Intersection of field extensions

abstract-algebra field-theory

Solution 1:

Hint: $[F(a):F]=[F(a):F(a)\cap F(b)][F(a)\cap F(b):F]$

$[F(b):F]=[F(a):F(a)\cap F(b)][F(a)\cap F(b):F]$

Thus $[F(a)\cap F(b):F]$ divides $[F(a):F]$ and $[F(b):F]$ so $[F(a)\cap F(b):F]=1$ since $[F(a):F]$ and $[F(b):F]$ are relatively prime.

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