Separability

Arun Ram
Department of Mathematics and Statistics
University of Melbourne
Parkville, VIC 3010 Australia
aram@unimelb.edu.au

Last update: 02 February 2012

Separability

Let $𝔽$ be a field.

An element $α \in \overline{𝔽}$ is separable over $𝔽$ if $m_{α, 𝔽}$ exists and all roots of $m_{α, 𝔽} (x)$ are multiplicity 1 (in $\overline{𝔽}$ ).
An extension $𝔼$ of $𝔽$ is separable if, for every $α \in 𝔼$ , $α$ is separable over $𝔽$ .

All roots of $m_{α, 𝔽} (x)$ have the same multiplicity.
If $char (𝔽) = 0$ of if $𝔽$ is finite then all elements of $\overline{𝔽}$ are separable.

(Theorem of the primitive element) If $𝔼$ over $𝔽$ is a finite separable extension of $𝔽$ then there exists an element $θ \in 𝔼$ such that $𝔼 = 𝔽 (θ)$ .

Let $𝔼$ be a finite extension of $𝔽$ . Then $𝔼 = 𝔽 (θ)$ for an element $θ \in 𝔼$ if and only if there are only a finite number of intermediate fields $𝔼 \supseteq 𝕂 \supseteq 𝔽$ .

All roots of $m_{α, 𝔽} (x)$ have the same multiplicity.
If $char (𝔽) = 0$ or if $𝔽$ is finite then all elements of $\overline{𝔽}$ are separable.

Proof.

Let $α$ and $β$ be roots of $m_{α, 𝔽} (x) .$ The isomorphism $\begin{array}{rccccc} σ : & 𝔽 (α) & ≅ & \frac{𝔽 [x]}{(m_{α, 𝔽} (x))} & ≅ & 𝔽 (β) \\ α & \mapsto & x & \mapsto & β \end{array}$ extends to an automorphism $\begin{array}{rrcl} σ : & \overline{𝔽} & \to & \overline{𝔽} \\ α & \mapsto & β \end{array}$ which is the identity on $𝔽$ . Thus $σ (m_{α, 𝔽} (x)) = m_{α, 𝔽} (x)$ and $σ ({(x - α)}^{k}) = {(x - β)}^{k} .$ So if $α$ is an root of $m_{α, 𝔽} (x)$ of multiplicity $k$ then ${(x - β)}^{k}$ is a factor of $m_{α, 𝔽} (x)$ in $\overline{𝔽} [x] .$ So $β$ has multiplicity $k$ .
Let $m_{α, 𝔽}$ be the minimal polynomial of $α \in \overline{𝔽}$ over $𝔽$ . Then $m_{α, 𝔽} (x) = (x - α) n (x)$ and $\frac{d}{d x} m_{α, 𝔽} (x) = (x - α) \frac{d n}{d x} + n (x) .$ So $m_{α, 𝔽} (x)$ has $α$ as a multiple root if and only if $m_{α, 𝔽}' (α) = 0 .$ Since $m_{α, 𝔽} (x)$ is the minimal polynomial which has $α$ as a root, $m_{α, 𝔽} (x)$ has a multiple root if and only if $m_{α, 𝔽}' (α) = 0 .$ If $m_{α, 𝔽} (x) = \sum_{i = 0}^{r} a_{i} x^{i} then m_{α, 𝔽}^{'} (x) = \sum_{i = 0}^{r} i a_{i} x^{i - 1} .$ So $m_{α, 𝔽}^{'} (x) = 0$ if and only if $p$ divides $i a_{i}$ for all $1 \leq i \leq r$ , where $p = char (𝔽) .$ Since $p$ divides $i a_{i}$ if and only if $p$ divides $a_{i}$ it follows that $m_{α, 𝔽} (x)$ has no multiple root when $char (𝔽) = 0$ . If $char (𝔽) = p$ then $m_{α, 𝔽} (x)$ has a multiple root if and only if $m_{α, 𝔽} (x) = g (x^{p})$ for some polynomial $g (x) \in 𝔽 [x] .$

$□$

(Theorem of the primitive element) If $𝔼$ over $𝔽$ is a finite separable extension of $𝔽$ then there exists an element $θ \in 𝔼$ such that $𝔼 = 𝔽 (θ) .$

		Proof.
	Case 2: $𝔽$ is infinite. Since $𝔼$ is a finite extension of $𝔽$ , $𝔼 = 𝔽 (γ_{1} γ_{2} ... γ_{k})$ for some $α_{1} ... α_{k}$ in $𝔼$ . The proof is by induction on $k$ . It is sufficient to show that if $𝔼 = 𝔽 (α, β)$ then there is an element $θ \in 𝔼$ such that $𝔼 = 𝔽 (θ)$ . Let $α = α_{1} α_{2} ... α_{n} ,$ and $β = β_{1} β_{2} ... β_{m}$ be the roots of $m_{α, 𝔽} (x)$ and $m_{β, 𝔽} (x),$ respectively. Since $𝔽$ is infinite there exists $a \in 𝔽$ such that $a \neq \frac{α_{i} - α}{β_{j} - β}, for any j such that β_{j} \neq β .$ Let $θ = α + a β$ . Let $p (x) = m_{α, 𝔽} (θ - a x) \in 𝔽 (θ) [x] .$ Then $p (β) = m_{α, 𝔽} (α + a β - a β) = 0, and p (β_{j}) = m_{α, 𝔽} (α + a β - a β_{j}) \neq 0, for all j \neq 1,$ since $α + a (β - β_{j}) \neq α_{i}$ for any $i$ . Now $m_{β, 𝔽 (θ)} (x)$ divides $m_{β, 𝔽} (x)$ and $m_{β, 𝔽 (θ)} (x)$ divides $p (x)$ . Since the only factor in common between $p (x)$ and $m_{β, 𝔽} (x)$ is $(x - β)$ , $m_{β, 𝔽 (θ)} (x) = (x - β) .$ So $β \in 𝔽 (θ)$ and $α = θ - a β \in 𝔽 (θ) .$ So $𝔼 = 𝔽 (α, β) = 𝔽 (θ) .$ $□$

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