Lemma 9.14.9 (Multiplicativity). Suppose given a tower of algebraic field extensions $K/E/F$. Then

$[K : F]_ s = [K : E]_ s [E : F]_ s \quad \text{and}\quad [K : F]_ i = [K : E]_ i [E : F]_ i$

Proof. We first prove this in case $K$ is finite over $F$. Since we have multiplicativity for the usual degree (by Lemma 9.7.7) it suffices to prove one of the two formulas. By Lemma 9.14.8 we have $[K : F]_ s = |\mathop{\mathrm{Mor}}\nolimits _ F(K, \overline{F})|$. By the same lemma, given any $\sigma \in \mathop{\mathrm{Mor}}\nolimits _ F(E, \overline{F})$ the number of extensions of $\sigma$ to a map $\tau : K \to \overline{F}$ is $[K : E]_ s$. Namely, via $E \cong \sigma (E) \subset \overline{F}$ we can view $\overline{F}$ as an algebraic closure of $E$. Combined with the fact that there are $[E : F]_ s = |\mathop{\mathrm{Mor}}\nolimits _ F(E, \overline{F})|$ choices for $\sigma$ we obtain the result.

We omit the proof if the extensions are infinite. $\square$

Comment #6526 by prime235711 on

https://math.stackexchange.com/questions/837018/can-an-algebraic-extension-of-an-uncountable-field-be-of-uncountable-degree/837020#837020 As you see here, a field can have uncountable extension. But the equation $[K:F]_s=\sup \{[K':F]_s\}$ makes the separable degree of K over F at most countable

Comment #6581 by on

OK, thanks for pointing this out. I think the first version of this chapter was written in a way where the degree of an extension by definition is either an integer or $\infty$. But as you ponit out it can be any cardinal and then you can still state this lemma. But for now I have removed the proof in the infinite case as we don't use it anyway. Anybody should feel free to latex up a detailed proof in the infinite case. See changes in this commit.

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• 3 comment(s) on Section 9.14: Purely inseparable extensions

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