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