The algebraic closure of a finite field and its Galois group

Question

$F$ is an extension field of a field $K$.
Let $F$ be an algebraic closure of $\mathbb{Z}_p $ ($p$ prime). Show that
$(i)$ $F$ is algebraic Galois over $\mathbb{Z}_p$
$(ii)$ The map $\alpha:F\rightarrow F$ given by $u\mapsto u^p$ is a nonidentiy $\mathbb{Z}_p$-automorphism of $F$.
$(iii)$ The subgroup $H=\langle \alpha \rangle$ is a proper subgroup of Aut$(F/\mathbb{Z}_p)$ where the fixed field is $\mathbb{Z}_p$, which is also the fixed field of Aut$(F/\mathbb{Z}_p)$ by $(i).$

So, here is my attempt for (i). Let $S \subset \mathbb{Z}_p[x])$ of monic polynomials of the form $x^{p^n}-x$. Then for all $f\in S$, gcd$(f,f^{\prime})=1$(i.e, the polynomials are separble) and $F=\mathbb{Z}_p(a\in F:f(a)=0)$. So $(F/\mathbb{Z}_p)$ is Galois. I'd be glad if I could get assistance for (ii) and (iii) as well. Thanks.

ADDED: Attempt at (iii). but I know that since $F/\mathbb{Z}_p$ is a finite Galois extension, its fixed field is $\mathbb{Z}_p$ hints...

The field $\mathbb{Z}_p$ must be contained in $F$. For $a\in \mathbb{Z}_p$, $\alpha(a)=a^p=a$. Thus the polynomial $x^p-x$ has $p$ zeros in $F$, namely, the elements of $\mathbb{Z}_p$. But the elements fixed under $\alpha$ are precisely the zeros in $F$ of $x^p-x$. Hence the fixed field of $\alpha$ is $\mathbb{Z}_p$ which is also the fixed field of Aut$(F/\mathbb{Z}_p)$.
Left to show that $H$ is a proper subgroup... If it helps I know that the order of $\langle \alpha \rangle $ is $n$...

Answer 1

I will use $\mathbf{F}_p$ to denote the field of $p$ elements (what you denote as $\mathbb{Z}_p$), $\mathbf{F}_{p^k}$ to denote the field of $p^k$ elements, and $\mathbf{F}$ for the algebraic closure of $\mathbf{F}_p$ (what the OP calls $F$).

Your work in (i) is very rough (and at point unintelligible). For instance, you write "$\gcd(f,f')$ (i.e., the polynomials are separable)". That is, prima facie, nonsensical. (Did you forget to say what the greatest common divisor was equal to?). Also, you are only dealing with certain kinds of polynomials. Finally, you did not say what you were going to do with $S$, it was just left hanging there.

Presumably, you were trying to say: "take $S$ to be the set of all polynomials of the form $x^{p^n}-x$, with $n$ a positive integer. These polynomials are separable, and $F$ is the splitting field of $S$. Thus, $F$ is Galois over $\mathbf{F}_p$, since it is the splitting field of a set of separable polynomials." That is okay as far as it goes, but how do you know that $\mathbf{F}$ is the splitting field of $S$? It is clear that the splitting field of $S$ is contained in $\mathbf{F}$, but you have provided not argument (or ghost of one) for showing that every element of $\mathbf{F}$ must be in the splitting field of $S$.

For that, you need to argue that given any $a\in\mathbf{F}$, you know that $[\mathbf{F}_p(a):\mathbf{F}]$ is finite (why?), hence $\mathbf{F}_p(a) = \mathbf{F}_{p^k}$ for some $k$ (why?), and so $a$ is a root of $x^{p^k}-x$ (why?), hence $a$ is in the splitting field of $S$. This will establish (i) correctly.

For (ii), proving that $u\mapsto u^p$ is indeed an automorphism is straightforward. To prove that it is not the identity... Think about $\mathbf{F}_{p^2}$. What do you know about its multiplicative subgroup? Can exponentiation by $p$ be the identity map on all those elements? Are they in $\mathbf{F}$?

