# Proof that a finite separable extension has only finite many intermediate fields

Let $E/F$ be finite separable extension. Is there any proof of the fact that there are only finitely many intermediate fields without using primitive element theorem or fundamental theorem of Galois theory?

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What's the motivation for not wanting to use either of those results? Considering the primitive element theorem is often stated as a finite extension has a primitive element if and only if there are a finite number of intermediate fields. –  JSchlather Feb 7 '13 at 14:48
If one cannot use galois-theory, then why tag this galois-theory? –  awllower Feb 7 '13 at 15:00
Because I thought it is related to galois theory or can be used in galois theory. –  Mohan Feb 7 '13 at 15:02
Indeed. Thanks for the clarification. –  awllower Feb 7 '13 at 15:11
@Mohan: Lang in his Algebra shows directly that a finite separable extension has a primitive element, and thus your result follows using the (stronger form of) The Primitive Element Theorem but no Galois Theory. (This argument is replicated in $\S 8$ of my field theory notes: math.uga.edu/~pete/FieldTheory.pdf.) I think this is not what you are looking for. Could you amplify on why you want to avoid the Primitive Element Theorem and Galois Theory? –  Pete L. Clark Mar 3 '13 at 17:57