Exercise 11.23

Question

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Let $p_1<p_2<p_3<\cdots$ denote the positive prime numbers arranged in order. Let $N_m = p_1^mp_2^m\cdots p_m^m$ and let $E_m$ denote the field with $q^{N_m}$ elements. Show that $E_m$ can be considered as a subfield of $E_{m+1}$ and that $E = \bigcup E_m$ is an extension of $E_0 = F$, a finite field with $q$ elements, with the following property; for every positive integer $n$, $E$ contains one and only one subfield $F_n$ with $q^n$ elements.

Richard Ganaye · Accepted Answer

\def\imp{\Rightarrow}
\def\gcro{\mbox{[\hspace{-.15em}[}}
\def\dcro{\mbox{]\hspace{-.15em}]}}

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\begin{proof}
Here $q = p^a$ is a power of $p$.

We build the family $E_m$ by induction.

For $m = 0$, $N_0 = 1$. Take $E_0 = F$, a finite field with $q$ elements, whose existence is proved in Theorem 3, Chapter 7. Then $|E_0| = q = q^{N_0}$.

Suppose that we know an extension $E_m$ of $F$ with $q^{N_m}$ elements, so that $[E_m : F] = N_m$.

Write $s = N_m$ and $t = N_{m+1}$. Then $t= \left (p_1\cdots p_m p_{m+1}^{m+1}ight) s$, so $s \mid t$, and $k = t/s$ is an integer. 
By Exercise 7.14, there exists a polynomial $p(x) \in E_m[x]$ of degree $k$, irreducible over $E_m$. Then $K = E_m[x]/(p(x))$ is a field, and the map $j : E_m 	o K$ defined by $j(\alpha) = \overline{\alpha} = \alpha + (p(x))$ is injective. This allows us to ``identify'' $E_m$ and $j(E_m)$.

More explicitly, if we define $E_{m+1} = (K \setminus j(E_m)) \cup E_m$ (that is, we replace the elements of $j(E_m)$ by the corresponding elements in $E_m$), then $E_m \subset E_{m+1}$ absolutely, and
$$
\varphi
\left\{
\begin{array}{ccl}
K &	o& E_{m+1}\
\alpha & \mapsto & 
	\left\{
	\begin{array}{ll}
	\beta &	ext{if } \alpha = j(\beta) \in j(E_m)\
	\alpha & 	ext{if } \alpha 
ot \in j(E_m)
	\end{array}
	ight.
\end{array}
ight.
$$
is a bijection. This bijection allows us to define a structure of field over $E_{m+1}$ by transport of structure, i.e. the laws $+, 	imes$ on $E_{m+1}$ are given by
$$u + v = \varphi(\varphi^{-1}(u) +\varphi^{-1}(v)),\quad u 	imes v = \varphi(\varphi^{-1}(u) 	imes \varphi^{-1}(v)), u,v \in E_{m+1}.$$
Then $E_{m+1}$ is a field for these laws, $\varphi$ is a field isomorphism, and $E_m$ is a subfield of $E_{m+1}$. Since the degree of $E_{m+1} \simeq K = E_m[x]/(p(x))$ is $k = \deg(p)$, $[E_{m+1} : E_m] = k = N_{m+1}/N_m$, therefore $[E_{m+1} : F ] = [E_{m+1} : E_m] [E_m:F] = k N_m = N_{m+1}$. Thus $|E_{m+1}| = q^{N_{m+1}}$.

To conclude this part, the sequence $(E_m)_{m\in \N}$ is an increasing sequence for inclusion, and for each $m\in N$, $|E_m| = q^{N_m}$.

\bigskip
Now consider the set union
$$E = \bigcup_{m \in \N} E_m.$$
We can define additive and multiplicative laws on $E$. If $\alpha, \beta \in E$, then $\alpha \in E_r, \beta \in E_s$. If $m = \max(r,s)$, then $\alpha, \beta \in E_m$, so that $\alpha + \beta$ is defined in $E_m$. Moreover, assume that $\alpha, \beta \in E_{m'}$ for another index $m'$. Then $E_m \subset E_{m'}$, or $E_{m'} \subset E_m$. If we suppose $E_m \subset E_{m'}$ (the other case is similar), $E_m$ is a subfield of $E_m'$, so that $\alpha + \beta$ is the same in $E_m$ or $E_{m'}$. This allows us to define $\alpha + \beta$ in $E$ as the sum of $\alpha , \beta$ in any field $E_m$ such that $\alpha, \beta$ are both in $E_m$. Similarly, we define the law $	imes$.

Then the axioms of a field are verified. For instance, if $\alpha, \beta, \gamma \in E$, there is some $m\in \N$ such that $\alpha, \beta,\gamma \in E_m$, where $\alpha(\beta + \gamma) = \alpha \beta + \alpha \gamma$. Thus this equality is true on $E$.

