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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 1X n-1 + ... + 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 Ef of f over F is given by
E f = F( 1,..., n).
It is clear that Ef 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 GalF(f ), is the finite group Gal(Ef /F).
By Theorem 4 of the section on Galois group of an extension, we have
Proposition 2: The order of GalF(f ) is m.
A typical element of GalF(f ) is an F-automorphism of Ef. But since Ef 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 F-conjugate of i, so that ( i) is a zero of IrrF( i,X) (by Proposition 2 of the section on conjugates). But since f F[X] has i as a zero, IrrF( 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)
is a homomorphism. Moreover, since
is completely determined by ( 1),..., ( n), we see that is an injection. Therefore, we have proved:
Theorem 4: GalF(f ) is isomorphic to a subgroup of the group of permutations of the zeros 1,..., n.
Corollary 5: GalF(f ) is isomorphic to a subgroup of Sn.
Corollary 6: GalF(f ) has order of at most n!
Note that the isomorphism of (1) will not generally be surjective, so that GalF (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 1-4, let F = Q
Example 1: f = X2 - 2. Then Ef = Q( and GalF(f ) consists of the mappings
 ,  .
Thus, GalF(f ) Z2.
Example 2: f = (X2 - 2)(X2 - 3). Here Ef = Q( , ) and 1 = , 2 = - , 3 = , 4 = - ,
Then GalF(f ) consists of the mappings
 ,  ,
 ,  .
Moreover, G1 = { 0, 1} and G2 = { 0, 2} are subgroups of GalF(f ) isomorphic to Z2 and
Example 3: f = X3 - 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 GalQ( f ) S3, so that all permutations of the roots 1, 2, 3 correspond to a Galois group. By Theorem 4,
GalQ( f ) is isomorphic to a subgroup of S3. Moreover, Ef/Q is a Galois extension, so that by Theorem 4 of the section the Galois group of an extension, the order of GalQ( f ) equals deg(Ef /Q). Therefore, it suffices to show that deg(Ef /Q) > 6. and this was proved in Example 3 of the section of examples of splitting fields.
Example 4: Let f = X5 - 6X + 3. We will show that GalQ( f ) S5. Let us begin by considering the function f(x) = x5 - 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) = x5 - 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 = 
= 
= 15 - 6 1 +3
= f( 1).
Therefore, 2 = 1 is a zero of f and Ef = Q( 1, 2, 3, 4, 5).
The mapping : Ef Ef defined by ( ) = is a Q-automorphism of Ef and therefore GalQ( f ). Moreover,
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 GalQ( 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(Ef /Q) is divisible by 5. Therefore, by Proposition 2 and Theorem 4, GalQ( f ) is a subgroup of the group of permutations of 1,... 5 and that the order of GalQ( f ) is divisible by 5. Moreover, by the first Sylow Theorem, GalQ( f ) contains a subgroup H of order 5. In particular, GalQ( 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 5-cycle, say
 = (1i 2i 3i 4i 5).
By replacing by some power of , we may assume has the form
 = (12i 3i 4i 5).
However, by permuting and renumbering the 3, 4, 5, we may assume that
 = (12345).
Therefore, we have shown that GalQ( 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, GalQ( f ) contains the transpositions (12), (23), (34), (45), and (51). And these transpositions generate all transpositions. Therefore, since S5 is generated by transpositions, GalQ( f ) = S5.
Example 5: Let be a primitive mth root of 1 and let F be a field containing . Let a F, f = Xm - 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 m-1,
so that f = (X - b)...(X - b m-1) in F[X], and E = F(b) since F. If GalF( f ), then there exists a unique integer a( ) such that
 (b) = b a( ), 0 < a(  ) < m - 1.
Let : GalF( f ) Zm be defined by
Then is an injection. Moreover,
=  (b) · a( ) (since  F and  is an F-automorphism of E)
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 GalF( f ) is isomorphic to a subgroup of Zm. In particular, since Zm is a cyclic group, GalF( f ) is cyclic. It is not usually true that GalF( f ) Zm.
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