1 First Theme: Sums of Squares


 Warren Lloyd
 4 years ago
 Views:
Transcription
1 I will try to organize the work of this semester around several classical questions. The first is, When is a prime p the sum of two squares? The question was raised by Fermat who gave the correct answer in 1640 and claimed he had a proof but never wrote it down; the first published proof, in 1747, is due to Euler. There is a beautiful article in Wikipedia which you can reach by Googling sum of two squares ; it is well worth reading. Today there is a very short exceedingly clever proof which requires practically no preparation but which also gives you no idea of why the theorem should be true. Our approach here, however, will not be to get to the result as speedily as possible but to bring in the many modern algebraic ideas that lead to a conceptual understanding of why Fermat s original assertion is correct. 1 First Theme: Sums of Squares When is an integer a sum of two squares, m = a 2 + b 2? First observation: if m, n are both sums of two squares then so is mn. For going over to complex numbers, m = a 2 + b 2 can be written as m = (a + bi)(a bi). Similarly, if n = c 2 + d 2 ) then we can write n = (c + di)(c di). It follows that mn = (a + bi)(c + di)(a bi)(c di) = [(ac bd) + (ad + bc)i][(ad bc) (ac bd)i], so we must have mn = (ac bd) 2 + (ad + bc) 2. An easy calculation shows that this is correct. This suggests that we look first at the question of when a prime p is a sum of two squares. Knowing which primes are sums of two squares won t completely answer the question, however. Certainly those numbers which can be written as products of primes each of which is a sum of two squares will, from what we have just seen, indeed be sums of two squares. It is still possible that some other numbers are, too. In fact, any number which is itself a square is trivially the sum of two squares, itself and zero. We will see that this really exhausts the possibilities: to be a sum of two squares a number must be a product of squares and of primes which are themselves a sum of two squares. So which primes qualify? Let s try out some small primes: 2 = , 3 fails, 5 = , 7 fails, 11 fails, 13 = , 17 = , 19 fails. So far, except for 2, those primes p which are congruent 1 modulo 4, i.e., which leave a remainder of 1 when divided by 4 are sums of two squares. This is written p 1 mod 4 or simply p 1(4). Those p with p 3(4) fail. And a number like 3 5 = 15 which has as a factor a prime that fails, namely 3, (but not 3 2 ), is not a sum of two squares. In fact we have the following Theorem 1 1. A prime p is a sum of two squares if and only if either p = 2 or p 1(4). 2. An integer m is a sum of two squares if and only if in the prime factorization of m those primes p 3(4) appear to an even power. There are classical computational proofs of this which do not involve any higher algebra (cf. the Wikipedia article). The first few weeks of our course will be devoted to developing enough modern algebra so that we can understand a conceptual proof. Part of this will be understanding when prime factorizations 1
2 exist and are unique; it is implicit in the statement of the theorem that this is the case for the ordinary integers. Passing to the complex numbers has been useful but before going further, let s introduce some terminology and some notation. You have probably already been introduced to the concept of a group, G but let me review it very briefly: G consists of a set of elements together with a multiplication map G G G which is associative; moreover (i) there is a unit element, denoted 1 or e (or 1 G if we want to emphasize that this is the unit element of the group G) with the property that 1 x = x 1 = x for all x G, and (ii) for every x G there is an inverse element x 1 such that x x 1 = 1 = x 1 x. This inverse is necessarily unique (as is the unit element). We say that x and y commute if xy = yx. A group in which all pairs of elements commute is called commutative or Abelian (in honor of N. H. Abel). In that case, multiplication is frequently written as addition and the unit element is denoted by 0 and called the zero element; the group is then usually called an additive group. The set of all permutations of an arbitrary set S forms a group; it is nonabelian whenever S has at least three elements. The most important Abelian group is the additive group of integers, here always denoted by Z. The integers, however, have more structure, there is an associative multiplication. This leads to the following definition: a ring R is an additive group with an associative multiplication that satisfies the distributive laws, i.e., where x(y + z) = xy + xz and (y + z)x = yx + zx. We need both of these statements since the multiplication need not be commutative. A good example of such a ring is set of all