alternative definition of algebraically closed
Proposition 1.
If $K$ is a field, the following are equivalent^{}:

(1)
$K$ is algebraically closed^{}, i.e. every nonconstant polynomial^{} $f$ in $K[x]$ has a root in $K$.

(2)
Every nonconstant polynomial $f$ in $K[x]$ splits completely over $K$.

(3)
If $LK$ is an algebraic extension^{} then $L=K$.
Proof.
If (1) is true then we can prove by induction^{} on degree of $f$ that every nonconstant polynomial $f$ splits completely over $K$. Conversely, (2)$\Rightarrow $ (1) is trivial.
(2)$\Rightarrow $ (3) If $LK$ is algebraic and $\alpha \in L$, then $\alpha $ is a root of a polynomial $f\in K[x]$. By (2) $f$ splits over $K$, which implies that $\alpha \in K$. It follows that $L=K$.
(3)$\Rightarrow $ (1) Let $f\in K[x]$ and $\alpha $ a root of $f$ (in some extension^{} of $K$). Then $K(\alpha )$ is an algebraic extension of $K$, hence $\alpha \in K$.
∎
Examples 1) The field of real numbers $\mathbb{R}$ is not algebraically closed. Consider the equation ${x}^{2}+1=0$. The square of a real number is always positive and cannot be $1$ so the equation has no roots.
2) The $p$adic field ${\mathbb{Q}}_{p}$ is not algebraically closed because the equation ${x}^{2}p=0$ has no roots. Otherwise ${x}^{2}=p$ implies $2{v}_{p}x=1$, which is false.
Title  alternative definition of algebraically closed 

Canonical name  AlternativeDefinitionOfAlgebraicallyClosed 
Date of creation  20130322 16:53:23 
Last modified on  20130322 16:53:23 
Owner  polarbear (3475) 
Last modified by  polarbear (3475) 
Numerical id  8 
Author  polarbear (3475) 
Entry type  Derivation^{} 
Classification  msc 12F05 