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Summa Summarum Mogens Esrom Larsen. Review quote " ""Throughout the book, the exposition is crisp and self-contained, and Weintraub manages to strike a very nice balance between explicit computations and abstract theory. There are a good number of exercises, and plenty of directions one could go in after reading this book. For that alone it should be rewarded, and this book has far more to offer. After explaining the main players integral domains and quadratic fields , the author proves that Euclidean rings are principal ideal domains, and that these have unique factorization.

This is followed up with a lengthy discussion of examples of nonunique factorization in quadratic rings. Learn more at Author Central. Popularity Popularity Featured Price: Low to High Price: High to Low Avg. Theory and Practice Feb 19, Available for download now. Only 3 left in stock more on the way. Only 1 left in stock more on the way. A Complement to Vector Calculus Aug 20, Galois Theory Universitext Oct 23, Available to ship in days. Only 1 left in stock - order soon. Representation Theory of Finite Groups: We envision several ways in which this book can be used.

In our investigations, we begin at the beginning, so this book is suitable for that purpose. Indeed, one of the themes of this book is that one can go a long ways with only elementary methods. Another way to use this book is for a more advanced course in number theory, and there is plenty of appropriate material here for such a course.

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If this book is used as a text for a more advanced course, these appendices will serve as background. Then the Fundamental Theorem of Arithmetic states that every positive integer can be factored into primes in an essentially unique way. Note that 1 is a special case. As its name implies, unique factorization is a fundamental property of the positive integers, a property that was known to the ancient Greeks. We will prove this property, and indeed our proof will follow that of Euclid. It turns out, and we shall prove, that numbers of this form also have unique factorization. For example, we have the following factorization 1 i i i i i i i i 2 Introduction into primes for numbers of this form: Numbers of this form do not have unique factorization.

For example, we have the following two factorizations of 6 into irreducibles: Numbers of this form also do not have unique factorization. For example, we have the following two factorizations of 10 into irreducibles: As we will see, instead of unique factorization being the norm and nonunique factorization the exception, the situation is reversed!

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It is really a very special property, though a crucially important one, of the ordinary integers that the Fundamental Theorem of Arithmetic holds for them. Here a and b are not always integers, but this is a technical point we will defer until later. Numbers of this form are called the Gaussian integers. As we i i i i i i i i Introduction 3 have remarked, in the Gaussian integers we do have unique factorization into primes, but we would like to know what the primes are.

Here we will show that the following is always true compare the factorizations above: In fact, this is closely related to a famous theorem of Fermat: Actually, we will give several proofs of this theorem. It uses a technique known as composition. It is a bit surprising that this abstract result gives such a concrete fact, but mathematics is full of beautiful surprises.

To describe our next objective, we have to get a bit more technical. The ancient Greeks considered the positive integers, but when we wish to generalize our investigations, we no longer have the idea of positivity. So we have to consider all integers. But when we do, we see that we have typical factorizations: We give a name to this situation. Then we also have the prime factorization in the Gaussian integers: In factoring numbers, we do not really care about units, but still it is an interesting question—indeed, a very interesting question—to ask what i i i i i i i i 4 Introduction the units are.

We have given the answer for the Gaussian integers, but we can ask the same question in other cases as well. But we can consider this equation for other values of D as well. Our proof is a variant of the cakravala method experimentally developed by Indian mathematicians between the ninth and twelfth centuries.

Our investigations in Chapters 1 through 4 can be considerably generalized. Indeed, our treatment here parallels the historical development of the subject. This subject is known as algebraic number theory. In Chapter 5 we survey some of the highlights of this subject. As we have seen, unique factorization of elements holds in the integers Z, but it does not always hold. While unique factorization of elements is the most straightforward generalization of the situation in the integers, it is not the right generalization.

The right generalization is unique factorization of ideals, which does hold. Therefore in Chapter 5 we will focus though not exclusively on ideals in general. For a precise description of the scope of our investigations in this chapter, we simply refer the reader to the table of contents. We have three appendices. Appendix A is a careful treatment of mathematical induction, an essential proof technique.

Appendix B is a treatment of congruences. Appendix C is a technical one, dealing with some of the more complicated cases of results in Chapter 2. First we introduce the general class of objects, and then we focus on the particular examples that will concern us. An integral domain is a set R equipped with two operations, addition and multiplication, that satisfy the following properties: Basic Notions 8 Multiplication is associative, i.

