# Constructing Galois extensions.

In Wiki Page I found the following statement.

A result of Emil Artin allows one to construct Galois extensions as follows: If $E$ is a given field, and $G$ is a finite group of automorphisms of $E$ with fixed field $F$, then $E$ over $F$ is a Galois extension.

Somebody please explain me, why the above statement is true.

Any hint is welcome. Thanks in advance.

• Read a little down that page, @Ga316: is because of the third characterization there. – DonAntonio May 26 '14 at 18:30
• @DonAntonio : But why would number of automorphisms of $E$ fixing $F$ would be same as $[E:F]$... I do not understand :O – user87543 May 26 '14 at 18:32
• @DonAntonio ya. but why $E$ over $F$ is an algebraic extension. To be Galois first of all it should be Algebraic no? – GA316 May 26 '14 at 18:33
• @PraphullaKoushik I am also having the same doubt. but if we know the extension is galois then they are equal. – GA316 May 26 '14 at 18:34
• @PraphullaKoushik why $[E:F]$ is finite? – GA316 May 26 '14 at 18:35

First we show that $E/F$ is finite, of degree less than or equal to $m = |G|$.

Let $\alpha \in E$. Let $\alpha=\alpha_1,\dots\alpha_r$ be the orbit of $\alpha$ under $G$, $r \leq m$. Then define $f(X) = \Pi(X-\alpha_i)$. $f$ is separable by definition and is fixed by every element of $G$, since any element of $G$ simply permutes the roots. Since $F = E^G$, this means $f \in F[X]$, so in particular $\alpha$ is algebraic over $F$. Also, the minimal polynomial of $\alpha$ over $F$ divides $f$, so is separable. Thus we have shown that $E/F$ is both algebraic and separable, and that for all $\alpha \in E$, $[F(\alpha):F]\leq m$.

Now let $\alpha$ be an element such that $[F(\alpha):F]$ is maximal. If $F(\alpha) \not = E$, take $\beta \in E\backslash F(\alpha)$. $E/F$ is separable, so by the primitive element theorem, $F(\alpha,\beta)= F(\gamma)$ for some $\gamma \in E$, with $[F(\gamma):F]>[F(\alpha):F]$. Contradiction.

Now all that remains to be shown is that $G = Aut(E/F)$. Certainly $G \subset Aut(E/F)$. But $E/F$ is finite, so $|Aut(E/F)|\leq[E:F]\leq|G|\leq|Aut(E/F)|$. Thus we have equivalence at each point in the equality, so in particular $G = Aut(E/F)$.

This is a great theorem, because it allows us to prove things like the fact that for in integral domain $R$, $R[X_1,\dots,X_n]/Sym_n$ where $Sym_n$ is the symmetric polynomials over $R$ with $n$ variables is finite and Galois. This is both hard to do otherwise, quite interesting, and useful in e.g. finding polynomials which have insoluble Galois groups. It also allows us to show that the characterisation of Galois extensions as those where $|E/F| = |Aut(E/F)|$ is equivalent to the definitions "$E/F$ is normal and separable" or "$E^{Aut(E/F)} = F$".

• This theorem is also true in much greater generality, in particular in any "galois category". For example, whenever you have a topological space X and a finite discrete group acting without fixed points on $X$, then the quotient map $X\rightarrow X/G$ is a galois covering map. – oxeimon May 26 '14 at 19:25
• @oxeimon Unfortunately I don't yet know any category theory, or what it means for a covering map to be Galois, but that sounds very interesting! Artin's theorem is actually one of my favourite theorems, so I was happy to see this question come up! I'll try and remember what you've said in the future! – Tom Oldfield May 26 '14 at 19:30
• @Tom_Oldfield a galois covering map $f : X\rightarrow Y$ is just a continuous map of topological spaces where for every $y\in Y$, there is a neighborhood $U$ such that $f^{-1}(U)$ is a disjoint union of copies of $U$. In particular, if $Y$ is connected, then the fibers of $f$ are all discrete, of size $n$. In this case $f$ is galois if the self homeomorphisms $\sigma$ of $X$ which "fix $Y$" in the sense that $f\circ\sigma = f$ act freely and transitively on the fibers (in fact you just need to check this for one particular fiber). – oxeimon May 26 '14 at 19:39
• @Tom_Oldfield The amazing thing about the theory of covering spaces is that it's a topological theory which in almost every respect is identical to traditional galois theory for perfect fields. Ie, every field extension $E/F$ can be thought of as a covering map $E$ over $F$. The absolute galois group of a field $F$ corresponds to the fundamental group "of $F$" (actually it's more like the profinite completion of the fundamental group). For simple extensions $F(\alpha)/F$, the roots of the minimal polynomial of $\alpha$ behave like the elements of the fiber of the corresponding covering space.. – oxeimon May 26 '14 at 19:43
• @oxeimon That's interesting, that's what I call a "regular" covering map, I know you can use a similar construction of a group acting on a topological space to identify fundamental groups. I have also seen what was called a "galois correspondence" between path connected (based) covering spaces modulo an equivalence if two spaces are homeomorphic by a map that commutes with the projections, and the subgroups of the (based) fundamental group of a space. I'm actually planning on doing a summer project soon where I hope to learn more about this kind of thing! – Tom Oldfield May 26 '14 at 19:45