n n matrices with real coefficients. We will always denote the real numbers by R and this ring by M n (R). (This notation differs from that in the text but is more common.) Notice that M n (R) has a unit element for multiplication. We will generally assume without mentioning it that the rings we deal with have a unit element but (unlike the text) not that they are commutative. A field is a commutative ring in which every nonzero element has a multiplicative inverse. The most important examples are the rational numbers Q, the real numbers, R, and the complex numbers C. There do exist noncommutative rings in which every nonzero element has a multiplicative inverse. These are called division rings, or skew fields, a term which the elder Artin has contracted to sfield. The most important example is the Hamiltonians, generally denoted by H after its discoverer. This algebra is a fourdimensional vector space over R with basis elements 1, i, j, k and multiplication defined by i 2 = j 2 = k 2 = 1, ij = k = ji. It follows that jk = i = kj, ki = j = ik. The general quaternion thus has the form q = a+bi+cj +dk, where a, b, c, d R. To see that this is a division ring, define the conjugate q = a bi cj dk. Then q q = q q = a 2 + b 2 + c 2 + d 2. Since a, b, c, d are real, this can t vanish unless all are zero, so we have q 1 = q/q q. A ring which is a vector space over a field is frequently called an algebra. This brings us to the concept of a morphism (older name homomorphism). Recall that a morphism between groups f : G H is a mapping which preserves the group multiplication, i.e., such that f(xy) = (f x)(f y). This implies that f(1 g ) = 1 H and that f(x 1 ) = (fx) 1. 2
3 Exercise 1 Prove the preceding assertions. In the newer terminology, a morphism which is onetoone, i.e., where x y implies f x f y, is called a monomorphism although I will frequently use the older word as well. A morphism which is onto is now usually called an epimorphism, but here, too, I will frequently use the older term. One which is both is frequently called a bijection. It is easy to check that if we have morphisms f g G H J then the composite G g f J is again a morphism. In elementary texts an isomorphism f : G H is usually defined to be a morphism which is onetoone and onto, a bijection. This will do for groups, rings, and all the structures you will encounter in this course, but the categorical definition is this: f is an isomorphism if there is a morphism g : H G such g f is the identity map of g, denoted id G, and such that f g = id H. The reason that the elementary definition works for groups and for rings is that if f : G H is a bijection then there is a set mapping g : H G sending every z H back to the unique x G with fx = z and this g is again a morphism. Exercise 2 Prove the preceding assertion. The concept of a morphism extends to rings: a morphism f : R S is a mapping which preserves both the addition and the multiplication, i.e., such that f(x + y) = fx + fy and f(xy) = (fx)(fy). While we certainly have f(0) = 0 it need no longer be the case that f(1 R ) = 1 S. For example, let R just be R and S = M 2 (R). We can ( then define ) a morphism R S by sending x 0 every real number x to the matrix. This is certainly a morphism but 0 0 the image of the unit element 1 R is not the unit element of S. The image of 1 is, however, an element whose square is itself; such an element is called an idempotent. The general concept of a morphism is that it preserves all the structure there is. Having a unit element is not part of the definition of a ring, even though the rings we deal with here generally do have units. To be more precise, we define a unital ring to be one with a unit element and a unital morphism to be one preserving the unit element. If you want a simple example where a morphism which is a bijection is not necessarily an isomorphism, considered partially ordered sets or posets, that is, sets in which for some but not necessarily all pairs of elements x, y there is a relation x y. This has to satisfy the axioms that x y and y z imply x z, x x, and x y together with y x imply x = y. A morphism f : S T of partially ordered sets is a set map such that x y implies fx fy. Now suppose that S carries a nontrivial partial order and that S is the same set with the partial order wiped out, i.e., in which there is no pair x, y with x y unless x = y. Now consider the identity map S S carrying every element to itself. This is trivially a morphism of partially ordered sets (since there are no conditions to satisfy) and is trivially a bijection, but the inverse map S S is not a morphism of partially ordered sets. 3
