Is $\# \text{Hom}(K, \mathbb{C}_p)=[K: \mathbb{Q}_p]$?

Question

Let $F$ be a finite extension of $\mathbb{Q}$, $\mathbb{C}$=complex numbers.

Let $K$ be a finite extension of $\mathbb{Q}_p$, $\mathbb{C}_p=\widehat{\overline{\mathbb{Q}_p}}$=$p$-adic complex numbers.

We know:

The number of field homomorphisms $F \to \mathbb{C}$(or $\bar{\mathbb{Q}_p}$) is equal to the degree of extension $[F:\mathbb{Q}]$.

Question:

Does the same hold for $p$-adic case ?

i.e., Is the number of field homomorphisms $K \to \mathbb{C}_p.$ equal to the degree of extension $[K: \mathbb{Q}_p]$ ?

Any comments please.

Answer 1

Yes. Recall that the separable degree $[L:K]_s$ of a finite algebraic field extension $L/K$ is defined as

$$ [L:K]_s=|\operatorname{Hom}_K(L,\overline K)| $$

for some fixed algebraic closure $\overline K$ of $K$. Moreover, we have $[L:K]=[L:K]_s$ iff $L/K$ is separable.

Since $\mathbb Q_p$ is of characteristic $0$ it is a perfect field implying that all field extensions are separable. This entails that $[K:\mathbb Q_p]=|\operatorname{Hom}_{\mathbb Q_p}(K,\mathbb C_p)|$ as desired.

Answer 2

More generally if $K$ is a field and $\Omega$ is an algebraically closed field extension of $K$, then for any separable finite extension $L$ of $K$, the number of $K$-algebra homomorphisms $L \to \Omega$ satisfies $|\mathrm{Hom}_K(L,\Omega)|=[L:K]$. Note that in the p-adic case, not any field homomorphism $L \to \Omega$ is necessarily $\Bbb Q_p$-linear, though. However, if we require the homomorphism to be continuous, then it is $\Bbb Q_p$-linear (the argument uses that $\Bbb Q$ is dense in $\Bbb Q_p$).

Answer 3

This is not true. There is a subfield $F\ne\Bbb Q$ of $\Bbb Q_p$ such that $F$ is finite and normal over $\Bbb Q$. Thus $F$ has a non-trivial automorphism. This extends to a field homomorphism $K\to\Bbb C_p$ wich is not the identity on $\Bbb Q_p$, hence $\#\operatorname{Hom}(K,\Bbb C_p)\geq\#\operatorname{Hom}_{\Bbb Q_p}(K,\Bbb C_p)+1=[K:\Bbb Q_p]+1$