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- $[LK^\prime : K^\prime] \geq e$:

$p$ is unramified in $K^\prime$ and $p$ is totally ramified in $L$ with ramification $e = [L : \mathbb{Q}]$ so $p$ ramified in $LK^\prime$ with ramification index $e$, by Theorem 2.1, $e \mid [LK^\prime:K^\prime]$, thus $[LK^\prime : K^\prime] \geq e$.

  • $e \geq [LK : K^\prime] \geq [LK^\prime : K^\prime]$:

$[LK : K^\prime] = |I^\prime_0|$ according to the setup. $\mathfrak{P}$ is tamely ramified since $p$ is tamely ramified in both $L$ and $K$. By Theorem 2.6, $I^\prime_0 \leq (k(p))^\times = (\mathbb{Z}_p)^\times$, so $I^\prime_0$ is a cyclic group. Similarly, $Gal(LK/\mathbb{Q})$ injects into $Gal(K/\mathbb{Q}) \times Gal(L/\mathbb{Q})$ and so $I^\prime_0$ does as well.

Let $\mathfrak{p}^\prime = \mathfrak{P} \cap K$. Then by definition, $I^\prime_0$ restricted to $K$ gives an element in the inertia group $I_0(\mathfrak{p}^\prime/p)$ but $\mathfrak{p}^\prime$ is conjugate to $\mathfrak{p}$ so the order of $I_0(\mathfrak{p}^\prime|p)$ is $|I_0| = e$.

Thus $I^\prime_0$ lives in the subgroup $I_0(\mathfrak{p}^\prime|p) \times Gal(L/\mathbb{Q})$, both of which are groups of order $e$ so $I^\prime_0$ has exponent $e$. Since it is cyclic and of exponent $e$, $|I^\prime_0| \leq e$.