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# proof of characterization of perfect fields

###### Proposition 1.

The following are equivalent:

- (a)
Every algebraic extension of $K$ is separable.

- (b)
Either $\operatorname{char}K=0$ or $\operatorname{char}K=p$ and the Frobenius map is surjective.

Proof. Suppose (a) and not (b). Then we must have $\operatorname{char}K=p>0$, and there must be $a\in K$ with no $p$-th root in $K$. Let $L$ be a splitting field over $K$ for the polynomial $X^{p}-a$, and let $\alpha\in L$ be a root of this polynomial. Then $(X-\alpha)^{p}=X^{p}-\alpha^{p}=X^{p}-a$, which has coefficients in $K$. This means that the minimum polynomial for $\alpha$ over $K$ must be a divisor of $(X-\alpha)^{p}$ and so must have repeated roots. This is not possible since $L$ is separable over $K$.

Conversely, suppose (b) and not (a). Let $\alpha$ be an element which is algebraic over $K$ but not separable. Then its minimum polynomial $f$ must have a repeated root, and by replacing $\alpha$ by this root if necessary, we may assume that $\alpha$ is a repeated root of $f$. Now, $f^{{\prime}}$ has coefficients in $K$ and also has $\alpha$ as a root. Since it is of lower degree than $f$, this is not possible unless $f^{{\prime}}=0$, whence $\operatorname{char}K=p>0$ and $f$ has the form:

$f=x^{{pn}}+a_{{n-1}}x^{{p(n-1)}}+\dots+a_{1}x^{p}+a_{0}.$ |

with $a_{o}\neq 0$. By (b), we may choose elements $b_{i}\in K$, for $0\leq i\leq n-1$ such that ${b_{i}}^{p}=a_{i}$. Then we may write $f$ as:

$f=(x^{n}+b_{{n-1}}x^{{n-1}}+\dots+b_{1}x+b_{0})^{p}.$ |

Since $f(\alpha)=0$ and since the Frobenius map $x\mapsto x^{p}$ is injective, we see that

$\alpha^{n}+b_{{n-1}}\alpha^{{n-1}}+\dots+b_{1}\alpha+b_{0}=0$ |

But then $\alpha$ is a root of the polynomial

$x^{n}+b_{{n-1}}x^{{n-1}}+\dots+b_{1}x+b_{0}$ |

which has coefficients in $K$, is non-zero (since $b_{o}\neq 0$), and has lower degree than $f$. This contradicts the choice of $f$ as the minimum polynomial of $\alpha$.

## Mathematics Subject Classification

12F10*no label found*

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