4 The kernel K of a group morphism f : G K is the set of all x G such that fx = 1. It is a normal subgroup of G, i.e., a subgroup such that xkx 1 = K, where xkx 1 is the set of all elements xkx 1, k K. The kernel of a ring morphism f : R S is similarly the set I of all x R such that fx = 0. This subset is an ideal of R, that is, (i) I is an additive subgroup of R (ii) with the additional property that if x R and z I then xz and zx are both in I. When we consider only the additive group structure of R (forget the multiplication for the moment) we generally write R +. Here is a simple but fundamental example. Suppose that R is commutative and pick an element a R. Then the set of all multiples of a, i.e. elements of the form ax with x R, forms an ideal. Such an ideal is called principal; it is the principal ideal generated by a, written ar or sometimes simply as (a) when the ring is understood. In particular, when R = Z if we pick an integer say 6, then the set of all multiples of 6 forms an ideal. As we will see, in Z every ideal is principal. Rings with this special property are generally called principal ideal rings. In any ring, the entire ring and the subring reduced to the zero element alone are ideals but we generally don t count these; all other ideals are called proper. It can happen that a ring has no proper ideals, in which case it is called simple. The rings M n (R) are simple. Exercise 3 (i) Prove this. (ii) Prove that if I is an ideal of a ring R then M n (I) is an ideal of M n (R) and (iii) that every ideal of M n (R) has this form. Recall that if we have a normal subgroup K of a group G then we can form the quotient group G/K (the set of cosets xk). Every normal subgroup K is actually the kernel of a group morphism, namely of the canonical morphism G G/K. Moreover, if we have a group morphism f : G H and if the kernel of f contains K then we can define a new morphism f : G/K H as follows: An element of G/K is a coset xk; set f(xk) = f(x). The representative x of the coset xk is not unique; it is just one of the elements of xk. However, if we have one representative x (that is, if we have in hand one element x of xk) then any other representative y must be a multiple xk of x. It follows that f(y) = f(x)f(k) = f(x), so the map f is well defined, and it is easy to see that it is a group morphism. With this, f can be factored: it can be written as the composite morphism can f G G/K H. The important thing now is that we can do the same for rings. Suppose that I is an ideal of R. Since I is an additive subgroup of R + its cosets are written in the form x + I and we define R/I to be the set of these cosets. This is again a ring with addition and multiplication defined by (x+i)+(y +I) = (x+y)+i; (x+i)(y +I) = xy +I. In our simple example of the multiples of 6 in Z, the ring Z/(6) (sometimes written even more simply as Z/6 has exactly six elements, 0, 1, 2,..., 5, the cosets of 0 through 5, respectively. Addition and multiplication in Z/6 are easy but there are some peculiarities: 2 3 = 0 so here we have a pair of elements, neither of which is the zero element of the ring, but their product is zero; such elements are called zero divisors. Also, ( 3) 2 = 3 3 = 3. This is another example of an idempotent. 4
5 The reason that Z/6 has zero divisors is that 6 is composite; 6 = 2 3. It is a basic theorem (a special case of a deeper one) that if p is a prime number then Z/p is a field. It is easy to check, for example, that Z/7 is a field: 2 4 = 3 5 = 6 6 = 1 (and, of course 1 is its own inverse), showing that every nonzero element of Z/7 has a multiplicative inverse. It follows that the nonzero elements form a group under multiplication. This group is cyclic; you can check that the powers of 3 give all the nonzero elements of Z/6. This is not an accident. We will prove the following Theorem 2 Any finite multiplicative subgroup of the multiplicative group of a field is cyclic. Exercise 4 Give examples to show that this need not hold for (i) a skew field or (ii) an infinite group. To prove this we shall need some information about the structure of Abelian groups. A group G (Abelian or not) is finitely generated if there is some finite subset of elements g 1,..., g n such that every element of G can be written as a product of these elements and their reciprocals. A group G which can be generated by a single element g is cyclic. If G is finite, say #G = n then G = {1, g, g 2,..., n 1}, where g n = 1. In this case G is isomorphic (as an a group) to Z/n, the isomorphism being given by g m m. If G is infinite then G = {..., g 2, g 1, 1, g, g 2,... }, where no power of G other then the zeroth is equal to the unit element. In this case, G = Z, the isomorphism being given by g m m. Recall that the direct product or simply the product of two groups, G H consists of all ordered pairs (g, h), g G, h H with group operation (g, h)(g, h ) = (gg, hh ). This construction extends in an obvious way to a product of any finite number of groups. With