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How Do Elements Really Factor in Z[ √ −5]?

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How Do Elements Really Factor in Z[ √ −5]? p How Do Elements Really Factor in Z[ −5]? Scott T. Chapman, Felix Gotti, and Marly Gotti p Abstract Most undergraduate level abstract algebra texts use Z[ −5] as an example of an integral domain which is not a unique factorization domain (or UFD) by ex- hibiting two distinct irreducible factorizations of a nonzero element. But such a brief example, which requires merely an understanding of basic norms, only scratches the surface of how elements actually factor in this ring of algebraicp integers. We offer here an interactive framework which shows that while Z[ −5] is not a UFD, it does satisfy a slightly weaker factorization condition, known as half-factoriality. The arguments involved revolve around the Fundamental Theorem of Ideal Theory in algebraic number fields. Dedicated to David F. Anderson on the occasion of his retirement. 1 Introduction Consider the integral domain p p Z[ −5] = fa + b −5 j a;b 2 Zg: Your undergraduate abstract algebra text probably used it as the base example of an integral domain that is not a unique factorizationp domain (or UFD). The Fundamen- tal Theorem of Arithmetic fails in Z[ −5] as this domain contains elements with Scott T. Chapman Sam Houston State University, Huntsville, TX 77341, e-mail: [email protected] Felix Gotti UC Berkeley, Berkeley, CA 94720 e-mail: [email protected] Marly Gotti University of Florida, Gainesville, FL 32611 e-mail: [email protected] 2010 AMS Mathematics Subject Classification: Primary 11R11, 13F15, 20M13. 1 2 Scott T. Chapman, Felix Gotti, and Marly Gotti multiple factorizations into irreducibles; for example, p p 6 = 2 · 3 = (1 − −5)(1 + −5) (1) p p even thoughp 2;3;1 − −5, and 1 + −5 are pairwisep non-associate irreducible el- ements in Z[ −5]. To argue this, the norm on Z[ −5], i.e., p N(a + b −5) = a2 + 5b2; (2) plays an important role, as it is a multiplicative function satisfying the following properties: • N(a) = 0 if and only if a = 0; p • N(ab) = N(a)N(b) for all a;b 2 [ −5]; Z p • a is a unit if and only if N(a) = 1 (i.e., ±1 are the only units of Z[ −5]); • if N(a) is prime, then a is irreducible. However, introductory abstract algebra books seldom dig deeper than what Equation (1) does. Thep goal of this paper is to use ideal theory to describe exactly how ele- mentsp in Z[ −5] factor into products of irreducibles. In doing so, we will show that Z[ −5] satisfies a nicep factorization property, which is known as half-factoriality. Thus, we say that Z[ −5] is a half-factorial domain (or HFD). Our journey will require nothing more than elementary algebra, but will give the reader a glimpse of how The Fundamentalp Theorem of Ideal Theory resolves the non-unique factoriza- tions of Z[ −5]. The notion that unique factorization in rings of integers could be recovered via ideals was important in the late 1800’s in attempts to prove Fermat’s Last Theorem (see [9, Chapter 11]). Our presentation is somewhat interactive, as many steps that follow from stan- dard techniques of basic algebra are left to the reader as exercises. The only back- ground we expect from the reader are introductory courses in linear algebra and abstract algebra. Assuming such prerequisites, we have tried to present here a self- contained and friendlyp approach to the phenomenon of non-uniqueness of factoriza- tions occurring in Z[ −5]. More advanced and general arguments (which apply to any ring of integers) can be found in [8] and [9]. 2 Integral Bases and Discriminants Although in this paper we are primarily concerned withp the phenomenon of non- unique factorizations in the particular ring of integers Z[ −5], it is more enlight- ening from an algebraic perspective to introduce our needed concepts for arbitrary commutative rings with identity, rings of integers, or quadratic rings of integers, depending on the most appropriate context for each concept being introduced. In what follows, we shall proceed in this manner while trying, by all means, to keep the exposition as elementary as possible. p How Do Elements Really Factor in Z[ −5]? 