Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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

share|cite|improve this question

marked as duplicate by Arturo Magidin, Gerry Myerson, Qiaochu Yuan Jun 2 '12 at 1:26

This question was marked as an exact duplicate of an existing question.

[K:F] is a finite extension – Biswarup Ray Apr 29 '12 at 16:03
Related:… – Zev Chonoles Apr 29 '12 at 16:13

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.

share|cite|improve this answer

Not the answer you're looking for? Browse other questions tagged or ask your own question.