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I am trying to understand what it means to have an extension that is an algebraic closure of the base field. I'm looking for someone who can help conceptually.

I understand how $C/R$ looks. The basis of the extension is $\{1,i\}$, so the degree must be $2$. Moreover, the extension is Galois, and $\mathrm{Gal}(C/R)$ is isomorphic to C2.

However, I am unclear about how this works when we are working with finite fields. If we let $K$ be the algebraic closure of $F_p$, then I know $K$ as equivalent to $F_p$ adjoin every single root of every single polynomial in $F_p[X]$. I know, given $\alpha \in K$ implies $\alpha$ is the root of some polynomial, $p(x)$ of deg $n$ in $F_p[X] \Rightarrow [F_p(\alpha):F_p]=n$, and $[K/F_p] = |N|$, as was kindly pointed out below.

This extension is Galois (finite fields are always separable, and $K$ is clearly the splitting field of each xˆ(p)ˆn - 1, for every $n \in N$). So $\mathrm{Gal} (K/F_p) = |N|$. I want to clarify: what does $\mathrm{Gal} (K/F_p)$ actually look like? Am I correct to write $\mathrm{Gal} (K/F_p) = \{\sigma: a \mapsto aˆpˆn | n \in N\}$ ?

Falko
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user138798
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  • On the $[K:F_p]$ point. Notice that $K$ must be countably infinite, and since $F_p$ is finite $[K:F_p]$ must be the cardinality of the natural numbers – Atticus Christensen Apr 04 '15 at 17:36
  • See also here for the Galois group. – Dietrich Burde Apr 04 '15 at 17:39
  • Thinking about this, this makes sense. I know that if α,b both have degree n over Fp, then Fp(α) and Fp(b) are isomorphic. Let [Fp(αn):Fp] = n. I can count the degree of K/Fp by [Fp(α1):Fp] x [Fp(α2):Fp] x ..... which is the cardinality of the natural numbers. – user138798 Apr 04 '15 at 17:45

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