Algebraic Inseparable Extension

 

Proposition. Suppose $\alpha$ is algebraic over $k$ and $f$ is its minimal polynomial over $k$. If $\text{char} k = 0$ then $f$ is separable. Otherwise, if $\text{char} k = p$ then there exists an integer $\mu$ such that every root of $f$ has multiplicity $p^{\mu}$. In other words
$$[k(\alpha): k] = p^{\mu}[k(\alpha): k]_s$$
and $\alpha^{p^{\mu}}$ is separable over $k$.

Proof. 

Comment: In particular, we see that if $\text{char} k = p$ then the minimal polynomial  $f$ of  $\alpha$ over $k$ must be a power of some polynomial $f = g^{p^{\mu}}$ where $g = \prod_{i=1}^m (x- \alpha_i).$
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We now assume $\text{char} k = p$.

Example. Consider $k(\alpha)/k$ where the minimal polynomial of $\alpha$ over $k$ is $(x- \alpha)^{p^{\mu}}.$ Then $[k(\alpha):k]_s = 1$ and $[k(\alpha): k]_i = p^{\mu}$ and we say that $k(\alpha)/k$ is purely inseparable.


Definition. An element $\alpha$ algebraic over $k$ is called purely inseparable if there exists $n$ such that $\alpha^{p^n}= c$ lies in $k$ (so the minimal polynomial of $\alpha$ over $k$ must divide $x^{p^n} - c$.)

Proposition/Definition. Let $E/k$ be an algebraic extension. TFAE:

1. $[E:k] = 1$;
2. Every element of $E$ is purely inseparable over $k$;
3. The irreducible polynomial over $k$ of every $\alpha \in E$ is of the form $X^{p^{\mu}}- c$ for some $\mu$ and some $c \in k$.
4. $E = k(\{\alpha_i\}_i)$ for some $\alpha_i$'s purely inseparable over $k$. 

If $E/k$ satisfies any of the above, we call it a purely inseparable extension.

Proposition. Purely inseparable extensions form a distinguished class of extensions.

Proposition. Let $E/k$ be an algebraic extension. Let $E_0$ be the compositum of all separable subextensions $F/k$. Then $E/E_0$ is purely inseparable and $E_0/k$ is separable.

Corollary. If an algebraic extension $E/k$ is both separable and purely inseparable then $E= k$.