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!


You work one simple extension at a time.

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.

Hence the inequality follows by grace of the two tower formulas for separability degree and degree.

  • $\begingroup$ what is $\#Hom_k(L,K)$? $\endgroup$ – Jorge Fernández Hidalgo Jun 15 '17 at 13:18
  • $\begingroup$ Number of $K$-embeddings. Which is the number of roots of the minimal polynomial $\endgroup$ – tomak Jun 15 '17 at 13:19
  • $\begingroup$ 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 $\sigma\vert_K = id_K$. $\endgroup$ – tomak Jun 15 '17 at 13:33
  • $\begingroup$ but wouldn't $Hom_K(L,K)$ consist of functions from $L$ to $K$? $\endgroup$ – Jorge Fernández Hidalgo Jun 15 '17 at 13:35
  • $\begingroup$ sorry I meant $Hom_K(L,K^a)$ from $L$ to the algebraic closure $\endgroup$ – tomak Jun 15 '17 at 13:44

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