this we have the following fundamental theorem on finitely generated Abelian groups (a special case of a slightly more general theorem which we will prove) Theorem 3 Every finitely generated Abelian group is isomorphic to a unique group of the form Z/d 1 Z/d 2 Z/d r Z Z where d 1 d 2 d r. The last condition means that d 1 divides d 2, d 2 divides d 3, etc.. If the group is finite, then the Z factors don t appear. There are generally other ways to decompose an Abelian group, but this is the shortest (least number of factors). The d i are called the principal divisors of the group. Note that if the group is finite then the largest one, here denoted d r is the exponent of the group, i.e., the smallest integer e such that g e = 1 for every g G. Notice that if a finite group is not cyclic, i.e., if r 2, then writing d r = e, the number of elements of the group satisfying the equation x e = 1 is greater than e. In a field the number of solutions to a polynomial equation can not exceed the degree of the equation. This is why a finite subgroup of the multiplicative group of a field must be cyclic. Sometimes, however, it is quite difficult to find a generator. For example, we will see that if p is a prime then Z/p is a field, usually denoted F p. Since this field is finite and has exactly p elements, its multiplicative group 5
6 is of order p 1 and cyclic, but the difficulty of finding a generator when p is a large prime has sometimes been used in coding schemes. (The existence of finite fields was discovered by E. Galois.) The concept of direct product can also be extended to rings R and S except that there it is frequently called the direct sum and denoted R S. It is the set of ordered pairs (r, s) with r R, s S with addition and multiplication defined by (r, s) + (r, s ) = (r + r, s + s ), (r, s)(r, s ) = (rr, ss ). Let s get back (in a sophisticated way) to our problem of determining which primes p (and more generally, which integers) are sums of two squares, but first, one elementary observation: Any prime p 3(4) can not be a sum of two squares. For observe that if n is an even integer then n 2 0(4), while if n is odd, say n = 2m + 1 then n 2 = 4m 2 + 4m + 1 1(4). Therefore, a sum of two squares can only be congruent to 0, 1, or 2 mod 4, hence never to 3. For deeper results we will look at a generalization of the concept of integer. A special case is the set of all complex numbers of the form a + bi where a and b are integers. These form a ring, the ring of Gaussian integers, denoted Z[i]. Ordinary primes (henceforth called rational primes because they are the primes in the field Q of rational numbers) may factor in Z[i]. For example, 2 = i(1+i) 2, 5 = (1+2i)(1 2i). If p is a sum of two squares, p = a 2 + b 2, a, b Z, then p = (a + bi)(a bi) in Z[i], so being a sum of two squares implies factorization in Z[i]. The converse is also true. Lemma 1 A rational prime p factors in Z[i] if and only if it is a sum of two squares, in which case it has exactly two nontrivial factors, i.e., factors other than ±1, ±i. Proof. Suppose that p is a prime which factors in a nontrivial way in Z[i] with p = αβ. Taking conjugates we also have p = ᾱ β. Multiplying gives p 2 = α 2 β 2, a factorization of p 2 in Z, but the only possibility for this is that α 2 = β 2 = p. It follows that neither α nor β can factor further since this would give a factorization of p in Z. Since both have the same absolute value and their product is real, one must be the conjugate of the other. So if α = a + bi, a, b Z then β = a bi and p = a 2 + b 2. A complex number α which is a root of some polynomial f(x) = c n x n + c n 1 x n c 1 x + c 0 with integer coefficients is called an algebraic number. We could make this equation monic, i.e., have leading coefficient equal to 1 by dividing by c n, but then the other coefficients would generally be rational numbers and not integers. If α is a root of a monic polynomial with integer coefficients then it is called an algebraic integer. If α is an algebraic number then there is a unique monic polynomial f(x) of minimum degree which it satisfies; that polynomial is called the minimum polynomial for α and its degree is called the degree of α. Before considering the general case, consider the special case where α = d, where d is some squarefree integer (i.e., not divisible by the square of any integer other than 1) but which may be negative, a most important case being d = 1. This is obviously algebraic and even an algebraic integer since it satisfies the equation x 2 d = 0. In fact, all complex numbers of the form α = a + b d with a, b Q are algebraic since α satisfies the equation 6
