# The algebraic closures of finite fields and their subfields

My first post is going to be about the algebraic closure of a finite field $\mathbb{F}_q$, where as usual $q$ stands for a power of a prime $p$. More exactly, I will describe what the Galois group $Gal(\overline{\mathbb{F}_q}/\mathbb{F}_q)$ is, what the subfields are of $\overline{\mathbb{F}_q}$ and the Galois group $Gal(\overline{\mathbb{F}_q}/F)$ with $F$ a field $\mathbb{F}_q \subseteq F \subseteq \overline{\mathbb{F}_q}$.

# The algebraic closure and $Gal(\overline{\mathbb{F}_q}/\mathbb{F}_q)$

First of all, what is the algebraic closure of $\mathbb{F}_q$ ? The easiest way to describe this is $$\overline{\mathbb{F}_q} = \bigcup_{n=1}^\infty \mathbb{F}_{q^n},$$ but there are other ways, like taking a direct limit, but luckily, they are all equal. It won’t matter how you calculate $Gal(\overline{\mathbb{F}_q}/\mathbb{F}_q)$, it’s always going to be the same group (isomorphism not withstanding).

We know that $Gal(\mathbb{F}_{q^n}/\mathbb{F}_q)$ is generated by the Frobenius automorphism $\sigma(x) = x^q$, so $Gal(\mathbb{F}_{q^n}/\mathbb{F}_q) = \mathbb{Z}/n\mathbb{Z}$. We also know that $\mathbb{F}_{q^n} \subseteq \mathbb{F}_{q^m}$ iff $n |m$, so we get a directed system of finite extensions of $\mathbb{F}_q$. We also get an inverse system of Galois groups.

Definition An inverse system of finite groups is a set of finite groups $G_i$, indexed by a directed set $I$ together with a set of group homomorphisms $\phi_{ij}:G_j \rightarrow G_i$, where as usual $\phi_{ij} \circ \phi_{jk} = \phi_{ik}$ when $i \geq j \geq k$ and $\phi_{ii} = 1_{G_i}$.

This inverse system of Galois groups gives us the opportunity to take the projective limit $$\varprojlim_{n \geq 1} Gal(\mathbb{F}_{q^n}/\mathbb{F}_q) = \{ (\sigma_{n})_{n \geq 1} \mid \sigma_m|_{\mathbb{F}_{q^n}} = \sigma_{n} \text{ if } n|m \} \subseteq \prod_{n \geq 1} Gal(\mathbb{F}_{q^n}/\mathbb{F}_q).$$ Since $Gal(\mathbb{F}_{q^n}/\mathbb{F}_q) \cong\mathbb{Z}/n\mathbb{Z}$, this is nothing but $\widehat{\mathbb{Z}}$, which in turn is isomorphic to $\prod_{p_{i} \text{ prime}} \mathbb{Z}_{p_{i}}$ (this is easily proved using the Chinese Remainder Theorem). It is easy to see that $\sigma:x \rightarrow x^q$ corresponds with $(1 \bmod n)_{n \geq 1}$.

$\widehat{\mathbb{Z}}$ has another nice property: if $S$ is any finite subset of $\mathbb{N}$, then $$\widehat{\mathbb{Z}} \subseteq \prod_{n \in \mathbb{N}-S} \mathbb{Z}/n\mathbb{Z},$$ so deleting any finite number of numbers to take the inverse limit $\varprojlim_{n \geq 1, n \notin S} \mathbb{Z}/n\mathbb{Z}$ doesn’t change a thing.

# Steinitz numbers and subfields of $\overline{\mathbb{F}_q}$

Steinitz numbers will help us to find all subfields of $\overline{\mathbb{F}_q}$, apart from the (obvious) finite ones.

