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First of all, my definition of separable degree of $K/F$ is $[K:F]_s=|\operatorname{Hom}_{F-alg}(K,\bar F)|$ where bar denotes algebraic closure.

It is well-known that in finite field extensions $[K:F]_s|[K:F]$. However, it is also known that in the infinite case, this is not even close to being true even for Galois extensions: consider $K/\mathbb Q$ where $K=\mathbb Q(\sqrt2,\sqrt3,\sqrt5,\sqrt7,\ldots)$, then any homomorphism $\sigma$ from $K$ into $\overline{\mathbb Q}$ (or rather the automorphism group) is determined by the sign of $\sigma$ on the generators (i.e. whether $\sigma(\sqrt{p_n})=\pm\sqrt{p_n}$), so $[K:\mathbb Q]_s=2^{\aleph_0}$ while $[K:\mathbb Q]$ is clearly $\aleph_0$ ($K$ lies inside $\overline{\mathbb Q}$ and the latter has countable degree).

Is there a deeper reason for this unexpected behavior in infinite extensions? (I can't even understand why this specific example behaves so counterintuitively.) What can we conclude about separable degree and extension degree in general?

Tesla Daybreak
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  • Here $\Bbb{Q}=K^G$ (the subfield fixed by $G$) with $G$ a group of cardinality $[K:\Bbb{Q}]$ (each of its elements leave invariant $\sqrt{p}$ for $p$ large enough) and $Gal(K/\Bbb{Q})$ is the profinite completion of $G$, having a much larger cardinality. Though it is not obvious if this generalizes well to arbitrary infinite Galois extensions. – reuns Mar 12 '21 at 17:57
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    When $K/F$ is Galois of countable degree then $K=\bigcup_{n\ge 1} K_n$ with $K_n/F$ finite Galois, $K_n\subset K_{n+1}$ and $$Gal(K/F)=\varprojlim_{n\to \infty} Gal(K_n/F)$$ has cardinality $2^{[K:F]}$. It "should" work the same way for uncountable Galois extensions. – reuns Mar 12 '21 at 18:16
  • @reuns Thanks, I get the idea of inverse limit (or profinite completion as you said). However is there any example of uncountable Galois extension? I mean even the algebraic closure has countable degree. – Tesla Daybreak Mar 12 '21 at 18:51
  • $\overline{\Bbb{C}(x)}/\Bbb{C}(x)$ is uncountable, it contains $\Bbb{C}(x,{ \sqrt{x+a},a\in \Bbb{C}})$ and for $b\not \in S$, $\Bbb{C}(x,{ \sqrt{x+a},a\in S})$ is a field of functions of $x$ meromorphic at $-b$ it doesn't contain $\sqrt{x+b}$ – reuns Mar 12 '21 at 19:01
  • @reuns Oh, I see. I implicitly assumed the field is countable. – Tesla Daybreak Mar 12 '21 at 19:05

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