7 x 2 2ax + (a 2 b 2 d) = 0. The set of all these numbers is denoted Q( d) and we claim that they form a field: it is clear that the sums, differences, and products of two numbers of the form a + b d is again a number of the same form. To see that the inverse is also, observe first that since d is not a square the rational number a 2 b 2 d can not vanish whenever either a or b is not 0. From x 2 2ax + (a 2 b 2 d) = 0 we have x(x 2a) = (a 2 b 2 d) whence x 1 = (x 2a)/(a 2 b 2 d), so Q d is indeed a field. It is also a vector space over Q with basis {1, d}, so it has dimension equal to 2 as a Qspace. Such a field is usually called a quadratic field. We will see that in fact the set of all algebraic numbers forms a field and that the set of all algebraic integers forms a subring of this field. When is a number a + b d; a, b Q an (algebraic) integer? Looking at the equation it satisfies, it is sufficient (and we will later show, necessary) that 2a and a 2 db 2 be integers. If a is an ordinary or rational integer then db 2 must also be an integer, and since d is squarefree it follows that b is an integer, too. The only other possibility is that a be a halfinteger, i.e., of the form m + 1/2, m Z, in which case a 2 = m 2 + m + 1/4. But then a 2 db 2 can only be an integer if b is also a half integer and d 1(4). So in this case the integers consist not just of all elements of the form m + n d, m, n Z but also of the elements (m ) + (n ) d. Exercise 5 Prove that the integers of Q d do form a ring. Since 1 1(4) we see finally that the ring of Gaussian integers is precisely the ring of algebraic integers inside the field Q(i). We are now close to understanding why a prime p 1(4) must be a sum of two squares. Observe that if p = a 2 + b 2 = (a + bi)(a bi), a, b Z then the rational prime p has factored inside the ring of Gaussian integers Z[i]. This raises the question of whether Z[i] behaves like Z in that it has primes which can not be factored and where every element can be factored uniquely into a product of primes. Here is the reason for the quotes. Even in Z factorization is not strictly unique because we could introduce factors of 1 so we make the following definition: In a commutative ring R (with unit element) an element u which has an inverse is called a unit. (This may not be the best terminology, but it is the historical one.) The units form a group under multiplication, usually denoted R. In Z the group of units consists only of {+1, 1}; in Z[i] it is {±1, ±i}. Elements x, y R with y = ux where u is a unit are called associates; this is obviously an equivalence relation. In a commutative ring it is meaningful to say that y divides x if x = yz for some z but it generally does not follow that if x and y divide each other that they are associates, for there may be zero divisors. A commutative ring R in which there are no zero divisors is called an integral domain or simply a domain. (The older name was domain of integrity.) In a domain, if x and y divide each other, x = yz and y = xw, then we have x = xwz or x(1 wz) = 0, so 1 wz = 0. Thus w and z are units and x and y are associates. An element which has no divisors except itself and associates is called irreducible. When we speak of unique factorization it means factorization into irreducibles which is unique up to the 7
8 order of the irreducible factors and multiplication by units (or replacement of factors by associates). It is quite possible in a domain for an element to have genuinely distinct factorizations into irreducibles. Consider, for example, the ring of integers of Q( 5); it consists of all m + n 5 with m, n Z. In this ring the elements 3, 7, , are all irreducible. Exercise 6 Prove this from first principles. (We will see more sophisticated reasons later.) Unfortunately unique factorization fails inside the ring of integers of Q( 5 for we have the two distinct factorizations into irreducible factors 21 = 3 7 = ( ) (1 2 5). A domain in which we have unique factorization into irreducibles is called a unique factorization domain, abbreviated UFD, or a factorial domain. In these we sometimes call irreducible elements primes, but bear in mind that with this definition +2 and 2 are both rational primes. It is a basic theorem (and not too difficult) that Z[i] is a factorial domain. Obviously in a factorial domain if an irreducible element divides a product then it must divide one of the factors; in fact, this is a crucial property that one must prove to show that a domain is factorial. If R is a factorial domain and π an irreducible element of R then the quotient ring R/π is again a domain and conversely. On the other hand Z[ 5]/3 is not a domain since (1+2 5) (1 2 5) = mod 3. We will prove that Z[i] is factorial, but for the moment let s accept it. Lemma 2 A finite domain R is a field Proof. We must show that every nonzero element x R has an inverse. Consider the set of all xy as y varies in R. No two of these can be identical, for if xy = xy then x(y y ) = 0, contradicting