# Proof of $\text{#Hom}_K (L,K) \leq [L:K]$

Let $L/K$ be finite extension, then how can I prove this? $$\text{#}\text{Hom}_K (L,K) \leq [L:K]$$

What I know is I can represent $L$ as $K(a_1, a_2,...,a_n)$ but I can't go any further. Thanks in advance!

On simple extension $Hom_K(L,K^a) \leqslant [L:K]$ as the left number counts the different roots of the minimal polynomial and the right number equals the degree of the minimal polynomial.
• what is $\#Hom_k(L,K)$? – Jorge Fernández Hidalgo Jun 15 '17 at 13:18
• Number of $K$-embeddings. Which is the number of roots of the minimal polynomial – tomak Jun 15 '17 at 13:19
• This is the definition I use: Let $L/K$ and $M/K$ be field extensions, a $K$-embedding of $L$ into $M$ is an (injecitive) field homomorphism $\sigma: L\rightarrow M$ such that $\sigma\vert_K = id_K$. – tomak Jun 15 '17 at 13:33
• but wouldn't $Hom_K(L,K)$ consist of functions from $L$ to $K$? – Jorge Fernández Hidalgo Jun 15 '17 at 13:35
• sorry I meant $Hom_K(L,K^a)$ from $L$ to the algebraic closure – tomak Jun 15 '17 at 13:44