Separable degree of an extension

Let $𝐹 : 𝐾$ be an algebraic extension. The separable degree of $𝐹 : 𝐾$ is given by the number of embeddings of this extension into the algebraic closure $―― 𝐾$ , #m/def/field i.e.

[𝐹 : 𝐾] s : = ∣ 𝖥 𝗅 𝖽 𝐾 (𝐹, ―― 𝐾) ∣

and is nonzero assuming Zorn's lemma, see Embedding an algebraic extension into an algebraically closed field.

Properties

Let $𝐾 (𝛼) : 𝐾$ be a simple algebraic extension. Then $[𝐾 (𝛼) : 𝐾] s$ equals the number of distinct roots in $―― 𝐾$ of the minimal polynomial $𝑚 𝛼 (𝑥) \in 𝐾 [𝑥]$ , and thus $[𝐾 (𝛼) : 𝐾] s \leq [𝐾 (𝛼) : 𝐾]$ , with equality iff $𝛼$ is separable.
If $𝐹 : 𝐿 : 𝐾$ is a tower of algebraic extensions then $[𝐹 : 𝐾] s = [𝐹 : 𝐿] s [𝐿 : 𝐾] s$ .

Proof of 1–2.

The proof of ^P1 is very similar to that of the Bound on the automorphism group of a finite simple extension. Associate with each $𝜄 \in 𝖥 𝗅 𝖽 𝐾 (𝐾 (𝛼), ―― 𝐾)$ the image $𝜄 (𝛼)$ , which must be a root of $𝑚 𝛼 (𝑥)$ . Since $𝜄 (𝛼)$ completely determines $𝜄$ , this correspondence is injective. For surjectivity, let $𝛽 \in ―― 𝐾$ be a root of $𝑚 𝛼 (𝑥)$ . Then by ^P1, there exists an isomorphism $𝐾 (𝛼) \to 𝐾 (𝛽)$ sending $𝛼 \mapsto 𝛽$ , and composing this with the embedding $𝐾 (𝛽) ↪ ―― 𝐾$ gives the corresponding $𝜄$ , proving ^P1.

For ^P2, suppose $𝐹 : 𝐿 : 𝐾$ is such a tower of extensions, and identify $―― 𝐿 = ―― 𝐾$ . It is not difficult to see that we have a bijection

𝜓 : 𝖥 𝗅 𝖽 𝐿 (𝐹, ―― 𝐾) \times 𝖥 𝗅 𝖽 𝐾 (𝐿, ―― 𝐾) \to 𝖥 𝗅 𝖽 𝐾 (𝐹, ―― 𝐾) .

by first extending $𝐾 ↪ ―― 𝐾$ to $𝐿 ↪ ―― 𝐾$ and then extending $𝐿 ↪ ―― 𝐾$ to $𝐹 ↪ ―― 𝐾$ .

Results

Separability of a finite extension

#state/tidy | #lang/en | #SemBr