Equivalence of Definitions of Purely Inseparable Extension

== Theorem ==
Let $E/F$ be an algebraic field extension.
=== Definition 1 ===
=== Definition 2 ===
=== Definition 3 ===
== Proof ==
=== Definition $3$ implies Definition $2$ ===
Suppose each $\alpha \in E$ has a minimal polynomial of the form $X^{p^n} - a$ where $n \in \N$. Then
:$\alpha^{p^n} = a$
By definition of minimal polynomial, $a \in F$. So
:$\alpha^{p^n} \in F$
=== Definition $1$ implies Definition $3$ ===
Let $p$ be the characteristic of $F$.
Suppose each $\alpha \in E$ is inseparable.
Let $f$ be the minimal polynomial of $\alpha$ over $F$.
Then $\map f X$ is irreducible in $F \sqbrk X$.
I claim that $f' = 0$ because,
By definition of inseparable element, $\map f X$ of $\alpha$ is not separable. So in the closure $\bar F$ of $F$, the roots of $f$ are not distinct, 
there is $a \in \bar F$ such that
:$\paren{X - a}^2 \divides \map f X$
Writing $\map f X = \paren{X - a}^2 \map h X$ gives that
:$\map {f'} X = 2 \paren {X - a} \map h X + \paren {X - a}^2 \map {h'} X$
which shows that $\map {f'} a=0$.
Then, since $f$ is irreducible, it divides every polynomial in $F \sqbrk X$ that has $a$ as a root. Thus
:$f \divides f'$
but since $\deg f > \deg f'$,
:$f' = 0$.
Write
:$\map f X = \sum_{i \mathop = 0}^n a_i X^i$ where $a_n \neq 0$
Then $\map {f'} X = 0$ means
$0 = \sum_{i \mathop = 1}^n i a_i X^{i - 1}$
and thus $i a_i = 0$ for $0 \leq i \leq N$.
Since $a_N \neq 0$ and $N a_N = 0$, we conclude that $p > 0$ and that $a_i = 0$ for all $i$ not a multiple of $p$.
Discarding terms that must be zero, we can now rewrite
:$\map f X = \sum_{j \mathop = 0}^{N / p} a_{p j} X^{p j}$
So
:$\map f X = \map g {X^p}$
where
:$\map g X = \sum_{j \mathop = 0}^{N / p} a_{p j} X^j$
To see that $g$ must be irreducible, we observe that a proper factorization
:$\map g X = \map h X \map k X$
would yield the proper factorization
:$\map f X = \map h {X^p} \map k {X^p}$
but $f$ is irreducible, so $g$ must be irreducible.
If $\deg g > 1$, we can repeat the above process. Using induction on the degree,
:$\map f X = \map h {X^{p^n}}$
for some polynomial $\map h X \in F \sqbrk X$ with $\deg h = 1$.
Since $\map f \alpha = 0$,
:$\map h {\alpha^{p^n}} = 0$
but $\deg h = 1$, its root must be in $F$, let $a = \alpha^{p^n} \in F$.
Since $f$ is monic, we can assume $h$ is monic, then
:$\map h X = X - a$ 
then
:$\map f X = \map h {X^{p^n}} = X^{p^n} - a$
=== Definition $2$ implies Definition $1$ ===
Suppose $F$ have positive characteristic $p$ and for each $\alpha \in E$ there exists $n \in \N$ such that $\alpha^{p^n} \in F$.
The polynomial
:$\paren{X - \alpha}^{p^n} = X^{p^n} - \alpha^{p^n}$
is in $F \sqbrk X$, since $\alpha^{p^n} \in F$.
Let $\alpha \in E \setminus F$. Then $p^n > 1$ and $\deg f > 1$.
To show that $\alpha$ is inseparable, we let $f$ be the minimal polynomial of $\alpha$ over $F$
Since $\alpha$ is a root of $\paren{X - \alpha}^{p^n} \in F \sqbrk X$,
:$f \divides \paren{X - \alpha}^{p^n}$
but $p^n > 1$ and $\deg f > 1$, we get
:$\paren {X - \alpha}^2 \divides f$
so $f$ does not have distinct roots,
so $\alpha$ is inseparable, as required.
== Sources ==
* : Chapter $19$ Separability and Inseparability $\S19$A Corollary $19.6$ -- $\S19$B Theorem $19.10$
Category:Definitions/Field Extensions