Equivalence of Definitions of Purely Inseparable Extension

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Theorem

Let $E/F$ be an algebraic field extension.

The following definitions of the concept of Purely Inseparable Field Extension are equivalent:

Definition 1

The extension $E/F$ is purely inseparable if and only if every element $\alpha \in E \setminus F$ is inseparable.


Definition 2

Let $F$ have positive characteristic $p$.


The extension $E/F$ is purely inseparable if and only if for each $\alpha \in E$ there exists $n \in \N$ such that $\alpha^{p^n} \in F$.


Definition 3

Let $F$ have positive characteristic $p$.


The extension $E/F$ is purely inseparable if and only if each element of $E$ has a minimal polynomial of the form $X^{p^n} - a$.


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

$\alpha^{p^n} \in F$

$\Box$

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 $\deg f > \deg f'$,

$f' = 0$.


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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 $p$.


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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

$\map g X = \sum_{j \mathop = 0}^{N / p} a_{p j} X^j$


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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

$\map f X = \map h {X^{p^n}} = X^{p^n} - a$

$\Box$

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.

$\blacksquare$

Sources

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