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Let $K \subset L$ finite (therefore algebraic) field extension. So not neccessary Galois. How to see that for automorphisms following inequility holds:

$$|Aut(L/K)|\le|L/K|$$?

user267839
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    Unless the extension is Galois, then it doesn't hold. – Angina Seng Aug 04 '18 at 16:04
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    Inequality? Do you want to prove $|\text{Aut}(L/K)| \le [L:K]$? – Stefan4024 Aug 04 '18 at 16:15
  • Yes, you are right, sorry, I fixed it. – user267839 Aug 04 '18 at 18:00
  • This follows from the linear independence of characters lemma. Have you heard that phrase? Locally you can start here or here. Keith Conrad's blurb is most likely a better resource. – Jyrki Lahtonen Aug 04 '18 at 18:06
  • @Jyrki Lahtonen: Linear independence of characters lemma says (if I found the correct one) that for any field $L$ and any group $G$ (here $G= L^$), the set* of homomorphisms from $L^$ into the multiplicative group $L^$ is linearly independent over $K$. I guess here your idea would be to show that the $K$-vector space $Aut_K(L,L)$ is finite dimensional? Or did I misunderstood your hint? – user267839 Aug 04 '18 at 18:36
  • Linear independence says that $Aut(L/K)$ is linearly independent over $L$ as a a subset of the space of $K$-linear transformations from $L$ to $L$. If you meant the latter with $Aut_K(L,L)$ then, correct. As a vector space over $L$ $End_K(L,L)$ has dimension $[L:K]$ and the claim then follows from basic linear algebra. – Jyrki Lahtonen Aug 04 '18 at 18:41

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