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A set of notes I am reading claims the following:

For $L/K$ and $M/K$ extensions of $p$-adic fields, if $L/K$ is unramified then the natural map $\{K$-embeddings $L \hookrightarrow M\} \to \{k$-embeddings $k_L \hookrightarrow k_M\}$ is a bijection, where $k, k_L, k_M$ are the residue fields.

Could anyone explain why this is to me? It's completely glossed over and I don't find the claim entirely trivial. I can certainly see what the natural map is, but not necessarily why it's a bijection in an unramified extension $L/K$. If anything in my question is unclear please just ask and I'll clear it up. Many thanks.

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2 Answers

Here's something that might help you solve your problem, and which is important to know in any case:

If $K/\mathbf{Q}p$ is finite, let $k_K$ denote its residue field. If $k/k_K$ is a finite extension then there exists a finite non-ramified extension $K(k)/K$ such that $k_{K(k)}=k$. This is proven using Hensel's lemma, if I recall correctly. In particular, if $M/K$ is any finite extension, it will decompose into $M/K(k_M)/K$ with $K(k_M)/K$ non-ramified of degree equal to $[k_M:k_K]$ and with $M/K(k)$ totally ramified. Inside $M$, the subfield $K(k_M)$ is unique (it's the maximal non-ramified extension of $K$ in $M$).

Note that if $L/K$ is non-ramified (i.e. $L=K(k_L)$) and if $L\to M$ is an embedding, then the image of this map must land inside $K(k_M)$ (by the above properties I listed). So, by the 'Hensel lifting' argument, any $k_K$-embedding of $k_L$ into $k_M$ gives a $K$-embedding of $K(k_L)=L$ into $k(k_M)\subset M$. This gives surjectivity of your natural map.

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An element in the unramified extension $L$ is (by definition of unramified) a Laurent series in the uniformizer $\pi_K$ with coefficients taken to be Teichmuller lifts of elements of $k_L$.

So a $K$-algebra map $L \to M$ is determined by where the Teichmuller lifts of $k_L$ go. Teichmuller lifts are functorial, and so it is determined by where the elements of $k_L$ go under the natural map $k_L \to k_M.$ (In the terminology of vgty6h7uij's answer, $L= K(k_L)$ must map into $K(k_M)$, and the map $K(k_L) \to K(k_M)$ is determined by the map $k_L \to k_M$.) This gives the claim.

If you want to see a general framework for discussing this sort of thing (and the theory of the $K(k_M)$s of the other answer) you can read about Witt vectors, e.g. in Serre's Local fields.

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