Note that, by taking the contrapositive, this condition may be equivalently rephrased as follows: An important property of integral domains is the cancellation property, which holds for both addition and multiplication. Let R be an integral domain and let a, b, and c be elements of R. Integral Domains 9 Here are some examples of integral domains.

The remaining properties of R follow directly from the corresponding properties of Z, except for the last one, the absence of zero divisors. We leave this for the exercises. We shall abbreviate this condition by saying that a and b have the same parity. Then R is also an integral domain. This requires more care.

Again, we leave the proof that R has no zero divisors to the exercises.

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Then R is an integral domain. Once again, properties 1 — 9 of an integral domain are easy to check and we defer the proof of property 11 to Example 1. The integral domains in Example 1. It turns out to be the case that, depending on the value of D, we sometimes want to consider the former and sometimes the latter. See the exercises for why this is the case.

An element a of an integral domain R is a unit if a has a multiplicative inverse. Let us consider the integral domains in Example 1. For this to make sense we need to know that the denominator is nonzero. This equation may or may not have a solution, but if it has a solution, that solution x is unique. We have imposed the restriction that D not be a perfect square, but now we want to impose a further restriction: This is purely to avoid duplication. For suppose D were not square-free, i. But we assumed D was not a perfect square, so this is impossible. For any element x of O D , N x is an integer.

Conversely, suppose that x is a unit. However, by Lemma 1.

But when D is positive the situation is much more involved. Then, by Lemma 1. We leave this proof for the exercises. Nevertheless, let us experiment a bit, by taking small values of D. Thus you will need four examples, one for each omitted condition. Show that the usual rules of signs hold in any integral domain R: Show that if this equation has a solution, that solution is unique.

Recall from Remark 1. Let R be an arbitrary integral domain. Show that R as in Example 1. But many of the statements we make have analogs for polynomials, and we leave the treatment of the polynomial situation to the exercises. Let R be an integral domain. In considering R[X] you may assume the usual properties of polynomial arithmetic. Show that R[X] is an integral domain. To be precise, our strategy will be as follows: Prove that certain integral domains are Euclidean domains.

Prove that every Euclidean domain is a principal ideal domain. Prove that every principal ideal domain is a unique factorization domain. Thus, putting all of these steps together, we see that certain integral domains are unique factorization domains. The obvious question now is: Unique Factorization We will not be able to settle the issue in all cases, and in fact, in complete generality the answer is unknown. In order to say what smaller is, we must have a notation of size. The following are norms: Certainly if a is an integer, a is a nonnegative integer.

Euclidean Domains 21 2 This follows from earlier work we have done. Let us see this. We showed in Lemma 1. This is the contrapositive of Lemma 1. Unique Factorization Lemma 2. The integers Z are a Euclidean domain. Euclidean Domains 23 As we go further, the details—though not the basic idea—get more complicated.

We do these cases here. The other cases are considerably more involved, and we defer them to Appendix C. Here the O-points are points with integer coordinates. The points apparently nearest the point s, t are the points in a square of side 1 centered at s, t , as in Figure 2. Here the O-points are points with both coordinates integers or half-integers, and the points apparently nearest the point s, t are the points in a diamond with diagonals of length 1 centered at s, t , as in Figure 2.

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A little thought shows that we have simply translated i. Thus, if we can show every Q-point apparently nearest the origin is within a distance of 1 from some O-point s, t , this will be true of all Q-points in the plane, and again we will be done. Thus, we have reduced our problem to considering the Q-points that are apparently nearest the origin. We shall denote this region by 0. Now the real work begins. Euclidean Domains 27 0. Now we bring in the analytic geometry. First, we consider the case when D 0. Actually, there is one exception.

Euclidean Domains 29 1 0. We state this result without proof. We shall not investigate this question. We shall save this for the next section, when will learn not only how to do it, but also what it is good for. We close this section by recording a lemma that will enable us to easily tell when an element of a Euclidean domain is a unit. This will provide a generalization of Lemma 1. Actually, using this lemma would enable us to simplify some of our earlier proofs slightly, but in the interest of directness we did not do so.

Let a and b be positive integers. We should point out that a priori the gcd may not exist. We are claiming that there is one and only one positive integer with a certain property, and a priori there may be no such integer or more than one such integer. But for the positive integers the gcd does indeed exist and can be found by taking the common prime factors of a and b. This is not really a satisfactory answer, however, because this assumes unique factorization, which we have not shown yet.

