The Galois correspondence on finite extensions

One way to exhibit the bijective nature of the Galois correspondence for finite Galois extensions involves a claim of the following sort:

If $L$ is a finite extension of $K$, and if $G$ is a (finite?) subgroup of $\operatorname{Gal}(L/K)$, then $\operatorname{Gal}(L/\operatorname{Fix}G)=G$.

Should the parenthesized "finite?" be stripped of its delimiters? Alternatively (I may have gotten the whole thing wrong), under what conditions does $\operatorname{Gal}(L/\operatorname{Fix}G)=G$ hold? For example, can the non-parenthesized "finite" be substituted with, say, "algebraic"?

• You should start by reading planetmath.org/infinitegaloistheory
– Ofir
Jan 27, 2015 at 7:41
• @Prometheus Thanks, but my problem is that I'm not sure if what I wrote is even correct. Jan 27, 2015 at 7:52
• @user210807 You should replace "finite" by "galois but not necessarily finite", see my answer. Jan 27, 2015 at 8:01
• @RobertGreen Is it really the case that you need the extension to be galois for the equality to hold? I'm not asking for the conditions under which the functions in the correspondence are mutual inverses - I'm asking about something a little less restrictive. Another way to put it: when is $\operatorname{Gal}$ a left-inverse of $\operatorname{Fix}$? Jan 27, 2015 at 8:05
• @user210807 As Fix(G) is always Galois, finite or not, you're bounded to use Galois extensions. Jan 27, 2015 at 8:13

Well-known fact. If $L/K$ is finite, then the group $Gal(L/K)$ is finite of order $\leq [L:K]$.

First I want to state a general lemma, that is really worth to be known.

Lemma. Let $L$ be a field, and $G$ be a group of field automorphisms of $L$. (Note that there is no base field for now.) Then the extension $L / L^G$ is algebraic if and only if all orbits of the $G$ action on $L$ are finite, in which case the extension is normal and separable. If $G$ is finite, $G =Gal(L/L^G)$.

Proof. Indeed, if $L/L^G$ is algebraic, any $x\in L$ is root of some $P\in L^G [X]$, and then for all $g\in G$ the element $g(x)$ is also a root of $P$ as $P$'s coefficients are fixed by $g$, and as $P$ as finitely many roots only, the orbit of $x$ is finite. Inversely, let $x\in L$, note $O(x)$ the orbit of $x$ under $G$ and set $P = \prod_{y\in O(x)} (X-y)$. Then you can express the coefficients of $P$ as symetric functions of the $y$'s in $O(x)$, which shows that $P$'s coefficients are fixed by every element of $G$, that is, that $P\in L^G [X]$, so that $x$, and then $L$, is algebraic over $L^G$. In this case the extension is obviously separable and normal, as the minimal polynomial of any $x\in L$ is divides the polynomial $\prod_{y\in O(x)} (X-y)$ which has all its distinct roots in $L$. Note that we have shown that $[L^G (x) : L^G] \leq Card(O(x))$. If $G$ is finite of order $n$, this is also $\leq n$. Let $x\in L$ such that $[L^G (x) : L^G]$ is maximal. For $y\in L$, the finite (as $y$ and $x$ are algebraic over $L^G$) extension $L^G (x,y)$ of $L^G$ is separable, as $L/L^G$ is. By the primite element theorem one can find $z\in L$ such that $L^G (x,y) = L^G (z)$. But $L^G (x) \subseteq L^G (x,y) \subseteq L^G (z)$ so that $[L^G (z) : L^G] \geq [L^G (x) : L^G]$. As $[L^G (x) : L^G]$ is maximal, one has in fact $[L^G (z) : L^G] = [L^G (x) : L^G]$ and as $L^G (x) \subseteq L^G (z)$ we have in fact $L^G (x) = L^G (z)$, which by definition of $z$ means $L^G (x) = L^G (x,y)$, showing that $y\in L^G (x)$. This is for all $y\in L$, so that $L = L^G (x)$, and $L/L^G$ is finite of degree $\leq n$. We can therefore write $$[L : L^G] \leq n = Card(G) \leq Card(Gal(L/L^G)) \leq [L : L^G]$$ where the first inequality comes from what we just did, the second from the fact that $G$ is a subgroup of $Gal(L/L^G)$ (obvious) and the third comes from the well-known fact recalled before the lemma. Thus, $Card(G) = Card(Gal(L/L^G))$ which implies $G = Gal(L/L^G)$. $\square$

About $\textrm{Gal}\circ\textrm{Fix}$

The previous lemma obviously implies that $\textrm{Gal}\circ\textrm{Fix} = \textrm{Id}$ if $L/K$ is finite, and answers your question.

