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### Dihedral quintic fields with a power basis Lavallee, Melisa Jean 2008

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```DIHEDRAL QUINTIC FIELDS WITH A POWER BASIS by Melisa Jean Lavallee B.Sc., The University of British Columbia, 2006 A THESIS SUBMITTED IN PARTIAL FULFILLMENT OF THE REQUIREMENTS FOR THE DEGREE OF MASTER OF SCIENCE in The College of Graduate Studies (Interdisciplinary Studies) THE UNIVERSITY OF BRITISH COLUMBIA (Okanagan) August 2008 c
1;
2; : : : ;
n) where
1;
2; : : : ;
n are algebraic numbers. That is, an algebraic number eld is a eld K obtained from the eld of rational numbers Q by adjoining nitely many algebraic numbers
1;
2; : : : ;
n. This eld is the smallest eld that contains Q and the algebraic numbers
1;
2; : : : ;
n. Example 1.6. Q( p 2; p 3) is an algebraic number eld as both p 2 and p 3 are algebraic numbers. Now we will state a classic theorem known as the Primitive Element Theorem to simplify the representation of elds like Q( p 2; p 3). The proof can be found on page 595, [9]. Theorem 1.3. If K = Q(
1;
2; : : : ;
n) is an algebraic number eld then there exists an algebraic number  such that K = Q(). This means that a composite eld can be expressed in the form Q(), where  is an algebraic number, and Q() is an algebraic number eld formed by constructing all possible numbers from  and the set of rationals using the operators +;;;. Example 1.7. To demonstrate the Primitive Element Theorem, we will show that Q( p 2; p 3) = Q() where  is the algebraic number p 2 + p 3. By Example, 1.3,  is an algebraic number with minimal polynomial f(x) = x4  10x2 + 1 2 Q[x]. Now, Q()  Q(p2;p3) as p2 +p3 2 Q(p2;p3) . Now, p 2 = 9 2 + 1 2 3 2 Q() and p3 = 11 2  1 2 3 2 Q() so that Q(p2;p3)  Q(): Thus Q() = Q( p 2; p 3) and so the theorem holds. Remark 1.1. The expression K = Q() is not unique as Q() = Q() = Q(1 + ) = :::etc. 9 1.3.1. Representation of Algebraic Number Fields We can also view K = Q() as an n-dimensional vector space over Q as seen in the following result (Theorem 6.1.3, pg. 110, [2]). Theorem 1.4. Let K = Q() be an algebraic number eld, where  is an algebraic number. Let A(x) be the minimal polynomial of  with degree n. Then every element of K is uniquely expressible in the form a0 + a1 + : : : + an1 n1, where a0; : : : ; an1 2 Q. Thus, K =  a0 + a1 + : : :+ an1 n1 j a0; : : : ; an1 2 Q  . Denition 1.9. K = Q() is an n-dimensional vector space over Q with basis 1; ; : : : ; n1  and the degree of K over Q is n. We denote the degree of K over Q as n = [K : Q]. Algebraic number elds are classied according to their degree. So K is called a quadratic eld if n = 2, cubic if n = 3, quartic if n = 4, quintic if n = 5, etc. We can also term algebraic number elds as being pure or non-pure. Denition 1.10. K = Q() is pure if  is a root of the polynomial xn  a where a 2 Q and n is the degree of the minimal polynomial of . If K is not pure, then we say K is non-pure. Algebraic number elds can also be classied using Galois theory. We will study this in detail in the following chapter. 10 Chapter 2 Introduction to Galois Theory We begin this chapter with some basic denitions that will be used to establish the foundation of Galois Theory. Denition 2.1. An isomorphism  of a eld K with itself is called an automor- phism of K. The collection of automorphisms of K is denoted Aut(K). Notice that every eld will have at least one automorphism - the identity map. Denition 2.2. An automorphism  2 Aut(K) is said to x an element  2 K if () = . If F  K then an automorphism  is said to x F if it xes all the elements of F . That is, (a) = a, 8a 2 F . Denition 2.3. Let K be an algebraic number eld. Let AutQ(K) be the collec- tion of automorphisms of K which x Q. Since Q  K, any automorphism  of K xes Q. This is because  xes 1, the identity, and Q can be built up from 1 using +;; and . Hence, Aut(K) = AutQ(K): IfK is not a eld extension of Q then AutQ(K) is a subgroup of the group Aut(K). Denition 2.4. An algebraic number eld K is separable if for each algebraic number  2 K, the minimal polynomial of  over K has no repeated roots. Denition 2.5. An algebraic number eld K is normal if for each algebraic num- ber  2 K, the minimal polynomial of  over K factors entirely in K[x]. That is, it completely factors into linear factors over K. Denition 2.6. An algebraic number eld is Galois if it is both separable and normal. The group AutQ(K) is called the Galois group of K and is denoted as Gal(K=Q). 