Equivalence of Definitions of Separable Degree

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Theorem

Let $E / F$ be a field extension.


The following definitions of the concept of Separable Degree are equivalent:

Definition 1

Let $S \subseteq E$ be the separable closure of $F$ in $E$.


The separable degree $\index E F_{\operatorname {sep} }$ of $E / F$ is the degree $\index S F$.


Definition 2

Let $\bar F$ be the algebraic closure of $F$.


The separable degree $\index E F_{\operatorname {sep} }$ of $E / F$ is the number of embeddings of $E$ into $\bar F$ that fix $F$.


Definition 3

Let $K$ be a normal extension of $F$ that contains $E$.


The separable degree $\index E F_{\operatorname {sep} }$ of $E / F$ is the number of embeddings of $E$ into $K$ that fix $F$.


Proof

Definition $1$ implies Definition $2$

Let $S \subseteq E$ be the separable closure of $F$ in $E$.

Let $\bar F$ be the algebraic closure of $F$.

We need to show that the number of embeddings $E \hookrightarrow \bar F$ fixing $F$ is

$n = \index E F_\text{sep} = \index S F$

Lemma

An embedding $\sigma: S \hookrightarrow \bar F$ fixing $F$ uniquely extends to an embedding $\tilde{\sigma}: E \hookrightarrow \bar F$ fixing $F$.

Proof

If characteristic of $F$ is $0$, we have $E = S$, nothing to prove.

If characteristic of $F$ is $p > 0$,

since $E / S$ is purely inseparable,

each element of $\alpha \in E \setminus S$ has $\alpha^{p^n} = a$ for some $a \in S$.

Then for all embedding $\tilde{\sigma}: E \hookrightarrow \bar F$ that extends $\sigma$,

$\map {\tilde{\sigma}} \alpha^{p^n} = \map {\tilde{\sigma}} {\alpha^{p^n}} = \map \sigma a$.

But $x^{p^n} = \map \sigma a$ has only one root in $\bar{F}$,

so $\map {\tilde {\sigma}} \alpha$ is uniquely determined by $\map \sigma a$.

Therefore $\sigma$ uniquely extends to $\tilde{\sigma}$.

$\Box$


By the lemma, it suffices to show that the number of embeddings $\sigma: S \hookrightarrow \bar F$ fixing $F$ is

$n = \index E F_\text{sep} = \index S F$

Since $S / F$ is separable, by Primitive Element Theorem, $S / F$ is generated by a single element $\alpha \in S$

$S = F \paren \alpha$

Let $P$ be the minimal polynomial of $\alpha$ over $F$.

Since $P$ is separable, $P$ has $n = \index S F$ distinct roots in $\bar F$. Let $\alpha_1, \ldots, \alpha_n$ be these roots.

By Simple Algebraic Field Extension consists of Polynomials in Algebraic Number, elements of $F$ are of the form

$\ds \map f \alpha$ for some polynomial $f \in F \sqbrk x$.

An embedding $\sigma: E \hookrightarrow \bar F$ is uniquely determined by the image of $\alpha$, since

$\forall f \in F \sqbrk x: \ds \map \sigma {\map f \alpha} = \map f {\map \sigma \alpha}$

Then we get $n$ distinct embeddings $\sigma_i: S \hookrightarrow \bar F$, $i = 1, \dots, n$ given by

$\map \sigma \alpha = \alpha_i$

And there can't be other embeddings, since the image of $\alpha$ has to be one of the roots $\alpha_i$ of $P$.

$\Box$


Definition $1$ implies Definition $3$


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