About $\textrm{Fix}\circ\textrm{Gal}$

Suppose $L/K$ be finite, and let $L'/K$ be a subextension such that $(\textrm{Fix}\circ\textrm{Gal}) (L') = L'$. You have $L^{Gal(L/L')} = L'$ and the easy part (the one about separability) of the previous general lemma shows that $L/L'$ is separable. This shows that if you want to violate the equation $\textrm{Fix}\circ\textrm{Gal} = \textrm{Id}$ in the framework of finite extensions $L/K$, you simply have to find a finite extension $L/K$ and an subextension $L'/K$ such that $L/L'$ is not separable. As in zero characteristic all extensions are separable, your fields $L,K,L'$ will have to be of characteristic $p>0$. But more simply, if $L/K$ already is not separable, we cannot have $L^{Gal(L/K)} = K$, so that we just have to find a finite not separable extension $L/K$ !

Let $p>0$ be a prime number and take $K = \mathbf F_p(T)=$ the field of rational functions in the formal variable $T$. The polynomial $P(X)=X^p-T\in\mathbf F_p(T)[X]$ is irreducible (just apply Eisenstein's Criterion in $\mathbf F_p[T]\subset \mathbf F_p(T)$ and the prime $\,T\,$ in it. Now let $\alpha$ be some root of $P$ in some field extension $L$ (for instance $L = \mathbf{F}_p [X] / (P)$, then in $L[X]$ one has : $P(X) = X^p - T = X^p - \alpha^p = (X-\alpha)^p$ so that obviously $L/K$ is not separable. Now, we have $Gal(L/K) = 1$ as the only root of $P$ in $L$ is $\alpha$, so that $L^{Gal(L/K)} = L \not= K$.

Actually, one can show that the previous extension is purely inseparable (radicielle in french). In general, one can show that $L'^{Gal(L/L')}$ is the purely inseparable closure of $L'$ in $L$.

Some remarks.

What happens in finite degree in non Galois case ? Consider the field $K = \mathbf{Q}(\sqrt[4]{2})$ and the extension $K / \mathbf{Q}$ of degree $4$ that I gave above in a comment to your answer. An element of $G = Gal(K/\mathbf{Q})$ is uniquely determined by its value on $\alpha = \sqrt[4]{2}$, and such value is a root of $T^4 - 2$, so that this value is $\pm \alpha$, as those are the only roots of $T^4 - 2$ in $K$. Thus $G \simeq \mathbf{Z}/2\mathbf{Z}$ has two elements. This group has only $2$ subgroups, and has in particular no non-trivial strict subgroup, whereas $K / \mathbf{Q}$ as a non trivial sub-extension, namely, the extension $\mathbf{Q}(\sqrt{2}) / \mathbf{Q}$. In this simple case, the extension $K/\mathbf{Q}$ is not normal, and the Galois correspondance fails to bijective, as the map $H\mapsto Fix(H)$ fails to be surjective, missing the field generated by $\sqrt{2}$.

What to do in the case of an infinite degree extension ? The more general statement (in the context of field extensions) is the following. Let $L$ be a Galois extension of $K$, but not necessarily a finite extension, and $G$ be the Galois group. Put the dicrete topology on $L$, the product to topology on $L^L$, and the induced topology on $G$. Let $A$ be the set of finite subextensions of $L/K$. For $\sigma\in G$ and $E\in A$, set $U_{\sigma}(E) :=$ the subset of $G$ consisting of elements $\tau$ having same restriction on $E$ that $\sigma$. One can show that $U_{\sigma}(E)$ is a filter basis of $\sigma$ for this topology, and that the restrictions $G \rightarrow Gal(L'/K)$ are continuous for any subextension $L'/ K$ that is Galois, but not necessarily finite. Moreover, one can show that $G$ is compact and totally disconnected. Actually, if $(L_i)_i$ is a filtered familly of galois subextensions such that $L = \cup_i L_i$, you have $G \simeq \varprojlim_i Gal(L_i / K)$. (As you can choose all $L_i$'s to be finite over $K$, you got in particular that $G$ is profinite - meaning by that that $G$ is inverse limit of finite groups.) Then the Galois correspondance $H \mapsto Fix(G)$ induces a bijection between normal closed subgroups of $G$ and galois subextensions of $L/K$. Ultimate (in this case readable) reference : Bourbaki, Algèbre, Chapitre V "Corps commutatifs", paragraph 10. One last remark : actually, every profinite group is a Galois group. (This is a bit harder, but there is a fun and easy proof of this if you assume the true following act : every finite group is a Galois group.)

