mathcal{O}}_L\hookrightarrow {\mathcal{O}}_M$ makes $\mathfrak{P}$ the inverse image of the prime $\mathfrak{Q}$, hence $\mathfrak{P}$ is prime. Since $p\in \mathfrak{Q}$, one has $p\in \mathfrak{P}$, so $\mathfrak{P}\mathrm{\ne }(0)$; moreover $\mathfrak{P}\cap \mathbb{Z}=(\mathfrak{Q}\cap {\mathcal{O}}_L)\cap \mathbb{Z}=\mathfrak{Q}\cap \mathbb{Z}=(p)$, so [F8] says $\mathfrak{P}$ lies above $p$. By [F4], the ramification exponents of these primes are $e(\mathfrak{P}\mathrm{/}p)=\phi (p^{v_p(r)})$ and $e(\mathfrak{Q}\mathrm{/}p)=\phi (p^{v_p(s)})$ (with $v_p(s)=0$ allowed and $\phi (1)=1$). By [F7], both $L\mathrm{/}\mathbb{Q}$ and $M\mathrm{/}\mathbb{Q}$ are Galois, so [F5] applies to $M\mathrm{/}L\mathrm{/}\mathbb{Q}$ and makes $I(\mathfrak{P}\mathrm{/}p)$ a quotient of $I(\mathfrak{Q}\mathrm{/}p)$; therefore $\phi (p^{v_p(r)})$ divides $\phi (p^{v_p(s)})$.

For every prime $p$ the exponents of $p$ in the reduced indices $r$ and $s$ lie in $\{0,1,2,\dots \}$ when $p$ is odd and in $\{0\}\cup \{k\ge 2\}$ when $p=2$; by [F6] the function $a\mapsto \phi (p^a)$ is strictly increasing on these sets and satisfies $\phi (p^0)=1<\phi (p^2)=2$ for $p=2$ and $1<\phi (p)=p-1$ for odd $p$, so $\phi (p^{v_p(r)})\le \phi (p^{v_p(s)})$ implies $v_p(r)\le v_p(s)$. Step 3.1 therefore gives $v_p(r)\le v_p(s)$ for every prime $p\mid r$ — vacuously when $r=1$ — so $r\mid s$; and $s=t(g)$ equals $g$ or $g\mathrm{/}2$, so $s\le g$ and hence $r\le g$.

Step 2.1 shows that $r$ is admissible and step 4.1 shows that every admissible $g$ satisfies $g\ge r$; hence the least admissible index — the conductor of $\mathbb{Q}({\zeta }_n)$ — equals $r=t(n)$. Consequently the conductor is $n$ when $n$ is odd or $4\mid n$ and is $n\mathrm{/}2$ when $n\equiv 2(\mathrm{m}\mathrm{o}\mathrm{d}4)$; in particular $\mathbb{Q}({\zeta }_{2m})=\mathbb{Q}({\zeta }_m)$ for odd $m$ by step 1.1, and for $n=2$ the conductor is $t(2)=1$, with $\mathbb{Q}({\zeta }_2)=\mathbb{Q}$.