Algebrically independent elements [duplicate]

Question

Possible Duplicate:
Why does K->K(X) preserve the degree of field extensions?

Suppose $t_1,t_2,\ldots,t_n$ are algebrically independent over $K$ containing $F$. How to show that $[K(t_1,\ldots,t_n):F(t_1,\ldots,t_n)]=[K:F]$?

Answer 1

Using the answer in link provided by Zev, your question can be answered by simple induction over $n$. For $n=1$ we proceed along one of the answers shown over there. Assume we have shown the theorem for some $n$. Then we have $[K(t_1,\ldots,t_n):F(t_1,\ldots,t_n)]=[K:F]$, and by the theorem for $n=1$ we have also $[K(t_1,\ldots,t_n,t_{n+1}):F(t_1,\ldots,t_n,t_{n+1})]=[K(t_1,\ldots,t_n)(t_{n+1}):F(t_1,\ldots,t_n)(t_{n+1})]=[K(t_1,\ldots,t_n):F(t_1,\ldots,t_n)]=[K:F]$ which completes the proof by induction.