# alternative definition of algebraically closed

###### Proposition 1.

If $K$ is a field, the following are equivalent:

1. (1)

$K$ is algebraically closed, i.e. every nonconstant polynomial $f$ in $K[x]$ has a root in $K$.

2. (2)

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

3. (3)

If $L|K$ 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 $L|K$ 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 $2v_{p}x=1$, which is false.

Title alternative definition of algebraically closed AlternativeDefinitionOfAlgebraicallyClosed 2013-03-22 16:53:23 2013-03-22 16:53:23 polarbear (3475) polarbear (3475) 8 polarbear (3475) Derivation msc 12F05