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Do all infinite fields of char p contain a subfield isomorphic to $F_{p}(x)$?

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2 Answers 2

up vote 11 down vote accepted

Perhaps this is an opportunity to walk through a mechanical process from the statement of the problem to its solution. Suppose $F$ is a field of characteristic $p$ which does not contain a subfield isomorphic to $\mathbb{F}_p(x)$. In other words, no element of $F$ is transcendental over $\mathbb{F}_p$. It follows that every element of $F$ must instead be algebraic over $\mathbb{F}_p$, hence that $F$ must be a subfield of $\overline{\mathbb{F}_p}$. As no finite field is algebraically closed, $\overline{\mathbb{F}_p}$ is an infinite field of characteristic $p$ which contains no transcendental elements, and we are done.

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Qiaochu, is there a typo in the last sentence? Don't you mean, "$F$ is an infinite field of characteristic $p$ which contains no transcendental elements..." ? –  Adrián Barquero Jul 14 '11 at 2:31
    
@Adrián: well, I just chose $F = \overline{\mathbb{F}_p}$ as it's fairly easy to prove that it's infinite. –  Qiaochu Yuan Jul 14 '11 at 2:32
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Oh, I see. ${}{}{}$ –  Adrián Barquero Jul 14 '11 at 2:37

HINT $\ $ Consider the algebraic closure of $\rm\:\mathbb F_p\:,\:$ recalling that algebraically closed fields are infinite, since for any finite field $\rm\:\mathbb F\:$ with elements $\rm\:a_1,\cdots,a_n,\:$ a la Euclid, we can construct the polynomial $\rm\:1 + (x-a_1)\cdots(x-a_n)\:$ coprime to all of the primes $\rm\: x-a_i\:,\:$ hence having no roots $\rm\:a_{\:i}\:$ in $\rm\:\mathbb F\:.$

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