alternative definition of algebraically closed
Proposition 1.
If is a field, the following are equivalent:
- (1)
is algebraically closed
, i.e. every nonconstant polynomial
in has a root in .
- (2)
Every nonconstant polynomial in splits completely over .
- (3)
If is an algebraic extension
then .
Proof.
If (1) is true then we can prove by induction on degree of that every nonconstant polynomial splits completely over . Conversely, (2) (1) is trivial.
(2) (3) If is algebraic and , then is a root of a polynomial . By (2) splits over , which implies that . It follows that .
(3) (1) Let and a root of (in some extension of ). Then is an algebraic extension of , hence .∎
Examples 1) The field of real numbers is not algebraically closed. Consider the equation . The square of a real number is always positive and cannot be so the equation has no roots.
2) The -adic field is not algebraically closed because the equation has no roots. Otherwise implies , which is false.