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I have it in my lectures notes that the claim: Let $K/F$ be a finite and separable extension then $K$ is a simple extension of $F$ follows immediately from the theorom : Let $K/F$ be a finite extension, then it is simple iff $K/F$ have a finite number of subfiels.

My question is why ? I know that the extension is finite and separable hence $K=F(a_1,...a_n)$ where $a_i$ are all separable, but why there are only finite number of subfields between $F$ and $K$ ?

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Whether this is circular or disallowed depends on how you've developed the theory, but one way is to take the normal closure, which will be a finite Galois extension, and use the Galois correspondence. (A finite group has but finitely many subgroups.) –  Dylan Moreland Jun 8 '12 at 19:37

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up vote 3 down vote accepted

Hint: Consider the Galois closure of $K$.

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