Page 294 - DMTH403_ABSTRACT_ALGEBRA

Unit 31: The Galois Group of a Polynomial

(ii) Find automorphisms of F that have fixed fields Q( 2 ), Q(i), and Q( 2 i), respectively. Notes

Solution: These subfields of Q( 2 , i) are the splitting fields of x -2, x +1, and x +2, respectively.
2
2
2
Any automorphism must take roots to roots, so if is an automorphism of Q( 2 , i), we must
have ( 2 ) = ± 2 , and (i) = ± i. These possibilities must in fact define 4 automorphisms of the
splitting field.
If we define ( 2 ) = 2 and  (i) = -i, then the subfield fixed by  is Q( 2 ). If we define
1
1
1
 ( 2 ) = – 2 and  (i) = i, then the subfield fixed by  is Q(i). Finally, for  =   we have
2
1
2
2
3
2
 ( 2 ) = – 2 and (i) = –i, and thus  ( 2 i) = 2 i, so  has Q( 2 i) as its fixed subfield.
3
3
3
Example: Find the Galois groups of x – 2 over the fields Z and Z .
3
5
11
Solution: The polynomial is not irreducible over Z , since it factors as x -2 = (x+2)(x -2x-1). The
2
3
5
quadratic factor will have a splitting field of degree 2 over Z , so the Galois group is cyclic of
5
order 2.
A search in Z for roots of x -2 yields one and only one: x = 7. Then x -2 can be factored as x -2 =
3
3
3
11
(x-7)(x +7x+5), and the second factor must be irreducible. The splitting field has degree 2 over
2
Z , and can be described as Z [x] / < x +7x+5 >. Thus the Galois group is cyclic of order 2.
2
11
11
Example: Find the Galois group of x -1 over the field Z .
4
7
Solution: We first need to find the splitting field of x -1 over Z . We have x -1 = (x-1)(x+1)(x +1).
2
4
4
7
A quick check of ±2 and ±3 shows that they are not roots of x +1 over Z , so x +1 is irreducible
2
2
7
over Z . To obtain the splitting field we must adjoin a root of x +1, so we get a splitting field
2
7
Z [x] / < x +1 > of degree 2 over Z .
2
7
7
The Galois group of x -1 over Z is cyclic of order 2.
4
7
Example: Find the Galois group of x -2 over the field Z .
3
7
Solution: In this case, x -2 has no roots in Z , so it is irreducible. We first adjoin a root a of x -2 to
3
3
7
Z . The resulting extension Z (a) has degree 3 over Z , so it has 7 = 343 elements, and each
3
7
7
7
element is a root of the polynomial x -x. Let b> be a generator of the multiplicative group of
343
the extension. Then (b ) = b = 1, showing that Z (a) contains a non-trivial cube root of 1. It
114 3
342
7
follows that x -2 has three distinct roots in Z (a): a, ab , and ab , so therefore Z (a) is a splitting
3
114
228
7
7
field for x -2 over Z . Since the splitting field has degree 3 over Z , it follows the Galois group of
3
7
7
the polynomial is cyclic of order 3.
Self Assessment
1. Galois considered ................... of the roots that leave the coefficient field fixed.
(a) polynomial (b) permutation
(c) combination (d) range
2. The modern approach is to consider ................... determined by permutation.
(a) homomorphism (b) automorphism
(c) isomorphism (d) ideal and subfield
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