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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 $[K(t_1,\ldots,t_n):F(t_1,\ldots,t_n)]=[K:F]$?

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[K:F] is a finite extension –  Jishnu Ray Apr 29 '12 at 16:03
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Related: math.stackexchange.com/questions/4581/… –  Zev Chonoles Apr 29 '12 at 16:13
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marked as duplicate by Arturo Magidin, Gerry Myerson, Qiaochu Yuan Jun 2 '12 at 1:26

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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.

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