finite field cannot be algebraically closed


The proof proceeds by the method of contradictionMathworldPlanetmathPlanetmath. Assume that a field F is both finite and algebraically closedMathworldPlanetmath. Consider the polynomialMathworldPlanetmathPlanetmathPlanetmath p(x)=x2-x as a function from F to F. There are two elements which any field (in particular, F) must have — the additive identity 0 and the multiplicative identityPlanetmathPlanetmath 1. The polynomial p maps both of these elements to 0. Since F is finite and the function p:FF is not one-to-one, the function cannot map onto F either, so there must exist an element a of F such that x2-xa for all xF. In other words, the polynomial x2-x-a has no root in F, so F could not be algebraically closed. ∎

Title finite field cannot be algebraically closed
Canonical name FiniteFieldCannotBeAlgebraicallyClosed
Date of creation 2013-03-22 16:29:09
Last modified on 2013-03-22 16:29:09
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 6
Author rspuzio (6075)
Entry type Theorem
Classification msc 12F05
Related topic AlgebraicClosureOfAFiniteField