In fact, we will use the gcd to prove unique factorization, not the other way around. Moreover, we will see that we have a gcd in any Euclidean domain. Actually, I have exaggerated here to make a point. I have no idea what the prime factorizations of and are. With these examples in mind, we get to work.

In general, a gcd may or may not exist. We shall soon explore the question of when it does. But for the moment, let us assume that a gcd does exist and explore the consequences of that assumption. Unique Factorization Property 1: This set is relatively prime, as it has a gcd of 1, but is not pairwise relatively prime. Looking at pairs of elements, we see that 6 and 10 have a gcd of 2, that 6 and 15 have a gcd of 3, and that 10 and 15 have gcd of 5.

Thus in this set, no two distinct elements are relatively prime. We have been proceeding a bit hypothetically, assuming a gcd exists and exploring the consequences of that assumption. Now let us turn to the question of when a gcd actually does exist. We shall formulate a stronger property than the mere existence of the gcd, and investigate that. This is a very important property, as we shall see, but GCD-L is not standard mathematical language. Here is our main theorem. First, we will give a very short and slick but nonconstructive proof of this theorem.

To see this, we must verify the properties of the gcd. We observe that D A is a nonempty set, as it contains the identity element of 1 of R. The argument in the other direction uses the identical logic. Note that at this point we do not want to assume that every set of elements of R, not all of which are zero, has a gcd. To be sure, we proved that in Theorem 2. In light of Remark 2. Again, we show that these two sets are equal by showing that every element of one of them is also an element of the other.

With these results in hand, we can now give our second proof of Theorem 2. Second Proof of Theorem 2. We prove the theorem by induction on n. This is the crucial case. So in this case we are done. This is a consequence of the Well-Ordering Principle. So it stops at some stage. Let that be stage k. Thus, we see we have a sequence of divisions: To prove this we use Lemma 2.

But it is easier to prove this instead by induction. Now for the inductive step. There are two easy cases: So suppose that neither of these is the case. But by Lemma 2. Now for the second condition. Here is the second: Unique Factorization In this example, we have followed the usual practice of using positive remainders. But there is another practice, namely, using remainders whose absolute value is as small as possible. Let us redo this example that way: Note, as a practical matter, that this computation can easily be performed with only the use of an arbitrary precision calculator.

Unique Factorization Example 2. The logic of this example depends on the proof of Theorem 2. Since we deferred the proof of that case of Theorem 2. Ideals and Principal Ideal Domains 2. Again, with a lot of mathematical hindsight, it is precisely these two properties that we wish to consider in general. An ideal I in an integral domain R is a nonempty subset of R with the following properties: In other words, an ideal I is a nonempty subset of R that is closed under addition and also is closed under multiplication by any element of R not just by elements of I.

Let us check that principal ideals are indeed ideals. Unique Factorization Here are two extreme cases. In this case, I is called an improper ideal. Otherwise, I is called a proper ideal. Every element of R is a multiple of 1. In this case, I is called the zero ideal. Otherwise, I is called a nonzero ideal.

The only multiple of 0 is 0. So you may ask: But we shall defer the answer to this question for a while. Right now, we continue with a general development of properties of ideals. Let I be an ideal of R.

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The following are equivalent: Then 1 is in I so, by Lemma 2. Thus, every nonzero element of R is a unit, so, by Remark 1. Now a multiple of an element is the same as a linear combination of that element. Let us see that IA is always an ideal, and in fact that we get every ideal this way. First, IA is closed under addition: Second, IA is closed under multiplication by any element of R: In fact, it may turn out that this is always the case.

This is a very important situation to which we give a name. Suppose that R is a PID. Consider the ideal IA. Let I be any ideal. This follows directly from Theorem 2. Not only did it show that every ideal i i i i i i i i 2. So we actually have the following result: Let R be a PID. One can ask whether, conversely, every PID is a Euclidean domain. The answer is no, but the examples are not particularly illuminating or useful , so we shall not give any.

The following is a classical and extremely useful result. By hypothesis, these two elements have a gcd. Thus, the question arises as to why we bothered to provide the second proof. Here is the answer. Unique Factorization Domains 51 had two conditions, 1 and 2 , and we used both of them. Hence the second proof works more generally. Although surprising, this turns out to be the right i i i i i i i i 52 2.

Unique Factorization thing to do. As we will show see Lemma 2. We will observe that 3 may be more practical to check. On the other hand, to check 4 , we must look at any number d and see if d is divisible by a. Actually, a prime element is always irreducible, although in general, an irreducible element may not be prime. We defer examples of the second. Thus, we must show that here in the case of a PID , an irreducible is prime.

