THE STRUCTURE OF UNIT GROUPS DANIEL KLINE Abstract. Since the time of ancient Greece and India, Pell's equation, a2 − 2 db = 1, has been studied and its solutions sought after. Thesep solutions are significant, because they correspond to the units of the rings Z[ d]. Along with prime numbers, units are vital in understanding the structure and especially the factorization within different rings. Dirichlect's Unit Theorem establishes the structure of the units of a number field as a finite abelian group. In addition to describing the nature of units, Dirichlect's Unit Theorem represents the interconnectedness of geometry (namely Minkowski's Theorem), algebra, and number theory. Contents p 1. Units of Z[ d] 1 2. Minkowski's Theorem in the Plane 3 3. Minkowski's Theorem in General 5 4. Basic Properties of Number Fields 6 5. Algebraic Integers 9 6. Real and Imaginary Embeddings 10 7. Dirichlet's Unit Theorem 10 Acknowledgments 14 References 14 p 1. Units of Z[ d] Definition 1.1. Given a ring R, u 2 R is a unit if there exists v 2 R such that u · v = 1. We denote the set of units of a ring R as R∗. Example 1.2. Consider the ring Z. It is easy to check that Z∗ = {±1g. Units are not generally as obvious as inp the case of Z. Throughout this section, we will considerp the ringsp of the form Z[ d] where d is a square-free integer. We define Z[ d] := fa + b d j a; b 2 Zg. For identifying units of rings in this form, we will rely on the useful property that the norm of a unit, which we will define below, is equal to ±1. First, wep will define what it means to be a conjugate and then the concept of a norm on Z[ d]. p p Definition 1.3. Let α 2 Z[ d]. This means that α is of the form α = pa + b d where a; b 2 Z. Then, we denote the conjugate of α as α, where α = a − b d. Date: August 29, 2014. 1 2 DANIEL KLINE p Definition 1.4. For the rings R = Z[ d], where d is a square-free integer, the norm of α 2 R is defined as N(α) = α · α 2 Z. Example 1.5. For example, for all z 2 Z, z = z. Therefore, the norm of z 2 Z is 2 NZ(z) = z . As observed above, the norms of the units of Z are equal to ±1. It is clear that the norm in this case cannot be −1 for any z 2 Z, so we consider the 2 elements whose norms are 1. We see that ±1 are units since NZ(±1) = (±1) = 1. 0 0 0 0 Also, note that for all z 2 Z such that z 6= ±1, z is not a unit and NZ(z ) 6= ±1. In this case, not only do the units of Z have a norm of 1 but also the non-units of Z do not have a normp of ±1. This is a special case of thep fact that the norm of an element α 2 Z[ d] is ±1 if and only if α is a unitp of Z[ d]. We will prove this below, but first we need to show that the norm on Z[ d] is multiplicative. p Lemma 1.6. The norm on Z[ d] is multiplicative. Proof. N(α · β) = α · β · α · β = α · β · α · β = α · α · β · β = N(α) · N(β) p p Proposition 1.7. Let α 2 Z[ d]. Then, N(α) = ±1 () α is a unit of Z[ d]. Proof. (=)) This direction is clear from the definitions. p ((=) Suppose that α is a unit. Then, there exists α−1 2 Z[ d] such that α·α−1 = 1. Now, we take the norm on both sides to get N(α · α−1) = N(1). As seen in 1.5, N(1) = 1 and by 1.6, N(α · α−1) = N(α) · N(α−1). Therefore, 1 = N(α) · N(α−1). By 1.5, the only values of N(α) that makes this statement true is N(α) = ±1. p Now,p we look more explicitly at the norm ofp the rings Z[ d]. Suppose that α 2 Z[ dp]. Then, α is of the form α = a + bp d, wherep a; b 2 Z. Note that 2 2 α = a − b d. Therefore, N(α) = αp· α = (a + b d)(a − b d) = a − b d. Then, we can find the units of the ring Z[ d] by setting N(α) = ±1. This equation for finding units, ±1 = a2 − db2, is commonly referred to as Pell's equation. We now look at two examples below to