About (wild) ramification of an abelian extension and its conductor

Question

Let $K/\mathbb{Q}$ be an abelian extension, and define the conductor of $K$ as the smallest integer $n$ such that $K\subset \mathbb{Q}(\zeta_n)$. How does one show the following two facts:

1. The extension is ramified at $p$ if and only if $p\mid n$.

2. The extension is wildly ramified if and only if $p^2\mid n$.

All I could see was that primes that ramify in $K$ also ramify in $\mathbb{Q}(\zeta_n)$ so if $p\mid \Delta_K$ then it must divide $\Delta_{\mathbb{Q}(\zeta_n)}$ whence it must divide $n$. I am not sure how to prove the converse of 1, and 2.

I am aware that one can find proofs of the general cases in Neukirch, but that is above my level (talking about ray class fields etc).

Answer 1

Let’s write $C_n=\mathbb{Q}(\mu_n)$ to ease notations.

Lemma: let $m \geq 1$ be an integer and $n$ be a multiple of $m$. The ramification index of $\mathbb{Q}(\mu_n)/\mathbb{Q}(\mu_m)$ at any prime ideal $\mathfrak{p}$ of $\mathbb{Q}(\mu_n)$ only depends on the residue characteristic $p$ of $\mathfrak{p}$. This ramification index is $p^{v_p(n)-v_p(m)}$ if $p \mid m$, $(p-1)p^{v_p(n)-1}$ if $p \nmid m$ and $p \mid n$, and $1$ otherwise.

Proof: Very standard. We first consider the case $n=mq^r$ with $q \neq p$ coprime to $m$. The extension is then given by taking a root of the integral polynomial $X^{q^r}-1$, which is separable mod $p$, hence $C_n/C_m$ is unramified above $p$. We then treat the case where $m=1,n=p^r$, which follows from Eisenstein’s criterion. Ramification indices are multiplicative, so we get the case where $m=1$ and $n$ is general, then the full general case.

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For 1, if $p \nmid n$, then $C_n/C_1$ isn’t ramified at $p$, so $K/\mathbb{Q}$ isn’t ramified at $p$ either.

Conversely, if $K/\mathbb{Q}$ is unramified at $p$ and contained in some $C_n$, write $n=mp^k$ with $m$ coprime to $p$ and $k \geq 0$ – we want to show $K \subset C_m$. If $k=0$, we’re done, so assume $k \geq 1$.

By the lemma, the inertia group at $p$ of $Gal(C_n/C_1)$ is $Gal(C_n/C_m)$, and we know that it projects trivially to $Gal(K/\mathbb{Q})$. So $Gal(C_n/C_m) \subset Gal(C_n/K)$ hence $K \subset C_m$.

For 2, the approach is very similar.

Assume that $K/\mathbb{Q}$ is wildly ramified at $p$. Then $C_n/C_1$ is wildly ramified at $p$ too: by the lemma, this forces $p^2\mid n$.

Now, assume that $K \subset C_n$ is only tamely ramified over $\mathbb{Q}$ at $p$. We want to show that $K \subset C_d$ with $v_p(d)\leq 1$. Clearly we can assume that $K/\mathbb{Q}$ is ramified at $p$ and $n=mp^k$ with $k \geq 2$ and $m$ coprime to $p$.

By the lemma, the wild inertia subgroup of $Gal(C_n/C_1)$ is $Gal(C_n/C_{mp})$, and it projects trivially to $Gal(K/\mathbb{Q})$. So, as above, $K \subset C_{mp}$, so we’re done.