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Thread: Is any algebraically closed field infinite ?

  1. #1
    Aki
    Aki is offline
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    Is any algebraically closed field infinite ?

    Is the following statement true ?
    If so, please explain why.

    If K is an algebraically closed field, then K is an infinite set.
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  2. #2
    Senior Member roninpro's Avatar
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    We know that all finite fields are extensions of fields $\displaystyle \mathbb{Z}_p$. And in particular, any finite extension of them will not be algebraically closed. So the algebraic closure of $\displaystyle \mathbb{Z}_p$ is necessarily infinite.
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  3. #3
    MHF Contributor Swlabr's Avatar
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    Quote Originally Posted by roninpro View Post
    We know that all finite fields are extensions of fields $\displaystyle \mathbb{Z}_p$. And in particular, any finite extension of them will not be algebraically closed. So the algebraic closure of $\displaystyle \mathbb{Z}_p$ is necessarily infinite.
    Every finite field has order $\displaystyle p^n$ for some n. The underlying multiplicative group is always cyclic (of order $\displaystyle p^{n}-1$). Thus, $\displaystyle g^{p^n-1}=1$ for all $\displaystyle g \in \mathbb{F}_{n}\setminus \{0\}$.

    Therefore, if $\displaystyle p \neq 2$, the equation $\displaystyle x^{p^n-1}+2$ will not have a root in $\displaystyle \mathbb{F}_n$.

    If $\displaystyle p=2$ then this trick doesn't work...however, if $\displaystyle F=\{a_1, a_2, \ldots, a_{m}\}$ is your finite field, then the polynomial $\displaystyle (x-a_1)(x-a_2)\ldots (x-a_{m})+1$ has no roots in $\displaystyle F$ (this works for all p).
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