If a and b are algebraic over F and F(a) = F(b), then a and b have the same degree over F

Let F be a field and E an extension of F. Suppose a,b \in E are algebraic over F and that F(a) = F(b). Prove that a and b have the same degree over F.


Note that b \in F(a). By Theorem 4.9 in TAN, \mathsf{deg}_F(b) \leq \mathsf{deg}_F(a). Similarly, \mathsf{deg}_F(a) \leq \mathsf{deg}_F(b). Hence \mathsf{deg}_F(a) = \mathsf{deg}_F(b).

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