Definition A Steinitz number is a symbol of the form $$\prod_{p_i \text{prime}}{p_i}^{e_{i}}$$ where $p_{i}$ denotes the $i$-th prime number and each $e_i$ belongs to $\mathbb{N} \cup \{\infty\}$. 2 Steinitz numbers are equal iff for all $p_i$ the corresponding exponents are the same. The set of Steinitz numbers is denoted $\mathbb{E}$.

We will use capital letters for abstract Steinitz numbers and lowercase letters for positive integers, who we will also treat as Steinitz numbers. For two Steinitz numbers $N=\prod_{i=1}^\infty {p_i}^{e_{i}}$ en $M=\prod_{i=1}^\infty {p_i}^{f_{i}}$, we define their product $NM$ as the Steinitz number $\prod_{i=1}^\infty {p_i}^{e_{i}+f_{i}}$ with the usual rules for addition with $\infty$.  We will say that a Steinitz number $N$ divides a Steinitz number $M$ iff for all $i$, $e_i \leq f_i$. In that case we define $\frac{M}{N} = \prod_{i=1}^\infty {p_i}^{f_{i}-e_{i}}$, with the rule $\infty – \infty = 0$. Otherwise, $\frac{M}{N}$ is not defined. It is also not difficult to prove that $N$ divides $M$ iff for all $n \in \mathbb{N}$ holds: $n|N \Rightarrow n|M$.  We also define the greatest common divisor $N \wedge M= \prod_{i=1}^\infty {p_i}^{\min(f_{i},e_{i})}$ and the least common multiple $N \vee M = \prod_{i=1}^\infty {p_i}^{\max(f_{i},e_{i})}$.

For a Steinitz number $N$, we define $$\mathbb{F}_{q^N} = \bigcup_{d \in \mathbb{N},d|N} \mathbb{F}_{q^d}.$$ It is easy to verify that this is a field (if $n$ and $m$ divide $N$, then so does $n \vee m$ and so the compositum $\mathbb{F}_{q^n}\mathbb{F}_{q^m} = \mathbb{F}_{q^{n\vee m}}$ is a part of $\mathbb{F}_{q^N}$, so the product of 2 elements is always defined, since all elements belong to finite fields). If $n \in \mathbb{N}$, then $\mathbb{F}_{q^n}$ is finite. It is also true that $\mathbb{F}_{q^N} \subseteq \mathbb{F}_{q^M}$ iff $N|M$. Moreover, one sees immediately that there is a bijection between Steinitz numbers and subfields of $\overline{\mathbb{F}_{q}}$. We shall proof this.

Theorem There exists a bijection $\phi: \mathbb{E} \rightarrow \{\text{ subfields of } \overline{\mathbb{F}_q} \}$ defined by $\phi(N) = \mathbb{F}_{q^N}$.

Proof To see that $\phi$ is injective, take 2 different Steinitz numbers $N=\prod_{i=1}^\infty {p_i}^{e_{i}}$ and $M=\prod_{i=1}^\infty {p_i}^{f_{i}}$. Since they are different, there exists a prime $p_i$ so that $e_i \neq f_i$, so we take $e_i < f_i$. But then we have that $\mathbb{F}_{q^{p^{e_i+1}}} \subseteq \mathbb{F}_{q^M}$, but $\mathbb{F}_{q^{p^{e_i+1}}} \nsubseteq \mathbb{F}_{q^N}$. So they aren’t equal and thus $\phi$ is injective.
To prove it is surjective, let $F$ be a subfield of $\overline{\mathbb{F}_{q}}$ that contains $\mathbb{F}_q$. We look for a Steinitz number $S$ so that $F = \mathbb{F}_{q^S}$. For each prime $p_i$, let $e_i$ be the greatest exponent so that $\mathbb{F}_{q^{p_i^{e_i}}} \subseteq F$. We let $e_i$ be $\infty$ if this is true for every $e_i \in \mathbb{N}$. Let $S$ be $\prod_{p_i \text{prime}} p_i^{e_i}$. First, it is easy to see that $\mathbb{F}_{q^S}$ is the compositum of all $\mathbb{F}_{p_i^{e_i}}$. Since all $\mathbb{F}_{q^{p_i^{e_i}}}$ are subsets of $F$, we have that $\mathbb{F}_{q^S} \subset F$. For the reverse inclusion, take $\alpha \in F$. Since $\alpha$ is algebraic over $\mathbb{F}_q$, it belongs to a certain $\mathbb{F}_{q^n}$ with $n \in \mathbb{N}$. If $n = \prod_{p_i \text{prime}} p_i^{f_i}$, then $f_i \leq e_i$ due to the maximality of $e_i$. So $n$ divides $S$ and so $\alpha \in S$. This concludes the proof.

