Galois group should not be confused with Galois fields.
In mathematics, in the area of abstract algebra known as Galois theory, the Galois group of a certain type of field extension is a specific group associated with the field extension. The study of field extensions and their relationship to the polynomials that give rise to them via Galois groups is called Galois theory, so named in honor of Évariste Galois who first discovered them.
For a more elementary discussion of Galois groups in terms of permutation groups, see the article on Galois theory.
Suppose that
E
F
E/F
E/F
E
F
E/F
\alpha:E\toE
\alpha(x)=x
x\inF
E/F
\operatorname{Aut}(E/F).
If
E/F
\operatorname{Aut}(E/F)
E/F
\operatorname{Gal}(E/F)
If
E/F
E/F
\operatorname{Aut}(K/F)
K
E
Another definition of the Galois group comes from the Galois group of a polynomial
f\inF[x]
K/F
f
f(x)=(x-\alpha1) … (x-\alphak)\inK[x]
over the field
K
f
K/F
K
One of the important structure theorems from Galois theory comes from the fundamental theorem of Galois theory. This states that given a finite Galois extension
K/k
k\subsetE\subsetK
H\subsetG.
E
K
H
E=KH=\{a\inK:ga=awhereg\inH\}
Moreover, if
H
G/H\cong\operatorname{Gal}(E/k)
E/k
\operatorname{Gal}(K/k)
Suppose
K1,K2
k
G1,G2.
K1K2
G=\operatorname{Gal}(K1K2/k)
G\toG1 x G2
K1\capK2=k
As a corollary, this can be inducted finitely many times. Given Galois extensions
K1,\ldots,Kn/k
Ki+1\cap(K1 … Ki)=k,
\operatorname{Gal}(K1 … Kn/k)\cong\operatorname{Gal}(K1/k) x … x \operatorname{Gal}(Kn/k).
In the following examples
F
\Complex,\R,\Q
One of the basic propositions required for completely determining the Galois groups[2] of a finite field extension is the following: Given a polynomial
f(x)\inF[x]
E/F
\left|\operatorname{Gal}(E/F)\right|=[E:F]
A useful tool for determining the Galois group of a polynomial comes from Eisenstein's criterion. If a polynomial
f\inF[x]
f=f1 … fk
f
fi
f
fi.
\operatorname{Gal}(F/F)
Another example of a Galois group which is trivial is
\operatorname{Aut}(\R/\Q).
\R
Consider the field
K=\Q(\sqrt[3]{2}).
\operatorname{Aut}(K/\Q)
K
2
\exp\left(\tfrac{2\pii}{3}\right)\sqrt[3]{2}
\exp\left(\tfrac{4\pii}{3}\right)\sqrt[3]{2},
are missing from the extension—in other words is not a splitting field.
The Galois group
\operatorname{Gal}(\Complex/\R)
The degree two field extension
\Q(\sqrt{2})/\Q
\operatorname{Gal}(\Q(\sqrt{2})/\Q)
\sigma
\sqrt2
-\sqrt2
p\in\N.
Using the lattice structure of Galois groups, for non-equal prime numbers
p1,\ldots,pk
\Q\left(\sqrt{p1},\ldots,\sqrt{pk}\right)/\Q
\operatorname{Gal}\left(\Q(\sqrt{p1},\ldots,\sqrt{pk})/\Q\right)\cong\operatorname{Gal}\left(\Q(\sqrt{p1})/\Q\right) x … x \operatorname{Gal}\left(\Q(\sqrt{pk})/\Q\right)\cong(\Z/2\Z)k
Another useful class of examples comes from the splitting fields of cyclotomic polynomials. These are polynomials
\Phin
\Phin(x)=\prod\begin{matrix1\leqk\leqn\ \gcd(k,n)=1\end{matrix}}
| ||||
\left(x-e |
\right)
whose degree is
\phi(n)
n
\Q
\Q(\zetan)
\sigmaa
\zetan\mapsto
a | |
\zeta | |
n |
1\leqa<n
n
n=
a1 | |
p | |
1 |
…
ak | |
p | |
k |
,
\operatorname{Gal}(\Q(\zetan)/\Q)\cong
\prod | |
ai |
\operatorname{Gal}\left
(\Q(\zeta | |||||||
|
)/\Q\right)
If
n
p
\operatorname{Gal}(\Q(\zetap)/\Q)\cong\Z/(p-1)\Z
In fact, any finite abelian group can be found as the Galois group of some subfield of a cyclotomic field extension by the Kronecker–Weber theorem.
