Let $K$ be a totally real Galois number field, and suppose there is only one prime above $p$, with ramification index $\leq p-1$. If $K_p$ is the completion of $K$ at the prime above $p$, the claim is that the $p$-part of the discriminant of $K$ is equal to the discriminant of $K_p$.
I came across this while reading Washington's 'Introduction to Cyclotomic Fields', where he mentions in the proof of Proposition 5.33 that "the $p$-part of the discriminant of $K$ is equal to the discriminant of $K_p$', where the set-up is as outlined above. It's not clear to me how bases for $\mathcal O_K$ and $\mathcal O_{K_p}$ are related, so I'm a bit unsure how to make sense of this. I've tried to unpack this a bit by looking at the example where $K=\mathbb Q(\zeta_p)^+$ is the maximal real subfield of $\mathbb Q(\zeta_p)$, in which case $(p)=(1-\zeta_p)^{p-1}$ is totally ramified in $\mathbb Q(\zeta_p)$, so $K$ satisfies the hypotheses. But even in this example, I'm having a tough time actually computing the relevant discriminants, let alone understanding this in general…
Best Answer
Let $\mathfrak{p}$ be the prime above $p$, $G = \text{Gal}(K/\mathbb{Q}), G_\mathfrak{p} = \text{Gal}(K_\mathfrak{p}/\mathbb{Q}_p)$, note that $G_\mathfrak{p} \hookrightarrow G$.
The assumption of ramification index $e\leq p-1$ says the (global version) of first ramification group $G_1$ is trivial: $$G_1 = \{\sigma\in G| v_\mathfrak{p} (\sigma(x)-x)\geq1 \quad \forall x\in \mathcal{O}_K\} = \{1\}$$ this implies the local version is also trivial since $G_\mathfrak{p} \hookrightarrow G$.
Therefore $p$ is tamely ramified in both $K$ and $K_\mathfrak{p}$, the $p$-part of their discriminant are both $f(e-1)$, with $f$ the inertial degree. (Note that we used the assumption there is only one prime lying above $p$)