# The algebraic closure of a finite field and its Galois group

$F$ 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 $n$...

• "Likely Exam Question" is a really bad title. Mar 15, 2011 at 6:28
• Sorry.first time on stackexchange.
– user8280
Mar 15, 2011 at 7:34
• @Eric: Is $\mathbb{Z}_p$ the integers modulo $p$, or the $p$-adic integers? This is a bad context in which to use the former notation, but in any case you might want to specify which you mean; for the field of $p$ elements, it is better to use $\mathbf{F}_p$, $\mathbb{F}_p$, or $\mathbb{Z}/p\mathbb{Z}$. Also, you might want to use \langle and \rangle instead of < and > in (iii). Mar 15, 2011 at 18:26
• @Eric: Then I urge you to use $\mathbf{F}_p$ instead of $\mathbb{Z}_p$. In the context of this question, a very natural interpretation for $\mathbb{Z}_p$ is that it refers to the $p$-adic integers. Mar 16, 2011 at 3:08
• @Eric: You are incorrect: $\mathbf{F}/\mathbb{F}_p$ is most definitely not a finite Galois extension. It contains subfields of degree $n$ for every $n\ge 0$, after all. Mar 16, 2011 at 5:19

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.

• Is $\mathbf{F}=\mathbf{F}_{p^n}$?
– user8280
Mar 16, 2011 at 6:13
• Somehow I'm failing to see that the subgroup generated by $\langle \alpha \rangle$ is proper.
– user8280
Mar 16, 2011 at 9:33
• @Eric: $\mathbf{F}$ is the algebraic closure of the field of $p$ elements. It contains $\mathbb{F}_{p^n}$ for every $n$. Mar 16, 2011 at 15:48
• Please could you edit the portion about the tower fields? Thanks.
– user8280
Mar 16, 2011 at 19:46
• As pointed out in the answer below, and in this thread there are no non-trivial automorphisms of a finite order of $\mathbf{F}$. The reason why $G=Aut(\mathbf{F}/\Bbb{F}_p)$ is not cyclic is a bit subtle. The Frobenius automorphism only generates a dense subgroup of $G$ that is isomorphic to the limit $\hat{\mathbf{Z}}=\lim_{n\to\infty}\mathbf{Z}_{n!}$. A more concrete way might be to show that for any prime $\ell$, the additive group of the $\ell$-adic integers embeds into $G$. So $G$ is uncountable, hence not cyclic. Jun 16, 2015 at 11:23

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.

• This is absolutely true, and for reference see Exercise 15 on page 71 of this book: amazon.com/Field-Galois-Theory-Graduate-Mathematics/dp/…
– user54992
Mar 26, 2014 at 6:54
• Another way to see this is that the given Galois group is $\hat{\Bbb Z}$, which is an infinite product (over the primes $p$) of the $p$-adic integers, thus is torsion-free. Jun 22, 2018 at 14:36

For (ii) we have the following: Let $a,b \in F$. Then $$\alpha(a+b) = (a+b)^{p} = a^p+b^p = \alpha(a)+\alpha(b)$$ Also $$\alpha(ab) = \alpha(a) \alpha(b)$$ Now $\alpha(a) = 0 \implies a = 0$ so that $\alpha$ is injective (i.e. the kernel of $\alpha$ is ${0}$). We also know that $\alpha$ is surjective since $F$ is finite. This is known as the Frobenius automorphism.

• thanks for your answer, but how does this show that $\alpha$ is a nonidentity?
– user8280
Mar 15, 2011 at 17:10
• @Eric: Why don't you try computing the fixed field of the Frobenius automorphism and see if you can answer this question yourself? Mar 15, 2011 at 18:10
• @ PEV: Your argument for showing α is NOT correct as $F$ is not a finite field.You know $F$ is algebraic closure of $F_{p}$ and hence $F$ is not a finite field.
– user54992
Mar 26, 2014 at 8:17