Unique Factorization Domains 53 Definition 2. Observe that essential uniqueness is the best we can hope for. We need the following technical lemma in our proof of Theorem 2. We show this in two stages: Unique Factorization We prove this claim: Thus to complete the proof, we need only show the process eventually stops. Then we can apply Lemma 2. Thus, we have a contradiction if the sequence goes on forever. This is impossible, and so we conclude that the sequence stops. Thus to complete the proof, we need only show that the process eventually stops.

Now we must prove the second stage: Unique Factorization Domains 55 Lemma 2. Continue in this fashion to conclude that p1 divides some qi. The following integral domains are unique factorization domains: Each of these integral domains is a Euclidean domain Lemma 2. It is natural to ask whether the converse of Theorem 2. The answer to this question is no, for general integral domains R. In the remainder of this section we will investigate UFDs more deeply. As we saw in Lemma 2. This is true in the more general situation of a UFD as well. Let R be a UFD. Thus, we must show that here in the case of a UFD an irreducible is prime.

Now we will reconsider this concept in the more general situation of UFDs. The key to our discussion is the following lemma, which is very useful in its own right. Then, for some primes p1 ,. Unique Factorization Proposition 2. Furthermore, also by Lemma 2. We use the same language in the more general situation of a UFD here. Note that Lemma 2. But in this more general situation we need a new proof. The following corollaries are word-for-word generalizations of Corollary 2.

But we will give a second albeit longer proof that uses prime factorization directly: Then, by Lemma 2. But again we will give a proof that uses prime factorization directly: As the reader will see, our results are much more complete for negative values of D than they are for positive values of D. To keep these straight, in Lemma 2. Note in Lemma 2. We will be using it or its consequences in all of the examples in this section and in the next section. So once again this provides an example of working hard enough with a single idea and being able to go a long way with it.

We use it so often that we give it a name although this name is not standard mathematical language. Now in a UFD, every irreducible is prime Lemma 2. We divide the proof into two cases: But if D 2 , it does not. Hence, by Corollary 2. Let p be the smallest prime dividing D.

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The Case D 65 by contradiction. We use Proposition 2. Actually, the methods we derive here will also apply to values of D with D i i i i i i i i 68 2. Once again, we divide the proof into two cases: But it is a result from number theory see Corollary B. We consider case 2b. We begin by noting that, by Lemma 2. The remainder of the proof is identical to the proof of Theorem 2. If b1 and b2 are small, trial and error is as good as any. We will use this in the statement of our results. We wish to apply Theorem 2. First, we consider condition 1 of that theorem.

We consider case b. As in the proof of Lemma 2. Let p2 be an odd prime and let q be any integer relatively prime to p. Note that p1 is odd. Thus, by Corollary 2. Once again we will formulate our results with the help of the Chinese Remainder Theorem.

Factorization : Unique and Otherwise

It states that if at least one of p1 and p2 is congruent to 1 mod 4 , then either both of these congruences have a solution or neither does, while if both p1 and p2 are congruent to 3 mod 4 , then exactly one of these congruences has a solution. Since in Theorem 2. We see how to use this in the next example. In order to satisfy hypothesis 2 of Theorem 6. This is the same as case 1a of Example 2. We proceed by exactly the same logic as before. The techniques in Example 6. Depending on circumstances, either one may be more convenient to use.

We shall show that O D does not have an element of norm p. We prove this by contradiction. For this we need primes congruent to 3 mod 8. Then the hypotheses of Theorem 2. For each value of D we give a pair q1 , q2 that provides the argument. In case q1 is an odd prime, we are using Theorem 2. For many values of D, there is more than one pair q1 , q2 that work.

In those cases, we have simply chosen one. In this section we will sum up our work and also report on some interesting results that are beyond our ability to prove here. Unique Factorization In Theorem 2. See also Example 2. We have the following conjecture, which has been open for years: Complete the proof of Lemma 2.

You must prove Lemma 2.

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Note by Lemma 2. Also by Lemma 2. Exercises 81 Exercise 2. Prove the following properties of a gcd of elements in an integral domain R: Do these exercises by using the results of Sections 2. Prove the analog of Lemma 2. Suppose that R is a UFD. In the notation of Proposition 2. We make the convention that in this case, we choose the gcd of a set of elements to be positive. Similarly, we choose the lcm of a set of elements to be positive. Exercises 83 Exercise 2. Show that in fact r is an integer.