Past exam of the mathematics course of the University of Cambridge / 2003 / iii / Paper 23 / 5 / i / Solution

Fix a prime $\mathfrak{q}$ of $E$ over $\mathfrak{p}$, and a prime $\mathfrak{r}$ of $LE$ over $\mathfrak{q}$. Let $\mathfrak{P}$ be its restriction to $L$. Inside the completed compositum, putThe assumption means $U/F$ is an unramified extension. Its residue degree is some $f$; the completed extension $(LE{)}_{\mathfrak{r}}/J$ is the corresponding compositum $UJ/J$. It suffices to prove that this is unramified, regardless of the ramification of $J/F$.

Here is a proof of unramified extensions remain unramified under base change using the classification already established. Choose an integral generator $\alpha$ of $U/F$ whose monic minimal polynomial $g\in {\mathcal{O}}_F[X]$ has irreducible separable reduction of degree $f$. Write $k_F={\mathbb{F}}_q$ and $k_J={\mathbb{F}}_{q^r}$. Let $J_f/J$ be the degree-$f$ unramified extension. Its residue field ${\mathbb{F}}_{q^{rf}}$ contains ${\mathbb{F}}_{q^f}$, so $\bar{g}$ has a simple root there. Hensel lemma lifts this root to a root of $g$ in $J_f$. This gives an $F$-embedding $U\hookrightarrow J_f$.

An unramified extension of a local field is Galois, so all $F$-embeddings of $U$ into a fixed algebraic closure have the same image. The original $U$ therefore lies in $J_f$, and $UJ$ is an intermediate field of the unramified extension $J_f/J$. Its ramification index is one. Since $\mathfrak{q}$ and $\mathfrak{r}$ were arbitrary,This proof also explains why no linear-disjointness assumption on $L$ and $E$, and no assumption that $E/K$ is unramified, is needed.