For (iii): the point here is to show that in the infinite extension case, the correspondence between subgroups of the Galois group and subfields of the extension no longer holds. You are being asked to prove that there is more than one subgroup whose fixed field is precisely the ground field (if you think about the finite Galois case, that never happens). Explicitly, you are being asked to show two things: that the fixed field of the subgroup generated by $\alpha$ is precisely the ground field; and that the subgroup generated by $\alpha$ is not all of the Galois group. This will show that there are at least two distinct subgroups (both $H$ and all of $\mathrm{Aut}(\mathbf{F}/\mathbf{F}_p)$) that have the same fixed field, which shows the correspondence you know from the finite extension case no longer holds here.

The first part should be easy if you have managed to understand part (ii). First, pick $a\in \mathbf{F}$, $a\notin \mathbf{F}_p$. Try thinking about $\mathbf{F}(a)$, and combining the ideas of (i) and (ii), use them to show that $\alpha(a)\neq a$, so that the fixed field of $\alpha$ is contained in $\mathbf{F}_p$.

The trickier part is to show that the powers of $\alpha$ are not the only automorphisms of the algebraic closure.

Added. Exhibing an explicit element of the Galois group that is not a power of $\alpha$ is fairly straightforward if you know what is going on behind the scenes, but it would likely be challenging when doing this problem, since it would seem to me that the point of the problem is to show you that there is something going on behind the scenes.

You can show that the automorphism group is not cyclic by exhibiting two elements of infinite order such that the subgroups they generate intersection trivially. One possibility: for distinct primes $q$ and $\ell$, consider the two towers \begin{align*} &\mathbf{F}_p \subseteq \mathbf{F}_{p^q}\subseteq \mathbf{F}_{p^{q^2}}\subseteq\cdots\\ &\mathbf{F}_p \subseteq \mathbf{F}_{p^{\ell}} \subseteq \mathbf{F}_{p^{\ell^2}}\subseteq\cdots \end{align*} The two towers are dijsoint: since $\mathbf{F}_{p^n}\subseteq \mathbf{F}_{p^m}$ if and only if $n|m$, any field from one tower intersects any field in the other tower at just $\mathbf{F}_p$.

Use the usual properties of extension of isomorphisms to show that there is an element in $\mathrm{Aut}(\mathbf{F}/\mathbf{F}_p)$ that acts like $\alpha$ on the first tower, but like the identity on the second tower. Then show that there is an element in $\mathrm{Aut}(\mathbf{F}/\mathbf{F}_p)$ that acts like $\alpha$ on the second tower and like the identity on the first tower. Call these $\beta_1$ and $\beta_2$. Now note that $\langle\beta_1\rangle\cap\langle\beta_2\rangle = \{1\}$. But if $\mathrm{Aut}(\mathbf{F}/\mathbf{F}_p) = \langle\alpha\rangle$, then any two nontrivial subgroups have a nontrivial intersection.

Added. I'm really at a loss on trying to give you enough without entirely spoiling the problem; based on the confusion that this (nontrivial) problem has engendered, I'm guessing that you are still very much a beginner with Galois groups in general and finite fields in particular...

Another thing you can do: since $\alpha$ has infinite order, it would be enough to exhibit an element of $\mathrm{Aut}(\mathbf{F}/\mathbf{F}_p)$ that has finite order. Again, you can try to do this by using extension properties of automorphisms.

Answer 2

The last statement recommended in Arturo Magidin's answer cannot work because it is not true at all: every nontrivial element is Galois group $\mathrm{Aut}_{F_{p}}F$ has infinite order (for reference see Exercise 15 on page no. 71 of this book). Let $K$ be the union of subfield in the first chain, i.e, union of $F_{{p}^{q^{n}}}$, then it is a field. Consider a homomorphism $h$ from $K$ to $F$ given by $x\mapsto x^p$; then by extension theorem there is an $F_p$ automorphism from $F$ to $F$. Since $H$ is not a proper subgroup, it implies that $h=α^n$ for some interger $n$ ( $n \neq0$). Apply $x$ both sides where $x$ is in $F_{p^{2^{n+1}}}$, it leads to $p^{2^{n+1}}$ less than roots of equation $h(x)=α^n(x)$, a contradiction.