Another useful class of examples of Galois groups with finite abelian groups comes from finite fields. If is a prime power, and if
F=Fq
E=F | |
qn |
q
qn
\operatorname{Gal}(E/F)
The field extension
\Q(\sqrt{2},\sqrt{3})/\Q
4
\sigma,\tau
\sigma(\sqrt{2})=-\sqrt{2}
\tau(\sqrt{3})=-\sqrt{3}.
4
Another example is given from the splitting field
E/\Q
f(x)=x4+x3+x2+x+1
Note because
(x-1)f(x)=x5-1,
f(x)
\exp\left(\tfrac{2k\pii}{5}\right).
\begin{cases}\sigmal:E\toE\ \exp\left(
2\pii | |
5 |
\right)\mapsto\left(\exp\left(
2\pii | |
5 |
\right)\right)l\end{cases}
generating a group of order
4
\sigma2
\Z/4\Z
Consider now
L=\Q(\sqrt[3]{2},\omega),
\omega
\operatorname{Gal}(L/\Q)
x3-2
\Q.
The Quaternion group can be found as the Galois group of a field extension of
\Q
\Q\left(\sqrt{2},\sqrt{3},\sqrt{(2+\sqrt{2})(3+\sqrt{3})}\right)
has the prescribed Galois group.[6]
If
f
p
f
Sp.
For example,
f(x)=x5-4x+2\in\Q[x]
f
S5
Given a global field extension
K/k
Q(\sqrt[5]{3},\zeta5)/Q
w
K
p
v
k
of local fields, there is an induced action of the Galois groupKw/kv
G=\operatorname{Gal}(K/k)
s\inG
Since we have taken the hypothesis thatsw:Kw\toKsw
w
v
Kw/kv
sw
kv
G
w
then there is a surjection of the global Galois group to the local Galois group such that there is an isomorphism between the local Galois group and the isotropy subgroup. Diagrammatically, this meansGw=\{s\inG:sw=w\}
where the vertical arrows are isomorphisms.[8] This gives a technique for constructing Galois groups of local fields using global Galois groups.\begin{matrix} \operatorname{Gal}(K/v)&\twoheadrightarrow&\operatorname{Gal}(Kw/kv)\\ \downarrow&&\downarrow\\ G&\twoheadrightarrow&Gw \end{matrix}
A basic example of a field extension with an infinite group of automorphisms is
\operatorname{Aut}(\Complex/\Q)
E/\Q
\Q(\sqrt{a})/\Q
a\in\Q
2
\operatorname{Aut}(\Complex/\Q).
One of the most studied classes of infinite Galois group is the absolute Galois group, which is an infinite, profinite group defined as the inverse limit of all finite Galois extensions
E/F
\operatorname{Gal}(\overline{F}/F):=\varprojlimE/F{\operatorname{Gal}(E/F)}
where
\overline{F}
F
\operatorname{Gal}(\overline{\Q}/\Q)
\operatorname{Gal}(\overline{F
Another readily computable example comes from the field extension
\Q(\sqrt{2},\sqrt{3},\sqrt{5},\ldots)/\Q
\operatorname{Gal}(\Q(\sqrt{2},\sqrt{3},\sqrt{5},\ldots)/\Q)\cong\prodp\Z/2
which can be deduced from the profinite limit
… \to\operatorname{Gal}(\Q(\sqrt{2},\sqrt{3},\sqrt{5})/\Q)\to\operatorname{Gal}(\Q(\sqrt{2},\sqrt{3})/\Q)\to\operatorname{Gal}(\Q(\sqrt{2})/\Q)
and using the computation of the Galois groups.
The significance of an extension being Galois is that it obeys the fundamental theorem of Galois theory: the closed (with respect to the Krull topology) subgroups of the Galois group correspond to the intermediate fields of the field extension.
If
E/F
\operatorname{Gal}(E/F)
\operatorname{Aut}(E/F)
E/F
\Q(\sqrt{2},\sqrt{3})=\Q ⊕ \Q ⋅ \sqrt{2} ⊕ \Q ⋅ \sqrt{3} ⊕ \Q ⋅ \sqrt{6}
\Q