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Let $K \subset L$ be two fields with ring of integers $\mathcal O_K$ and $\mathcal O_L$.

If a prime $p$ is totally ramified in $\mathcal O_K$, is it true that $p$ is also ramified in $\mathcal O_L$?

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If $p$ is (not necessarily totally) ramified in $K$ then there is a prime ideal $\mathfrak p$ of $\mathcal O_K$ s.t. $p \in \mathfrak p^2$. Let $\mathfrak P$ be any prime ideal of $\mathcal O_L$ lying over $\mathfrak p$. Then $p \in \mathfrak p^2 \subseteq \mathfrak P^2$, so $p$ is also ramified in $\mathcal O_L$.

In general, $(p) = p\mathcal O_K$ can be uniquely written as a product of prime ideals of $\mathcal O_K$: $$(p) = \mathfrak p_1^{e_1} \cdots \mathfrak p_r^{e_r}.$$ The number $e(\mathfrak p_i | p) := e_i$ is called the ramification index of $\mathfrak p_i$ over $p$. Similarly, each $\mathfrak p_i$ can be written as a unique product of prime ideals of $\mathcal O_L$. Substitute this into the product of $\mathfrak p_i$'s to find $$e(\mathfrak P | p) = e(\mathfrak P | \mathfrak p) \cdot e(\mathfrak p | p)$$ whenever $\mathfrak P \subset\mathcal O_L$ lies over $\mathfrak p$ and $\mathfrak p \subset \mathcal O_K$ lies over $p$. Now $p$ is ramified in $K$ iff there is a prime ideal $\mathfrak p $ of $K$ s.t. $e(\mathfrak p | p) > 1$. By the multiplicativity of the ramification index, this implies $e(\mathfrak P|p) > 1$ for a prime ideal $\mathfrak P$ of $L$ lying over $\mathfrak p$, so $p$ is also ramified in $L$.

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  • $\begingroup$ Thank you for your answer. $\endgroup$ – Lisa Mainhard Jul 6 '12 at 12:05
  • $\begingroup$ I have another question: If $L/K$ is a totally ramified extension of fields. Does it mean that every prime ideal $\mathfrak p \neq 0$ in $\mathcal O_K$ is totally ramified in $\mathcal O_L$? I'm not sure, because I've read that there are only finitely many prime ideals which ramifies. $\endgroup$ – Lisa Mainhard Jul 6 '12 at 14:12
  • $\begingroup$ In an extension $L/\mathbb Q$ every ramified prime must divide the discriminant of $L$, so there are only finitely many ramified primes. However, if $L$ and $K$ are local fields they have only one prime ideal and the extension $L/K$ is called totally ramified if the prime ideal $\mathfrak p$ of $K$ is totally ramified in $L$ (or equivalently, if their residue fields are equal). $\endgroup$ – marlu Jul 7 '12 at 12:23

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