11 2.1. Galois Groups of Polynomials - Overview Denition 2.7. Consider a polynomial f(x) 2 Q[x]. The splitting eld L of the polynomial f(x) over Q is the smallest subeld of C (set of complex numbers) containing Q and all the roots of f(x). Denition 2.8. A polynomial f(x) over Q is called separable if it has no multiple roots (i.e. all its roots are distinct). We now state an important property of the splitting eld of a separable polynomial (Corollary 6, pg. 562, [9]). Corollary 2.1. If L is the splitting eld over Q of a separable polynomial f(x) then L is Galois. Because L is Galois, we are able to nd its Galois group. This leads us to the following denition. Denition 2.9. If f(x) is a separable polynomial over Q, then the Galois group of f(x) over Q is the Galois group of the splitting eld of f(x) over Q and is denoted as Gal(f(x)). Now it is important to note that the above corollary implies the splitting eld of any polynomial over Q is Galois. Why? Clearly, the splitting eld of f(x) is the same as the splitting eld of the product of the irreducible factors of f(x) (i.e. the polynomial obtained by removing multiple factors). Now this latter polynomial is separable as it is the product of distinct irreducible polynomials (and so the factors never have zeros in common). Then using the above corollary on this polynomial, we get the result. From this, we can also conclude that every polynomial over Q has a Galois group. Now we consider Galois theory from a permutational perspective. Given a poly- nomial, we wish to nd its Galois group. First we introduce a short denition. Denition 2.10. An algebraic equation is one of the form P = 0 where P is a polynomial with coe¢ cients in Q. 12 The main idea of Galois theory is to consider those permutations of the roots of our polynomial having the property that any algebraic equation satised by the roots is still satised after the roots have been permuted. These permutations form a group and is called the Galois group of the polynomial over Q. Thus, the elements of Gal(f(x)) are permutations that act on the roots of f(x). So, Gal(f(x)) is a subgroup of Sn - the group of all permutations of the set f1; 2; : : : ; ng. Now, we say subgroup because we may not have all permutations of the roots of f(x) as elements of Gal(f(x)). So jGal(f(x))j j jSnj: How does this correspond to the splitting eld of the polynomial f(x)? If f(x) is a separable polynomial over Q and if L denotes its splitting eld over Q, then we know that Gal(f(x)) = Gal(L=Q). But Gal(L=Q) consists of automorphisms from L to itself that x Q. Thus, any permutation of the roots which respects algebraic equations as described above gives rise to an automorphism of L over Q, and vice versa. That is, the automorphisms of L correspond to elements of Gal(f(x)): We will now do a small example to illustrate the above concept. This example parallels the "second example" in [33]. Example 2.1. Consider the polynomial x4  5x2 + 6 = (x2  3)(x2  2). We wish to describe the Galois group of this polynomial over the eld of rational numbers. The polynomial has four roots: (1) A = p 2 (2) B = p2 (3) C = p 3 (4) D = p3 There are 24 possible ways to permute these four roots, but not all of these permu- tations are members of the Galois group. The members of the Galois group must preserve any algebraic equation with rational coe¢ cients involving A, B, C and D. One such equation is: A+B = 0: 13 Therefore the permutation (A;B;C;D) ! (A;C;B;D) is