Can a field of characteristic $p$ be algebraically closed?

I know finite fields cannot be algebraically closed, but there are also infinite fields of characteristic $p$, so can they be algebraically closed? If not, what is a good\easy counterexample.

  • $\begingroup$ algebraic closure K of $F_p$ $\endgroup$ – Bhaskar Vashishth Jun 6 '15 at 9:49
  • $\begingroup$ What about equation $px+1=0$? $\endgroup$ – Michael Galuza Jun 6 '15 at 9:50
  • $\begingroup$ @MichaelGaluza What about it? That equation is simply $\;0=1\;$ over a field of characteristic $\;p\;$ , and thus false in any case. $\endgroup$ – Timbuc Jun 6 '15 at 9:59
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    $\begingroup$ @Timbuc, and therefore have no roots $\endgroup$ – Michael Galuza Jun 6 '15 at 10:00
  • $\begingroup$ @MichaelGaluza The very same thing can be said about the equation $\;0=1\;$ over any field...but there is no polynomial of degree greater than zero here! Are you implying there are no closed fields of positive characteristic? Because that'd be false: there are. $\endgroup$ – Timbuc Jun 6 '15 at 10:02

Every field has an algebraically closed extension called the algebraic closure. Now, the algebraic closure of a field of characteristic $p$ is again of characteristic $p$.


Any field has an algebraic closure, whatever its characteristic. Also, a finite field cannot be algebraically closed.


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