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# 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 $L|K$ 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.

## Mathematics Subject Classification

12F05*no label found*

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## Comments

## allignment

do you know why (1) is not alligned with its line?

it has the same effect with itemize or enumerate