Answer 3

My answer is an alternative approach which avoid using certain results that is used in Professor Arturo Magidin’s answer, for instance, $\mathbf{F}_p(a)=\mathbf{F}_{p^k}$ and $a$ is root of $x^{p^k}-x$ and multiplicative group of $\mathbf{F}_{p^2}$. Though you should known about these results.

You can kill two bird with one stone if you use perfect field notion. Following are equivalent definition of perfect field $K$:

$(1)$ Either $\text{char }K=0$, or, when $\text{char }K=p$, the Forbenius endomorphism $\varphi :K\to K$ given by $u\mapsto u^p$ is surjective.

$(2)$ Every irreducible polynomial over $K$ is separable.

$(3)$ Every algebraic extension of $K$ is separable.

Proof: $(1)\Rightarrow (2)$ If $\text{char }K=0$, then every irreducible polynomial in $K[x]$ is separable. Suppose $\text{char }K=p$ and $f\in K[x]$ be irreducible polynomial. We have to show $f$ is separable, which is equivalent to $f’\neq 0$. Assume towards contradiction, $f’=0$. Then $f=\alpha_0 +\alpha_1 x^p+\cdots +\alpha_n x^{np}$ because $rx^{r-1}\neq 0$ if $r$ is not multiple of $p$. Since $\varphi$ is surjective, $\exists \beta_i\in F$ such that $\varphi (\beta_i)=\alpha_i =\beta_i^p$. By freshman dream, $$ f = \beta_0^p +\beta_1^p x^p+\cdots +\beta_n^px^{np} = (\beta_0 +\beta_1 x+ \cdots +\beta_n x^n)^p.$$ Which contradicts our initial assumption of $f$ is irreducible. Thus $f’=0$ and $f$ is separable.

$(2)\Rightarrow (1)$ Let $v\in K$ and $f=x^p-v$. Let $F$ be splitting of $f$ over $K$. By freshman dream, $f= x^p-v=(x-u)^p$ for some $u\in F$. Let $g\in K[x]$ be irreducible factor of $f$. Since $F[x]$ is unique factorization, $g=(x-u)^k$ where $1\leq k\leq p$. By hypothesis, $g$ is separable. So $k=1$. Thus $u\in K$ and $\varphi (u)=u^p=v$.

$(2)\Rightarrow (3)$ is trivial. For $(3)\Rightarrow (2)$ use algebraic closure or splitting field.

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(i) It is easy to see finite field is perfect. Since $F$ is algebraic over $\Bbb{Z}_p$ and algebraically closed, we have $F$ is separable over $\Bbb{Z}_p$ and $F$ is splitting field of $\Bbb{Z}_p[x]$ over $\Bbb{Z}_p$. Thus $F$ is algebraic Galois over $\Bbb{Z}_p$.

(ii) Only algebraic extension of $F$ is $F$ itself, since $F$ is algebraically closed. So $F$ is perfect field. Thus $\alpha$ is $\Bbb{Z}_p$-automorphism of $F$. Assume towards contradiction, $\alpha$ is identity. Then every $u\in F$ is root of $x^p-x\in \Bbb{Z}_p[x]$. Which implies $|F|\leq p$. But a finite field can’t be algebraically closed (here is detail). Thus we reach contradiction and $\alpha$ is non identity.

Obviously you can show $\alpha$ is surjective straightforwardly. Let $u\in F$. Then $x^p-u \in F[x]$ has root in $F$ since $F$ is algebraically closed.

(iii) Let $H=\langle \alpha \rangle$. We have to show $\Bbb{Z}_p=H’=\{u\in F\mid \sigma (u)=u,\forall \sigma \in H\}$ (fixed field of $H$). Inclusion $\subseteq$ is trivial. Let $u\in H’$. Then $\alpha (u)=u^p=u$. So $u$ is root of $x^p-x$. Since $\alpha$ is identity on $\Bbb{Z}_p$, we have $\Bbb{Z}_p$ is the set of all roots of $x^p-x$. Thus $u\in \Bbb{Z}_p$. Hence $\Bbb{Z}_p=H’$.

Proof of $H$ is proper subgroup of $\text{Gal}(F/\Bbb{Z}_p)$ is in Magidin’s and Nguyen’s answer, for more details see this post.