see the differing behavior of Pell's equation for d < 0 and d > 0. Example 1.8. First, consider an example in which d < 0. For instance, consider the ring Z[i], where d = −1. The units of this ring then are those elements of the form a + bi which satisfy ±1 = a2 + b2. It is clear that either a = ±1 and b = 0 or a = 0 and b = ±1. Therefore, the set of units in this casep is {±1; ±ig. Similarly, for any square-free integer d < −1, the set of units of Z[ d] is {±1g. Example 1.9. Now,p we will consider an example in which d > 0. To find the units of the ring Z[ 2], we again use Pell's equation, which in this case, looks like 2 2 a − 2b = ±1. We quickly seep that ±1 isp a solution. With a little effort, we can find other units, such as 1 + 2 and 3p + 2 2. We will soon see that there arep an m infinite number of units of the ring Z[ 2], which each have the form ±(1 + p2) 0 for mp 2 Z. Thisp agree with thep units wep have already found as 1 = (1 + 2) , 1 + 2 = (1 + 2)1, and 3 + 2 2 = (1 + 2)2. Now, we will use the conceptp of the fundamental unit to describe the behavior of the units of the ring Z[ 2] as well as others. THE STRUCTURE OF UNIT GROUPS 3 Definition 1.10. A fundamental unit u is a unit of infinite order such that every unit is of the form ζum, where ζ is a root of unity and m 2 Z. p For all rings of the form Z[ d], there exists a fundamental unit. Ireland [3] provides a theoremp to find the fundamental units in these cases,p which we will use to identify 1 + 2 as the fundamental unit of the ring Z[ 2]. First, we need a proposition. Later, we will define a notion of the ring of algebraic integers for finite-dimensional field extensions of Q. For quadratic field extensions, though, we know what the set of algebraic integers looks like. Proposition 1.11. For d 2 and d square-free, the set of algebraic integers O p pN p 1+ d of Q( d) is either the set Z[ d] when d 6≡ 1 mod 4 or the set Z[ 2 ] when d ≡ 1 mod 4. p Theorem 1.12. Let d 2 N such that d is squarep free. Consider the set Q( d) ⊂ R. Let O be the ring of algebraic integers of Q( d). There exists a fundamental unit u 2 O, such that u > 1 and all the units of O are in the form of ±um, where m 2 Z. We will prove the above theoremp in section 7, but for now, we will assumep the theorem in order to show that 1 + 2 must be the fundamental unit of Z[ 2]. Example 1.13. By the previous result,p there exists a fundamental unit, which without loss of generality is > 1,p of Z[ 2]. First, we will show that the fundamental unit must bep of the form a + b 2, where a; b 2 N. Assume by contradiction that m u = ±(a − b 2) > 1,p where a; b 2 N. Then, by induction, for all m 2 N, u is n of the form ±(x −py 2), where x; y 2 N.p Also, for all np < 0, u ≤ 1. However, we know that 1 + 2 > 1 is a unitp of Z[ 2]. Since 1 + 2 is clearly not plus or minus an integer power of ±(a − b 2), then the fundamental unit cannot be of this form. Also, note that for the fundamental unit, a; b 6= 0, because the only only units in this case would be ±p1, which cannot be the fundamental unit as we have seen other unitsp such as 1 + 2. Therefore, the fundamentalp unit must be of the form u = a + b 2, wherep a; b 2 N. Thus, because 1 + 2 is thep unit of this form that is closest to 1, 1 + 2 must be the fundamental unit of Z[ 2]. The above reasoning fails to say anything about the unit group of other rings. For this reason, we will prove Dirichlet's Unit Theorem so as to build up an answer to the structure of unit groups in general as well as prove 1.12. 2. Minkowski's Theorem in the Plane To provide an intuition for Minkowski's Theorem, we will first show this theorem for the specific case of the integer lattice in R2.