
The Galois Group of a Polynomial
In the introduction we promised that we would associate to a polynomial a finite group which reflects the algebraic properties of the polynomial. We have now reached the point where that promise is very simple to fulfill.
Let F be a field and let
f = X ^{n} + a _{1}X ^{n1} + ... + a _{n} F[X]
be a nonconstant polynomial. Let us fix an algebraic closure F of F. Then, in F[X], we can write
f = (X  _{i})
where _{1},...,_{n} are the zeros of f in F. By Theorem 1 of the section on the restrictive assumption, _{1},...,_{n} are all distinct. The splitting field E_{f} of f over F is given by
E _{f} = F( _{1},..., _{n}).
It is clear that E_{f} is a normal extension of F and is therefore a Galois extension of degree m, say.
Definition 1: The Galois group of f over F denoted Gal_{F}(f ), is the finite group Gal(E_{f }/F).
By Theorem 4 of the section on Galois group of an extension, we have
Proposition 2: The order of Gal_{F}(f ) is m.
A typical element of Gal_{F}(f ) is an Fautomorphism of E_{f}. But since E_{f} is generated over F by _{1},...,_{n}, is completely determined once (_{i}) (i = 1,...,n) is specified. Let us try to pin down what (_{i}) can be.
Lemma 3: (1) The image of _{i} is an _{j} for some j depending on i. In other words (_{i}) = _{j(i)} for some j(i) (1 < j(i) < n),
(2) If _{j(i)} = _{j(i')}, then i = i'.
Proof: (1) Clearly (_{i}) equals some Fconjugate of _{i}, so that (_{i}) is a zero of Irr_{F}(_{i},X) (by Proposition 2 of the section on conjugates). But since f F[X] has _{i} as a zero, Irr_{F}(_{i},X)f. Therefore, (_{i}) is a zero of f and (_{i}) = _{j(i)} for some j(i) (1 < j(i) < n).
(2) If _{j(i)} = _{j(i')}, then (_{i}) = (_{i'}), But since is an injection, this implies that _{i} = _{i'}. Therefore, i = i', since all _{i} are distinct.
Lemma 3 asserts that the set {(_{1}),...,(_{n})} coincides with the set {_{1},...,_{n}}. For, indeed {(_{1}),...,(_{n})} {_{1},...,_{n}} by (1); and {(_{1}),...,(_{n})} contains n distinct elements by (2). In other words,
P _{} =
is a permutation of the n distinct zeros {_{1},...,_{n}}. It is trivial to verify that the mapping
: Gal _{F}( f ) S _{n},
(1)
( ) = P _{}
is a homomorphism. Moreover, since
is completely determined by (_{1}),...,(_{n}), we see that is an injection. Therefore, we have proved:
Theorem 4: Gal_{F}(f ) is isomorphic to a subgroup of the group of permutations of the zeros _{1},...,_{n}.
Corollary 5: Gal_{F}(f ) is isomorphic to a subgroup of S_{n}.
Corollary 6: Gal_{F}(f ) has order of at most n!
Note that the isomorphism of (1) will not generally be surjective, so that Gal_{F} (f ) will not generally contain all the permutations on _{1},...,_{n}. The reader is referred to the examples below for instances of this phenomenon. In Examples 14, let F = Q
Example 1: f = X^{2}  2. Then E_{f} = Q( and Gal_{F}(f ) consists of the mappings
, .
Thus, Gal_{F}(f ) Z_{2}.
Example 2: f = (X^{2}  2)(X^{2}  3). Here E_{f} = Q(,) and _{1} = , _{2} =  , _{3} = , _{4} =  ,
Then Gal_{F}(f ) consists of the mappings
, ,
, .
Moreover, G_{1} = {_{0},_{1}} and G_{2} = {_{0},_{2}} are subgroups of Gal_{F}(f ) isomorphic to Z_{2} and
Example 3: f = X^{3}  2. Then _{1} = , _{2} = , _{3} = ^{2}, where denotes a real cube root of 2 and is a primitive cube root of unity. Let us show that Gal_{Q}( f ) S_{3}, so that all permutations of the roots _{1}, _{2}, _{3} correspond to a Galois group. By Theorem 4,
Gal_{Q}( f ) is isomorphic to a subgroup of S_{3}. Moreover, E_{f}/Q is a Galois extension, so that by Theorem 4 of the section the Galois group of an extension, the order of Gal_{Q}( f ) equals deg(E_{f }/Q). Therefore, it suffices to show that deg(E_{f }/Q) > 6. and this was proved in Example 3 of the section of examples of splitting fields.
