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§ 8.1 The Galois Group of a Polynomial: solved problems

7. Determine the group of all automorphisms of a field with 4 elements.

Solution: The automorphism group consists of two elements: the identity mapping and the Frobenius automorphism.

Read on only if you need more detail. By Theorem 6.5.2 and Theorem 6.4.5, up to isomorphism there is only one field with 4 elements. Since x2+x+1 is irreducible over Z2, by Theorem 4.4.6 this field with 4 elements can be constructed as F = Z2[x] / < x2+x+1 >. Letting a be the coset of x, we have F = {0, 1, a, 1+a}. Any automorphism of F must leave 0 and 1 fixed, so the only possibility for an automorphism other than the identity is to interchange a and 1+a. Is this an automorphism? Since x2+x+1 0, we have x2 -x-1 x+1, so a2 = 1+a and (1+a)2 = 1+2a+a2 = a. Thus the function that fixes 0 and 1 while interchanging a and 1+a is in fact the Frobenius automorphism of F.

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8. Let F be the splitting field in C of x4+1.

(i) Show that [F:Q] = 4.

Solution: The polynomial x8-1 factors over Q as x8-1 = (x4-1)(x4+1) = (x-1)(x+1)(x2+1)(x4 +1). The factor x4 +1 is irreducible over Q by Eisenstein's criterion. The roots of x4+1 are thus the primitive 8th roots of unity, ± / 2 ± / 2 i, and adjoining one of these roots also gives the others, together with i. Thus the splitting field is obtained in one step, by adjoining one root of x4+1, so its degree over Q is 4.

It is clear that the splitting field can also be obtained by adjoining first and then i, so it can also be expressed as Q(, i).

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

Solution: These subfields of Q(, i) are the splitting fields of x2-2, x2+1, and x2+2, respectively. Any automorphism must take roots to roots, so if is an automorphism of Q(, i), we must have () = ± , and (i) = ± i. These possibilities must in fact define 4 automorphisms of the splitting field.

If we define 1 () = and 1 (i) = -i, then the subfield fixed by 1 is Q(). If we define 2 () = - and 2 (i) = i, then the subfield fixed by 2 is Q(i). Finally, for 3 = 2 1 we have 3 () = - and (i) = -i, and thus 3 ( i) = i, so 3 has Q( i) as its fixed subfield.

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9. Find the Galois group over Q of the polynomial x4+4.

Solution: Problem 6.4.1 shows that the splitting field of the polynomial has degree 2 over Q, and so the Galois group must be cyclic of order 2.

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10. Find the Galois groups of x3-2 over the fields Z5 and Z11.

Solution: The polynomial is not irreducible over Z5, since it factors as x3-2 = (x+2)(x2-2x-1). The quadratic factor will have a splitting field of degree 2 over Z5, so the Galois group is cyclic of order 2.

A search in Z11 for roots of x3-2 yields one and only one: x = 7. Then x3-2 can be factored as x3-2 = (x-7)(x2+7x+5), and the second factor must be irreducible. The splitting field has degree 2 over Z11, and can be described as Z11[x] / < x2+7x+5 >. Thus the Galois group is cyclic of order 2.

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11. Find the Galois group of x4-1 over the field Z7.

Solution: We first need to find the the splitting field of x4-1 over Z7. We have x4-1 = (x-1)(x+1)(x2+1). A quick check of ±2 and ±3 shows that they are not roots of x2+1 over Z7, so x2+1 is irreducible over Z7. To obtain the splitting field we must adjoin a root of x2+1, so we get a splitting field Z7[x] / < x2+1 > of degree 2 over Z7.

It follows from Corollary 8.1.7 that the Galois group of x4-1 over Z7 is cyclic of order 2.

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12. Find the Galois group of x3-2 over the field Z7.

Solution: In this case, x3-2 has no roots in Z7, so it is irreducible. We first adjoin a root a of x3-2 to Z7. The resulting extension Z7(a) has degree 3 over Z7, so it has 73 = 343 elements, and each element is a root of the polynomial x343-x. Let b> be a generator of the multiplicative group of the extension. Then (b114)3 = b342 = 1, showing that Z7(a) contains a nontrivial cube root of 1. It follows that x3-2 has three distinct roots in Z7(a): a, ab114, and ab228, so therefore Z7(a) is a splitting field for x3-2 over Z7. Since the splitting field has degree 3 over Z7, it follows from Corollary 8.1.7 that the Galois group of the polynomial is cyclic of order 3.


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