One can calculate the Galois group $Gal(\overline{\mathbb{F}_q}/\mathbb{F}_{q^S})$ by using $S$. It will come as no surprise that if $S = \prod_{p_i \text{prime}} p_i^{e_i}$ then $$Gal(\overline{\mathbb{F}_q}/\mathbb{F}_{q^S}) = \prod_{p_i \text{prime}} p_i^{e_i} \mathbb{Z}_{p_i}$$ with $p_i^{e_i}\mathbb{Z}_{p_i} = 0$ if $e_i=\infty$. As an example, one can calculate $Gal(\overline{\mathbb{F}_q}/\mathbb{F}_{q^n})$ with $n \in \mathbb{N}$. Since $q^n$ is just another power of a prime $p$ and we started with any power of $p$, we should get $\widehat{\mathbb{Z}}$ again. So if $n = \prod_{p_i \text{prime}} p_i^{f_i}$ with all but finitely $f_i$ equal to zero and all $f_i < \infty$, we get $$Gal(\overline{\mathbb{F}_q}/\mathbb{F}_{q^n})=\prod_{p_i \text{prime}} p_i^{f_i} \mathbb{Z}_{p_i}$$. But $p_i^{f_i} \mathbb{Z}_{p_i} \cong \mathbb{Z}_{p_i}$ as a group (certainly not as a ring) if $f_i < \infty$, so we get $$\prod_{p_i \text{prime}} p_i^{f_i} \mathbb{Z}_{p_i} \cong \prod_{p_i \text{prime}} \mathbb{Z}_{p_i} = \widehat{\mathbb{Z}}$$ again. Calculating $Gal(\mathbb{F}_{q^n}/\mathbb{F}_{q})$, we get $$\widehat{\mathbb{Z}}/\prod_{p_i \text{prime}} p_i^{f_i} \mathbb{Z}_{p_i} = \prod_{p_i \text{prime}}\mathbb{Z}_{p_i}/\prod_{p_i \text{prime}} p_i^{f_i} \mathbb{Z}_{p_i} \cong \prod_{p_i \text{prime}} \mathbb{Z}/p_i^{f_i}\mathbb{Z},$$ which by the Chinese Remainder Theorem is isomorphic to $\mathbb{Z}/n\mathbb{Z}$, like we want.

## 4 thoughts on “The algebraic closures of finite fields and their subfields”

1. bruno says:

Best interessant, wel wat verwarrend omdat er enerzijds een p wordt vastgezet (de p van het lichaam F_q) en anderzijds in de exponenten alle priemen opnieuw ‘vrij’ liggen. Desalniettemin denk ik dat ik het begrepen heb…

2. Kevin De Laet says:

Bruno, ik heb alleen in het begin p vastgelegd, het is daarom dat ik heel de tijd met F_q zit te werken ipv met F_p om verwarring tegen te gaan. Niettemin, ik denk dat ik maar 1 keer p als index heb gebruikt, voor de rest heb ik p_i gebruikt. Ik heb het veranderd, als er nog dingen fout of verwarrend zijn, hoor ik het wel.

3. Interesting read! Seems like I stumble upon a lot of good math blogs these days.

One comment: it is not “to proof” but “to prove” in English

4. Omri F says:

It seems like the first union is wrong. Your index is i, but it is not used inside.