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If $K_1 = F(a_1,..., a_n)$ and $K_2 = F(b_1,..., b_m)$.

Claim: Composite $K_1 K_2$ is equal to $F(a_1,... ,a_n,b_1,... ,b_m)$.

What we have to prove here is unclear to me.

$K_1 K_2 = F(a_1,..., a_n)(b_1,..., b_m) = F(a_1,..., a_n,b_1)(b_2,..., b_m) = ... \\ = F(a_1,..., a_n,b_1,..., b_{m-1})(b_m) =F(a_1,... ,a_n,b_1,... ,b_m). $

Hints to the problem are welcome. Thank You.

1 Answers 1

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The composite $K_1K_2$ is the smallest field which contains $K_1$ and $K_2$. First, prove $F(a_1,\ldots,a_n,b_1,\ldots,b_m)$ contains $K_1$ and $K_2$. Then, take an arbitrary field $F$ containing both $K_1$ and $K_2$ and prove that $ F(a_1,\ldots,a_n,b_1,\ldots,b_m) \subset F$.

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    proving both the facts is trivial...\ – 2017-01-23
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    Of course, but thats the proof. – 2017-01-23
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    ok...then I am done...the fact was so easy....I thought If I am missing something – 2017-01-23