Let E/F be a finite Galois extension with Galois group G. If H is a subgroup of G, let F(H) be the fixed field of H,and if K is an intermediate field,let G(K) be Gal(E/K), the fixing group of K.

(1) F is a bijective map from subgroups to intermediate fields,with inverse G. Both maps are inclusion-reversing,that is,if H1 ≤ H2 then F(H1) ≥ F(H2),and if K1 ≤ K2, then G(K1) ≥ G(K2).

(2) If the intermediate field K corresponds to the subgroup H and σ is any automorphism in G,then the field σK = {σ(x): x ∈ K} corresponds to the conjugate subgroup $σHσ^{−1}$. For this reason, σK is called a conjugate subfield of K.


(1) First,consider the composite mapping H →F(H) → GF(H). If σ ∈ H then σ fixes F(H) by definition of fixed field,and therefore σ ∈ GF(H) = Gal(E/F(H)). Thus H ⊆ GF(H). If the inclusion is proper,then by (6.1.2) part (ii) with F replaced by F(H), we have F(H) > F(H),a contradiction. [Note that E/K is a Galois extension for any intermediate field K,b y (3.4.7) and (3.5.8).] Thus GF(H) = H. Now consider the mapping K → G(K) → FG(K) = F Gal(E/K). By (6.1.2) part (i) with F replaced by K,we have FG(K) = K. Since both F and G are inclusion-reversing by definition,the proof of (1) is complete.

(2) The fixed field of $σHσ^{−1}$ is the set of all x ∈ E such that $στσ{−1}(x)$ = x for every τ ∈ H. Thus F($σHσ^{−1}$) = {x ∈ E: $σ{−1}(x)$ ∈ F(H)} = σ(F(H)).

My Questions

(1) I have a question about the composite mapping H→F(H)→GF(H). So we have two functions in the composite: H→F(H) and F(H)→GF(H), right? But I don't really understand this. In other words, H is a set of permutations and F(H) is the set of elements that are fixed by those permutations. So when they say H→F(H), does it mean that each permutation corresponds to one element that is fixed by all permutations? Also, for F(H)→GF(H), I'm guessing that this is the same as F(H)→G(F(H))...so now each element that is fixed by all permutations of H corresponds to one permutation that fixes all elements of F(H) (which is the opposite of the first one, right)? My main concern is that...how can they "assume" that there is exactly one permutation that corresponds to each element of F(H) and vice versa? And also with K → G(K) → FG(K), how can they "assume" that each element of K corresponds to exactly one permutation that fixes all elements of K?

(2) I understand why F($σHσ^{−1}$) = σ(F(H)), but I don't understand how this proves that the field σK corresponds to the conjugate subgroup $σHσ^{-1}$.

Thank you in advance


(1) No: it is rather a mapping $H \mapsto F(H)\mapsto G(F(H))$. That is, $F$ is a function defined on the subgroups (and not on the elements) of $Gal(E/F)$. Similarly, $G$ inputs intermediate fields and not field elements.

All behind of this there is the Galois connection induced by the relation 'fixes' between automorphisms and field elements (i.e. $\ \phi\sim a \overset{def}\iff \phi(a)=a$).

In general, if we have any relation $r\subseteq A\times B$ between sets $A,B$, then it induces two mappings between $\mathcal P(A)$ and $\mathcal P(B)$: $$F:\mathcal P(A)\to \mathcal P(B):\ U\mapsto \{b\,\mid\,\forall u\in U:\,u\,r\,b\} \\ G:\mathcal P(B)\to \mathcal P(A):\ V\mapsto \{a\,\mid\,\forall v\in V:\,a\,r\,v\}$$ One can easily show that both $F$ and $G$ reverses the inclusion ($U\subseteq U'\implies F(U)\supseteq F(U')$), and $G(F(U))\supseteq U$ and $F(G(V)\supseteq V$ for all $U\in\mathcal P(A)$ and $V\in\mathcal P(B)$. Then using these, show that $F\circ G\circ F=F$ and $ G\circ F\circ G=G$.

A subset $V\subseteq B$ is called closed w.r.t. the Galois connection induced by $r$, if $V=F(U)$ for some $U\subseteq A$, or equivalently, if $V=F(G(V))$. It follows, that, restricted to closed elements, the maps $F$ and $G$ are inverses of each other. This is the Galois correspondence: a closed set on one side has a correspondent closed set on the other side.

In the case when $r=$'fixes' between automorphisms and field elements, we have that the closed subsets are exactly the intermediate fields on the side of elements and the subgroups on the side of automorphisms.

(2) The intermediate field $K$ corresponds to the subgroup $G(K)=Gal(E/K)$, and the subgroup $H$ corresponds to $F(H)=Fixed(H)$. So, $F(\sigma H\sigma^{-1})=\sigma\,F(H)=\sigma\,K$ where $K$ is the correspondent intermediate field of $H$ (that means $F(H)=K$, or alternatively, $G(K)=H$).

  • $\begingroup$ Thanks a lot. But there is still something that's confusing me. The beginning of the theorm states "let F(H) be the fixed field of H". On another page (which this theorem relates to in the textbook), it says that "F(H) = {x ∈ E: σ(x) = x for every σ ∈ H}". So does F(H) have two seperate meanings? $\endgroup$ – user58289 May 7 '13 at 11:18
  • $\begingroup$ @Artus: The terms "fixed field of $H$ (in $E$)" and "$\{x \in E : \sigma(x) = x {\rm \ for \ every \ } \sigma \in H\}$" mean exactly the same thing. You'd better look at some concrete examples if you don't realize those have identical meanings. $\endgroup$ – KCd May 7 '13 at 12:00
  • $\begingroup$ @KCd Thanks, but I do realize that the fixed field of H is {x∈E:σ(x)=x for every σ∈H}. My question was that...if F(H) means the "fixed field of H" which is "{x∈E:σ(x)=x for every σ∈H}", then how does it become a function of subgroups (as it is in Berci's answer and also probably in the proof). That was my question. Because the theorem defines F(H) as the fixed field of H, but then uses F(H) to mean a function of subgroups. $\endgroup$ – user58289 May 7 '13 at 12:34
  • 2
    $\begingroup$ @Artus: The two ways of thinking of $F(H)$ are exactly the same as defining $f(x)$ as $x^3 + x$ for each real number $x$ and then saying $f(x)$ is a function on the real numbers, where $x$ is no longer a specific number. Do you have a problem with shifting back and forth between the two viewpoints of "$f(x)$" as a value of a function at the number $x$ and a function of a variable $x$? Think of $H$ as a "variable subgroup" if it helps, I guess. $\endgroup$ – KCd May 7 '13 at 14:07
  • $\begingroup$ How we show that functions $F$ and $G$ are inclusion reversing? $\endgroup$ – Idonknow Nov 11 '13 at 17:12

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