the assumption that R is a domain. Since R is finite, the set of all xy with fixed x must be all of R, so there is a y such that xy = 1. Exercise 7 We really don t need the commutativity in the preceding Lemma or even the existence of a unit element. Prove that if S is a finite set with an associative multiplication with both the left and right cancelation property, i.e., xy = xy implies y = y and yx = y x also implies y = y then S is in fact a group (and in particular, there is a unit element for multiplication). (Hard) Suppose only one of the two cancelation properties holds. Is S still a group? (Give a proof or a counterexample, but don t spend too much time on it.) Finally (assuming some of the things we have not yet proven) we have the proof of Theorem 1: Proof. To prove assertion 1. we must show that if p 1(4) then p factors in Z[i]. Suppose to the contrary that it remained irreducible. Since Z[i] is a factorial domain it would follow that Z[i]/p is again a domain, with exactly p 2 elements, and being finite it would be a field. The equation x 4 = 1 could then have no more than four roots in Z[i]/p. But Z[i]/p Z/p and the latter is a field with exactly p elements. Its multiplicative group is therefore a cyclic 8
9 group with p 1 elements, and this is a multiple of 4, say p 1 = 4m. If a is any generator of this group then 1, a m, a 2m and a 3m are four distinct elements satisfying x 4 = 1. But (the classes of) i and i are not amongst these and also satisfy the equation, so there are too many roots. Therefore Z[i]/p can not be a field, so p must factor. For 2., suppose that m is a sum of two squares, hence of the form m = (a + bi)(a bi) for some a, b Z and that a prime p 3(4) divides m. Since p is still irreducible in Z[i] it must divide one of the two factors. Suppose p k (a + bi) (meaning that k is the precise power to which p divides a+bi). Since p k is real, taking conjugates we see that also p k (a bi) so p 2k m. So one way to understand the if (hard) part of Fermat s original assertion, that an odd prime p is a sum of two squares if and only if p 1(4), is to say that such a prime must factor in Z[i]. There are many details that must be filled in; that will be our next project. Exercise 8 Show that if a, b are integers then Z/(a + bi) always has exactly a 2 + b 2 elements, and if a, b are relatively prime then there is an isomorphism Z/(a 2 +b 2 ) Z/(a+bi) but not otherwise. What is the structure of the additive group of Z/(a + bi) when a and b are not relatively prime? (Hint: You might want to do the relatively prime case first.) 9
Rings. Chapter 1. Definition 1.2. A commutative ring R is a ring in which multiplication is commutative. That is, ab = ba for all a, b R.
Chapter 1 Rings We have spent the term studying groups. A group is a set with a binary operation that satisfies certain properties. But many algebraic structures such as R, Z, and Z n come with two binary
More informationHonors Algebra 4, MATH 371 Winter 2010 Assignment 3 Due Friday, February 5 at 08:35
Honors Algebra 4, MATH 371 Winter 2010 Assignment 3 Due Friday, February 5 at 08:35 1. Let R 0 be a commutative ring with 1 and let S R be the subset of nonzero elements which are not zero divisors. (a)
More informationALGEBRA II: RINGS AND MODULES OVER LITTLE RINGS.
ALGEBRA II: RINGS AND MODULES OVER LITTLE RINGS. KEVIN MCGERTY. 1. RINGS The central characters of this course are algebraic objects known as rings. A ring is any mathematical structure where you can add
More informationHomework 10 M 373K by Mark Lindberg (mal4549)
Homework 10 M 373K by Mark Lindberg (mal4549) 1. Artin, Chapter 11, Exercise 1.1. Prove that 7 + 3 2 and 3 + 5 are algebraic numbers. To do this, we must provide a polynomial with integer coefficients
More informationSUPPLEMENTARY NOTES: CHAPTER 1
SUPPLEMENTARY NOTES: CHAPTER 1 1. Groups A group G is a set with single binary operation which takes two elements a, b G and produces a third, denoted ab and generally called their product. (Mathspeak:
More informationChapter 3. Rings. The basic commutative rings in mathematics are the integers Z, the. Examples
Chapter 3 Rings Rings are additive abelian groups with a second operation called multiplication. The connection between the two operations is provided by the distributive law. Assuming the results of Chapter
More informationAlgebraic structures I
MTH5100 Assignment 110 Algebraic structures I For handing in on various dates January March 2011 1 FUNCTIONS. Say which of the following rules successfully define functions, giving reasons. For each one
More informationA Generalization of Wilson s Theorem
A Generalization of Wilson s Theorem R. Andrew Ohana June 3, 2009 Contents 1 Introduction 2 2 Background Algebra 2 2.1 Groups................................. 2 2.2 Rings.................................
