For an assignment, I am trying to determine the separability degree of some algebraic field extension $L/K$. The definition of the separability degree of polynomial is not difficult to find at all, namely it is the degree of the unique irreducible, separable polynomial we can associate with any polynomial. As of yet, I have been unable to find the definition of the separability degree of a field extension.

Could someone give this definition or point me in the right direction to a definition?

Based on the fact that if $L$ is the splitting field for $K$, then $|Aut(L/K)|\leq [L:K]$ with equality if and only if $L$ is separable over $K$, I am tempted to guess that the separability degree of $L/K$ has something to do with $Aut(L/K)$. Is this justified?

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    $\begingroup$ "Separable degree". You can define it as the number of distinct $K$-embeddings of $L$ into a separable (or algebraic) closure of $K$, or as the degree of the maximal separable subextension of $L \mid K$. $\endgroup$ – Zhen Lin May 9 '12 at 10:32
  • $\begingroup$ @ZhenLin Just to make sure, does $k$-embedding mean that $K$ is fixed? $\endgroup$ – Holdsworth88 May 9 '12 at 12:47
  • $\begingroup$ Yes. The elements of $K$ are sent to their corresponding elements in the image. $\endgroup$ – Zhen Lin May 9 '12 at 13:05

The definition can be found in this Wikipedia page, in the paragraph "Separable extensions within Algebraic extensions". I will synthesize it here.

Given an algebraic extension $L/K$ we consider the field:

$S=\{ \alpha\in L : \alpha \; \text{is separable over} \;K\}$

It is clearly an algebraic (separable) extension of $K$, and the separable degree of $L/K$ is simply $[S:K]$, the degree of the field extension $S/K$.

  • $\begingroup$ Based on your definition $S \subseteq K$, whereas it should be an extension of the base field. I think you mean to write $S=\{\alpha \in L : \alpha$ is separable over $K\}$ $\endgroup$ – Holdsworth88 May 9 '12 at 15:36
  • $\begingroup$ @Holdsworth88, For sure! I'm sorry for the typo, I'm going to edit it now! $\endgroup$ – Giovanni De Gaetano May 9 '12 at 15:44

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