Example 4: Let f = X^{5}  6X + 3. We will show that Gal_{Q}( f ) S_{5}. Let us begin by considering the function f(x) = x^{5}  6x + 3 for real x. From elementary calculus, we see that this function has a maximum for x = (6/5)^{1/4} and a minimum at (6/5)^{1/4}. Moreover, these are the only extrema of f(x). Thus, we may sketch the graph of f(x) as in figure 1, and wee see immediately that f has exactly three real zeros, _{3}, _{4}, and _{5}. Let _{1} = a + b C be a nonreal zero of f.
Figure 1: Graph of f(x) = x^{5}  6x + 3.
If = x + y (x,y R) is a complex number, let us define the complex conjugate of of by
= x  y .
The mapping is an automorphism of C. From this fact it follows that
0 = 0 =
=
= _{1}^{5}  6 _{1} +3
= f( _{1}).
Therefore, _{2} = _{1} is a zero of f and E_{f} = Q(_{1},_{2},_{3},_{4},_{5}).
The mapping : E_{f}E_{f} defined by () = is a Qautomorphism of E_{f} and therefore Gal_{Q}( f ). Moreover,
( _{i}) = _{i}, (i = 3,4,5)
Therefore, when is viewed as a permutation, is just the permutation (12), which interchanges _{1} and _{2} and leaves all other _{i} fixed. Therefore, we have located one specific element of Gal_{Q}( f ).
By the Eisenstein irreducibility criterion, f is irreducible in Q[X], so that deg(Q(_{1})/Q) = 5 by Theorem 3 of the section on algebraic numbers. Therefore, since
deg(E _{f }/ Q) = deg(E _{f }/ Q( _{1}))·deg( Q( _{1})/ Q),
we see that deg(E_{f} /Q) is divisible by 5. Therefore, by Proposition 2 and Theorem 4, Gal_{Q}( f ) is a subgroup of the group of permutations of _{1},..._{5} and that the order of Gal_{Q}( f ) is divisible by 5. Moreover, by the first Sylow Theorem, Gal_{Q}( f ) contains a subgroup H of order 5. In particular, Gal_{Q}( f ) contains an element of order 5. Let
= (i _{1}...i _{r})(j _{1}...j _{s})(k _{1}...k _{t})...
be the decomposition of into a product of disjoint cycles. The order of is the least common multiple of r,s,t, ..., Therefore, since has order 5, must be a 5cycle, say
= (1i _{2}i _{3}i _{4}i _{5}).
By replacing by some power of , we may assume has the form
= (12i _{3}i _{4}i _{5}).
However, by permuting and renumbering the _{3},_{4},_{5}, we may assume that
= (12345).
Therefore, we have shown that Gal_{Q}( f ) contains the permutations (12), (12345). However, it is easy to check that for r = 0,1,2,3,
(12345)^{r}(12)(12345)^{r} = (r + 1, r + 2),
(12345)^{4}(12)(12345)^{4} = (51).
Therefore, Gal_{Q}( f ) contains the transpositions (12), (23), (34), (45), and (51). And these transpositions generate all transpositions. Therefore, since S_{5} is generated by transpositions, Gal_{Q}( f ) = S_{5}.
Example 5: Let be a primitive mth root of 1 and let F be a field containing . Let a F, f = X^{m}  a F[X], and let E = the splitting field of f over F. If b is a zero of f in F, then the zeros of f ar given by
b,b ,...,b ^{m1},
so that f = (X  b)...(X  b^{m1}) in F[X], and E = F(b) since F. If Gal_{F}( f ), then there exists a unique integer a() such that
(b) = b ^{a()}, 0 < a( ) < m  1.
Let : Gal_{F}( f ) Z_{m} be defined by
Then is an injection. Moreover,
= (b) · ^{a()} (since F and is an Fautomorphism of E)
= b ^{a()+b()}.
On the other hand, ()(b) = b^{a()+b()}, so that ^{a()+b()a()} = 1 and a() +b()  a() 0 (mod m). Thus,
and is an isomorphism. We have proved that Gal_{F}( f ) is isomorphic to a subgroup of Z_{m}. In particular, since Z_{m} is a cyclic group, Gal_{F}( f ) is cyclic. It is not usually true that Gal_{F}( f ) Z_{m}.