More informationExercises on chapter 1
Exercises on chapter 1 1. Let G be a group and H and K be subgroups. Let HK = {hk h H, k K}. (i) Prove that HK is a subgroup of G if and only if HK = KH. (ii) If either H or K is a normal subgroup of G
More informationAlgebra Review. Instructor: Laszlo Babai Notes by Vincent Lucarelli and the instructor. June 15, 2001
Algebra Review Instructor: Laszlo Babai Notes by Vincent Lucarelli and the instructor June 15, 2001 1 Groups Definition 1.1 A semigroup (G, ) is a set G with a binary operation such that: Axiom 1 ( a,
More informationPRACTICE FINAL MATH , MIT, SPRING 13. You have three hours. This test is closed book, closed notes, no calculators.
PRACTICE FINAL MATH 18.703, MIT, SPRING 13 You have three hours. This test is closed book, closed notes, no calculators. There are 11 problems, and the total number of points is 180. Show all your work.
More informationMATH 326: RINGS AND MODULES STEFAN GILLE
MATH 326: RINGS AND MODULES STEFAN GILLE 1 2 STEFAN GILLE 1. Rings We recall first the definition of a group. 1.1. Definition. Let G be a non empty set. The set G is called a group if there is a map called
More informationCHAPTER I. Rings. Definition A ring R is a set with two binary operations, addition + and
CHAPTER I Rings 1.1 Definitions and Examples Definition 1.1.1. A ring R is a set with two binary operations, addition + and multiplication satisfying the following conditions for all a, b, c in R : (i)
More informationDIHEDRAL GROUPS II KEITH CONRAD
DIHEDRAL GROUPS II KEITH CONRAD We will characterize dihedral groups in terms of generators and relations, and describe the subgroups of D n, including the normal subgroups. We will also introduce an infinite
More informationGroup, Rings, and Fields Rahul Pandharipande. I. Sets Let S be a set. The Cartesian product S S is the set of ordered pairs of elements of S,
Group, Rings, and Fields Rahul Pandharipande I. Sets Let S be a set. The Cartesian product S S is the set of ordered pairs of elements of S, A binary operation φ is a function, S S = {(x, y) x, y S}. φ
More information1. Group Theory Permutations.
1.1. Permutations. 1. Group Theory Problem 1.1. Let G be a subgroup of S n of index 2. Show that G = A n. Problem 1.2. Find two elements of S 7 that have the same order but are not conjugate. Let π S 7
More information1 Fields and vector spaces
1 Fields and vector spaces In this section we revise some algebraic preliminaries and establish notation. 1.1 Division rings and fields A division ring, or skew field, is a structure F with two binary
More informationφ(xy) = (xy) n = x n y n = φ(x)φ(y)
Groups 1. (Algebra Comp S03) Let A, B and C be normal subgroups of a group G with A B. If A C = B C and AC = BC then prove that A = B. Let b B. Since b = b1 BC = AC, there are a A and c C such that b =
More informationMATH 403 MIDTERM ANSWERS WINTER 2007
MAH 403 MIDERM ANSWERS WINER 2007 COMMON ERRORS (1) A subset S of a ring R is a subring provided that x±y and xy belong to S whenever x and y do. A lot of people only said that x + y and xy must belong
More informationφ(a + b) = φ(a) + φ(b) φ(a b) = φ(a) φ(b),
16. Ring Homomorphisms and Ideals efinition 16.1. Let φ: R S be a function between two rings. We say that φ is a ring homomorphism if for every a and b R, and in addition φ(1) = 1. φ(a + b) = φ(a) + φ(b)
More informationMath 120 HW 9 Solutions
Math 120 HW 9 Solutions June 8, 2018 Question 1 Write down a ring homomorphism (no proof required) f from R = Z[ 11] = {a + b 11 a, b Z} to S = Z/35Z. The main difficulty is to find an element x Z/35Z
More informationMath 2070BC Term 2 Weeks 1 13 Lecture Notes
Math 2070BC 2017 18 Term 2 Weeks 1 13 Lecture Notes Keywords: group operation multiplication associative identity element inverse commutative abelian group Special Linear Group order infinite order cyclic
More informationGroups. 3.1 Definition of a Group. Introduction. Definition 3.1 Group
C H A P T E R t h r e E Groups Introduction Some of the standard topics in elementary group theory are treated in this chapter: subgroups, cyclic groups, isomorphisms, and homomorphisms. In the development