not permitted because it transforms the valid equation A+B = 0 into the equation A+C = 0, p2+p3 = 0, which is invalid. Another equation that the roots satisfy is (A+B)2 + (C +D)2 = 0. This will exclude further permutations, such as (A;B;C;D)! (A;C;B;D): Continuing in this way, we nd that the only permutations satisfying both equations simultaneously are: (1) (A;B;C;D)! (A;B;C;D) (trivial permutation) (2) (A;B;C;D)! (B;A;C;D) (3) (A;B;C;D)! (A;B;D;C) (4) (A;B;C;D)! (B;A;D;C). Thus, the Gal(f(x)) = V = Z2  Z2 (the Klein 4-group). Now showing the above four permutations satisfy every possible algebraic equa- tion satised by A;B;C and D is less obvious and requires the theory of symmetric polynomials. Also, as we increase the number of roots of our given polynomial, this method becomes less e¢ cient. We can see that calculating the Galois group of a polynomial is a nontrivial task. However, with technology, the task no longer seems daunting. For example, MAPLETM will calculate the Galois group of a polynomial up to degree 9. We use the galois() command to do so. We now introduce an important property needed in order to examine the Galois group of polynomials, in particular, those of degree 3. For proof, see Theorem 14, pp. 523-524, [9]. Proposition 2.1. The multiplicative property of elds states that if E  L  F where E, L and F are elds and "" denotes "subeld", then [F : E] = [F : L]  [L : E]. Recall [F : E] denotes the degree of F over E. The next corollary is important when we examine the Galois group of polynomials. We refer the reader to pp. 51 of [3] for more details and the proof. 14 Corollary 2.2. Let K be a nite extension of the eld Q. If K is Galois then jAutQ(K)j = [K : Q]: As a result, we know that AutQ(L) = Gal(L=Q) and since L is Galois (Corollary 2.1), [L : Q] = jGal(L=Q)j . Lastly, we now present a proposition (Proposition 34, pp. 611, [9]) that will be encompassed in the theorems presented in these next sections. Proposition 2.2. The Galois group of f(x) 2 Q[x] is a subgroup of An if and only if the discriminant D(f) is a square in Q. 2.2. Irreducible Polynomials Denition 2.11. A polynomial p(x) in Q[x] is called irreducible over Q if it is non-constant and cannot be represented as the product of two or more non-constant polynomials from Q[x]. A useful test for irreducibility is Eisensteins Criterion. Proposition 2.3. Suppose we have the following polynomial with integer coe¢ - cients: f(x) = anx n + an1xn1 + : : :+ a1x+ a0. Assume that there exists a prime number p such that  p divides each ai for i 6= n  p does not divide an  p2 does not divide a0. Then f(x) is irreducible over Q. Proof. Suppose f(x) = anxn + an1xn1 + : : : + a1x + a0 2 Z[x] satises the above conditions for some prime p. By way of contradiction, assume f is reducible over Q: That is, f(x) = g(x)  h(x) 15 where g(x) and h(x) are non-constant polynomials over Q. The reduction map mod p is given by ' : Z[x] ! Zp[x] and is a homomorphism. Thus, '(f) = '(g)  '(h): But '(f) = '(anx n + an1xn1 + : : :+ a1x+ a0) = xn as p divides each ai for i 6= n and where an  (mod p). So '(g)  '(h) = xn and so '(g) = xk and '(h) =
xnk for some 0 < k < n, where  
=  1 2 3 3 2 1 ! = (1 3)  =  1 2 3 2 1 3 ! = (1 2): We will now show the multiplication table for S3 (Table 2.3). Table 2.3. Group Table for the Symmetric Group of Order 6  
  
        

 

 
   Note that
=  and  = 2 (recall for permutations you read right to left). Now we want to nd all the possible subgroups of S3. Since the order of S3 is 6, any subgroup must have order dividing 6. Thus, the possible 44 orders for the subgroups of S3 are 1; 2; 3 and 6: The subgroup of order 1 is the trivial subgroup fg. Remembering that each subgroup has to contain the identity permutation  and must be closed under function composition, we get the following subgroups of S3: hi = fg = 1 hi = f; ; g = hi hi = f; g , 2 = f; 2g h
i = f;
g , hi = f; g hi = f; g h; i = f; ; ; ;

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