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Math Help - Trancendence degree infinite

  1. #1
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    Trancendence degree infinite

    Let K be algebrically closed field, over F(base field).Trancendence Degree of K over F is infinite.Please give an example of
    a F-homomorphism from K->K which is not surjective.
    (I proved the theorem that if Trancendence Degree of K over F is finite,K bein algebrically closed, then any F-homomorphism from K->K is surjective.)
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  2. #2
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    Re: Trancendence degree infinite

    let L=F(S), where S=\{x_1,x_2, \ldots \} is an infinite set of algebraically independent elements over F. let K be the algebraic closure of L and define the F-homomorphism \varphi: L \longrightarrow K by \varphi(x_n)=x_{n+1}, \ n \geq 1. since K is algebraically closed, a Zorn's lemma argument shows that \varphi can be extended to an F-homomorphism \psi: K \longrightarrow K. now \psi is not surjective because, for example, if \psi(u)=x_1 for some u \in K, then, since u is algebraic over L, x_1 would be algebraic over \psi(L)=\varphi(L) = F(S \setminus \{x_1\}). but that is impossible because then x_1 and S \setminus \{x_1\} would be algebraically dependent.
    Last edited by NonCommAlg; April 17th 2012 at 06:20 PM.
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