This shows that $(E,+,	imes)$ is a field, and $E_m$ is a subfield of $E$ for every $m\in \N$. In particular, $E$ is an extension of $F = E_0$.

Now we verify that $E$ has the expected property. Let $n \in \N^*$ be a positive integer. Consider
$$F_n = \{\alpha \in E \mid \alpha^{q^n} = \alpha\}.$$
Then $F_n$ is a subfield of $E$. Indeed, $1 \in F_n$, and if $\alpha, \beta \in F_n$, and $\gamma \in F_n^*$, then
\begin{align*}
(\alpha + \beta)^{q^n} &= \alpha^{q^n} + \beta^{q^n} = \alpha + \beta,\
(\alpha  \beta)^{q^n} &= \alpha^{q^n}  \beta^{q^n} = \alpha  \beta,\
(\gamma^{-1})^{q^n} &=( \gamma^{q^n})^{-1} = \gamma^{-1}
\end{align*}
thus $\alpha + \beta, \alpha \beta, \gamma^{-1} \in F_n$.        
Let $n = p_1^{a_1}\cdots p_k^{a_k}$ be the decomposition of $n$ in prime factors, for some $k\in \N$,  and $a_i \geq 0, \ i= 1,2,\ldots,k$. If $m = \max \{a_1,\ldots,a_k\}$, then $n \mid N_m$. By Lemma 2 and 3 of Chapter 7,  this shows that $q^n - 1 \mid q^{N_m} - 1$, thus $x^{q^n - 1} - 1 \mid x^{q^{N_m} - 1} - 1$, thus $x^{q^n} - x \mid x^{q^{N_m}} - x$. By proposition 7.1.1, since $E_m$ is a field with $q^{N_m}$ elements,
$$x^{q^{N_m}} - x =\prod_{\alpha \in E_m} (x- \alpha).$$
therefore the factor $x^{q^n} - x$ of $x^{q^{N_m}} - x$ splits completely over $E_m$, a fortiori over $E$, and Corollary 2 shows that all the roots of $x^{q^n} - x$, which are in $E_m$, are simple roots.

This proves that
$$ x^{q^n} -  x =\prod_{\alpha \in A} (x- \alpha),$$
where $A \subset E_m \subset E$.

By definition of $F_n$, for all $\alpha \in E$, $\alpha$ is a root of $x^{q^n}- 1$ if and only if $\alpha \in F_n$, thus $A = F_n$, and
$$ x^{q^n} -  x =\prod_{\alpha \in F_n} (x- \alpha),$$
The comparison of the degrees gives
$$q^n = |F_n|.$$
This proves that $E$ contains a field $F_n$ with $q^n$ elements.

Suppose that $E$ contains another field $F'_n$ with $q^n$ elements, then the preceding argument shows that
$$x^{q^n} - x =\prod_{\alpha \in F_n} (x - \alpha) =\prod_{\alpha \in F'_n} (x - \alpha),$$
therefore $F_n = F'_n$.

For every positive integer $n$, $E$ contains one and only one subfield $F_n$ with $q^n$ elements.
\end{proof}

\bigskip
Note: We can show a little more, that $E$ is an algebraic closure of $F$.

First, $E$ is algebraic over $F$, since every $\alpha \in E$ is in some $E_m$, which is a finite extension of $F$, thus $\alpha$ is algebraic over $F*$.

Next, we show that $E$ is algebraically closed. Let $p(x) = \sum_{k=0}^l a_k x^k \in E[x]$ be any non constant polynomial with coefficients in $E$. There is a $m\in \N$ such that all the coefficients $a_i$ are in $E_m$, so that $p(x) \in E_m[x]$. Let $f(x)$ be an irreducible factor of $p(x)$ over $E_m$, with $\deg(f) = d \geq 1$.

%The generalization of Theorem 2, Chapter 7, shows that
%$$x^{q^{dN_m}} - x = \prod_{\delta \mid d N_m} F_\delta(x),$$
%o $F_\delta(x)$ is the product of the monic irreducible polynomials in $E_m[x]$ of degree $\delta$. 
Then $f(x)$ has a root $\gamma$ in the field $K = E_m[x]/(f(x))$, where $|K| = q^{dN_m}$, so $\gamma$ is a root of $x^{q^{dN_m}} - x$. Since $f(x)$ is the minimal polynomial of $\gamma$ over $E_m$, this proves that $f(x) \mid x^{q^{dN_m}} - x$. If $n = dN_m$, we have seen that if $F_n$ is the subfield of $E$ with $q^n$ elements, then  $x^{q^n} - x =\prod_{\alpha \in F_n} (x - \alpha)$, so that $f(x)$ splits completely over $F_n = F_{dN_m} \subset E$. Therefore $f(x)$ has a root in $E$, and also $p(x)$.

We have proved that $E$ is an algebraic closure of $F$ (with a concrete construction, without the axiom of choice, used in the general proof of the existence of algebraic closure).

Exercise 11.23

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Exercise 11.23

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