3 An element a 2 C is said to be algebraic provided that it is a root of a nonzero polynomial with rational coefficients, while a is said to be an algebraic integer provided that it is a root of a monic polynomial with integer coefficients. It is not hard to argue that every subfield of C contains Q and is a Q-vector space. Definition 2.1. A subfield K of C is called an algebraic number field provided that it has finite dimension as a vector space over Q. The subset OK := fa 2 K j a is an algebraic integerg of K is called the ring of integers of K. The ring of integers of any algebraic number field is, indeed, a ring. The reader is invited to verify this observation. If a is a complex number, then Q(a) denotes the smallest subfield of C containing a. It is well known that a subfield K of C is an algebraic number field if and only if there exists an algebraic number a 2 C such that K = Q(a) (see, for example, [6, Theorem 2.17]). Among all algebraic number fields, we are primarily interested in those that are two-dimensional vector spaces over Q. Definition 2.2. An algebraic number field that is a two-dimensional vector space over Q is called a quadratic number field. If K is a quadratic number field, then OK is called a quadratic ring of integers. For a 2 C, let Z[a] denote the set of all polynomial expressions in a having integer coefficients.p Clearly, Z[a] is a subring of Q(a). It is also clear that, for d 2 Z, the field Q( d) has dimension at most two as a Q-vector space and, therefore, it is an algebraic number field. Moreover, if d 2= f0;1g and d is squarefree (i.e.,pd is not divisible by the square of any prime), then it immediately follows that Q( d) is a two-dimensional vector space over Q and, as a result, a quadratic number field. As we are mainly interested in the case when d = −5, we propose the following exercise. Exercise 2.3.pLet d 2 Z n f0;1g be a squarefree integer such that d ≡ 2;3p(mod 4). Prove that Z[ d] is the ring of integers of the quadratic number field Q( d). Remark. When d 2 nf0;1g is a squarefree integer satisfying d ≡ 1 (mod 4), it is Z p p 1+ d not hard to argue that the ring of integers of Q( d) is Z[ 2 ]. However, we will not be concerned with this case as our case of interest is d = −5. p For d asp specified in Exercise 2.3, the elementsp of Z[ d] can be written in the form a + b d for a;b 2 Z. The norm N on Z[ d] is defined by p N(a + b d) = a2 − db2 p (cf. Equation (2)). The norm N on Z[ d] also satisfies the four properties listed in the introduction. 4 Scott T. Chapman, Felix Gotti, and Marly Gotti Let us now take a look at the structure of an algebraic number field K with linear algebra in mind. For a 2 K consider the function ma : K ! K defined via multipli- cation by a, i.e., ma (x) = ax for all x 2 K. One can easily see that ma is a linear transformation of Q-vector spaces. Therefore, after fixing a basis for the Q-vector space K, we can represent ma by a matrix M. The trace of a, which is denoted by Tr(a), is defined to be the trace of the matrix M. It is worth noting that Tr(a) does not depend on the chosen basis for K. Also, notice that Tr(a) 2 Q. Furthermore, if a 2 OKp, then Tr(a) 2 Z (see [10, Lemma 4.1.1], or Exercise 2.5 for the case when K = Q( d)). Definition 2.4. Let K be an algebraic number field that has dimension n as a Q- vector space. The discriminant of a subset fw1;:::;wng of K, which is denoted by D[w1;:::;wn], is det T, where T is the n × n matrix Tr(wiw j) 1≤i; j≤n. With K as introduced above, if fw1;:::;wng is a subset of OpK, then it follows that D[w1;:::;wn] 2 Z (see Exercise 2.5 for the case when K = Q( d)). In addition, the discriminantp of any basis for the Q-vector space K is nonzero; we will prove this for K = Q( d) in Proposition 2.10. Exercise 2.5. Let d 2 Z n f0;1g be a squarefree integer such that d ≡ 2;3 (mod 4). p p 1. If a = a + a d 2 [ d], then Tr(a) = 2a . 1 2 Z p p 1 2. If, in addition, b = b1 + b2 d 2 Z[ d], then 2 a s(a) D[a;b] = det = 4d(a b − a b )2; b s(b) 1 2 2 1 p p where s(x + y d) = x − y d for all x;y 2 Z. Example 2.6. Let d 2= f0;1g be a squarefree integerp such that d ≡ 2;3 (modp 4).
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