finite field cannot be algebraically closed
Theorem.
Proof.
The proof proceeds by the method of contradiction. Assume that a field
is both finite and algebraically closed
. Consider the polynomial
as a function from to . There are two elements
which any field (in particular, ) must have — the additive identity
and the multiplicative identity
. The polynomial maps both of
these elements to . Since is finite and the function is not one-to-one, the function cannot map onto either, so
there must exist an element of such that for
all . In other words, the polynomial has no root
in , so 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 |