Why is this necessary in infinite degree ? Take $K=\mathbf{F}_p$ and let $L$ be an algebraic closure of $K$. $L$ is the union of fields $\mathbf{F}_{p^n}$ for $n\geq 1$. (Remember that $\mathbf{F}_{p^n}$ is the unique field with $p^n$ elements contained in $L$.) Consider $\varphi \in G$ defined by $\varphi(x)=x^p$, the so-called Frobenius. Let $H$ be the (infinite) subgroup of $G$ generated by $\varphi$. For all $n$ set $N_n = 1! + 2! +\ldots+(n-1)!$ and define $\sigma_n \in Gal(\mathbf{F}_{p^n} / \mathbf{F}_{p})$ by $\sigma_n (x) = x^{p^{N_n}}$ for all $x\in \mathbf{F}_{p^n}$. Note that if $x\in L$ is in $\mathbf{F}_{p^n}$ and $\mathbf{F}_{p^m}$ then $\sigma_n (x) = \sigma_m (x)$, so that there exists a unique $\sigma\in G$ whose restriction to $\mathbf{F}_{p^n}$ is $\sigma_n$ for all $n$. From this you see that $H$ is a strict subgroup of $G$. You can then see that $Fix (H) = K$, so that by the Galois "would-be correspondence" map, $Fix (H)$ is mapped to $Gal(L/K) = G$, instead of being mapped to $H$. This shows that the Galois "would-be correspondence" map is not bijective for Galois extensions with finite degree. (I let you work out whether $H$ is a closed subgroup of $G$ or not.) This shows the necessity to put some constraints on sugroups of $G$ to make the Galois map bijective in infinite degree, already when $L/K$ is Galois.

What's next ? If you want to generalize even more, you will have to leave the context of fields extensions, and pass first to separable algebras, and then to schemes and to Galois categories, read "SGA I : Revêtement étales et groupe fondamental" from Grothendieck, first exposé's. But as a starter, you could already think of the Galois theory of coverings in the case of compact Riemann surfaces... (Grothendieck's theory allows you to treat equally "fields" and Riemann surfaces as they are all instances of the same "concept".)

• My question isn't about when the Galois correspondence is bijective. It's about when one is a left-inverse of the other. With your example, the subgroups of $\operatorname{Gal}(\mathbf{Q}(\sqrt[4]{2})/\mathbf{Q})$ are itself, and the trivial group. If $G$ is either of those subgroups, $\operatorname{Gal}(\mathbf{Q}(\sqrt[4]{2})/\operatorname{Fix}G)=G$. So for your example, $\operatorname{Gal}$ is a left-inverse of $\operatorname{Fix}$. I want to know when is this true in greater generality. Jan 27, 2015 at 21:14
• @user210807 Ok, thx for finally stating (now for the first time) what your real question is... :-/ Anyway, for a finite extension $L/K$ and for $L'/K$ a subextension of $L'/K$, the field $L'' = L^{Gal(L/L'}$ (equal to $L'$ if $L/K$ is Galois, as we know) is simply the purely inseparable closure ("cloture radicielle" in french) of $L'$ in $L$. Obviously this closure always contains $L'$, and we are looking for an example (with $L/K$ non-galois) of $L'$ with strict inclusion. Intuitively, any (if any ;-)) $L'$ not unseparably closed in $L$ would do the job. Jan 27, 2015 at 21:46
• @user210807 Let me just mention that if $L/K$ is separable (which is the case if $K$ is of characteristic $0$) then, with previous notations, one has always $L' = L^{Gal(L/L')}$ for all subextensions $L'/L$, that is : $\textrm{Fix}\circ \textrm{Gal} = Id$. So my example will necessarily be in positive characteristic. Jan 27, 2015 at 21:54
• @user210807 I will update my answer later (tomorrow if you don't mind) with an example of such $K,L,L′$. Jan 27, 2015 at 21:54
• I'm sorry if my question wasn't clear, and I appreciate your post, which answers some other interesting questions, but in my defense, I never asked for the conditions under which the Galois correspondence are mutual inverses. My original (and current) question asks for when the equality $\operatorname{Gal}(L/\operatorname{Fix}(G))=G$ holds for every subgroup $G$ of the Galois group. Also, I'm not exactly asking for an example in which the equality does not hold, but such an example might help to answer the question, and it would be interesting to find one. Jan 27, 2015 at 22:13