Separability of a finite extension

Let $𝐹 : 𝐾$ be a finite field extension. Then the separable degree satisfies¹

[𝐹 : 𝐾] s \leq [𝐹 : 𝐾]

and the following are equivalent: #m/thm/field

$𝐹 = 𝐾 (𝛼 𝑖) 𝑟 𝑖 = 1$ for separable elements ${𝛼 𝑖} 𝑟 𝑖 = 1 \subset 𝐹$ ;
$𝐾 : 𝐹$ is a separable extension;
$[𝐹 : 𝐾] s = [𝐹 : 𝐾]$ .²

Proof

$𝐹 : 𝐾$ is automatically a Finitely generated field extension, so $𝐹 = 𝐾 (𝛼 𝑖) 𝑟 𝑖 = 1$ for some ${𝛼 𝑖} 𝑟 𝑖 = 1 \subset 𝐹$ . Applying ^P1 and iterating on ^P1 and ^P2, we have

[𝐹 : 𝐾] s = 𝑟 - 1 \prod 𝑖 = 0 [𝐾 (𝛼 𝑗) 𝑖 𝑗 = 1 (𝛼) : 𝐾 (𝛼 𝑗) 𝑖 𝑗 = 1] s = 𝑟 - 1 \prod 𝑖 = 0 [𝐾 (𝛼 𝑗) 𝑖 𝑗 = 1 (𝛼) : 𝐾 (𝛼 𝑗) 𝑖 𝑗 = 1] \leq [𝐹 : 𝐾]

proving the inequality.

To show ^S1 implies ^S3: If ${𝛼 𝑖} 𝑟 𝑖 = 1$ are separable over $𝐾$ , then they are separable over all algebraic extensions, so equality follows by ^P1.

To show ^S3 implies ^S2: Suppose $[𝐹 : 𝐾] s = [𝐹 : 𝐾]$ , and for some $𝛼 \in 𝐹$ consider the intermediate extension $𝐹 : 𝐾 (𝛼) : 𝐾$ . By ^P2, we have

[𝐹 : 𝐾 (𝛼)] s [𝐾 (𝛼) : 𝐾] s = [𝐹 : 𝐾] s = [𝐹 : 𝐾] = [𝐹 : 𝐾 (𝛼)] [𝐾 (𝛼) : 𝐾],

which can only hold (by the inequality of ^P1) if in particular

[𝐾 (𝛼) : 𝐾] s = [𝐾 (𝛼) : 𝐾] .

Therefore, again by ^P1, every $𝛼 \in 𝐹$ is separable, whence by definition $𝐹 : 𝐾$ is separable.

That ^S2 implies ^S1 is clear, since a finite extension is finitely generated.

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

Actually, one can show that $[𝐹 : 𝐾] s = [𝐹 s e p : 𝐾]$ where $𝐹 : 𝐹 s e p : 𝐾$ . ↩
2009. Algebra: Chapter 0, §VII.4.3, pp. 437–438 ↩

Separability of a finite extension

Footnotes