More informationSolutions of exercise sheet 8
DMATH Algebra I HS 14 Prof. Emmanuel Kowalski Solutions of exercise sheet 8 1. In this exercise, we will give a characterization for solvable groups using commutator subgroups. See last semester s (Algebra
More informationALGEBRA PH.D. QUALIFYING EXAM September 27, 2008
ALGEBRA PH.D. QUALIFYING EXAM September 27, 2008 A passing paper consists of four problems solved completely plus significant progress on two other problems; moreover, the set of problems solved completely
More informationMATH 3030, Abstract Algebra Winter 2012 Toby Kenney Sample Midterm Examination Model Solutions
MATH 3030, Abstract Algebra Winter 2012 Toby Kenney Sample Midterm Examination Model Solutions Basic Questions 1. Give an example of a prime ideal which is not maximal. In the ring Z Z, the ideal {(0,
More informationRINGS: SUMMARY OF MATERIAL
RINGS: SUMMARY OF MATERIAL BRIAN OSSERMAN This is a summary of terms used and main results proved in the subject of rings, from Chapters 1113 of Artin. Definitions not included here may be considered
More informationAbstract Algebra II. Randall R. Holmes Auburn University. Copyright c 2008 by Randall R. Holmes Last revision: November 7, 2017
Abstract Algebra II Randall R. Holmes Auburn University Copyright c 2008 by Randall R. Holmes Last revision: November 7, 2017 This work is licensed under the Creative Commons Attribution NonCommercialNoDerivatives
More informationTHE JOHNS HOPKINS UNIVERSITY Faculty of Arts and Sciences FINAL EXAM  FALL SESSION ADVANCED ALGEBRA I.
THE JOHNS HOPKINS UNIVERSITY Faculty of Arts and Sciences FINAL EXAM  FALL SESSION 2006 110.401  ADVANCED ALGEBRA I. Examiner: Professor C. Consani Duration: take home final. No calculators allowed.
More informationA field F is a set of numbers that includes the two numbers 0 and 1 and satisfies the properties:
Byte multiplication 1 Field arithmetic A field F is a set of numbers that includes the two numbers 0 and 1 and satisfies the properties: F is an abelian group under addition, meaning  F is closed under
More informationAbstract Algebra II. Randall R. Holmes Auburn University
Abstract Algebra II Randall R. Holmes Auburn University Copyright c 2008 by Randall R. Holmes Last revision: November 30, 2009 Contents 0 Introduction 2 1 Definition of ring and examples 3 1.1 Definition.............................
More informationCourse 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra
Course 311: Michaelmas Term 2005 Part III: Topics in Commutative Algebra D. R. Wilkins Contents 3 Topics in Commutative Algebra 2 3.1 Rings and Fields......................... 2 3.2 Ideals...............................
More informationPh.D. Qualifying Examination in Algebra Department of Mathematics University of Louisville January 2018
Ph.D. Qualifying Examination in Algebra Department of Mathematics University of Louisville January 2018 Do 6 problems with at least 2 in each section. Group theory problems: (1) Suppose G is a group. The
More information(January 14, 2009) q n 1 q d 1. D = q n = q + d
(January 14, 2009) [10.1] Prove that a finite division ring D (a notnecessarily commutative ring with 1 in which any nonzero element has a multiplicative inverse) is commutative. (This is due to Wedderburn.)
More informationAbstract Algebra II Groups ( )
Abstract Algebra II Groups ( ) Melchior Grützmann / melchiorgfreehostingcom/algebra October 15, 2012 Outline Group homomorphisms Free groups, free products, and presentations Free products ( ) Definition
More information9. Integral Ring Extensions
80 Andreas Gathmann 9. Integral ing Extensions In this chapter we want to discuss a concept in commutative algebra that has its original motivation in algebra, but turns out to have surprisingly many applications
More informationNOTES ON FINITE FIELDS
NOTES ON FINITE FIELDS AARON LANDESMAN CONTENTS 1. Introduction to finite fields 2 2. Definition and constructions of fields 3 2.1. The definition of a field 3 2.2. Constructing field extensions by adjoining
More information18. Cyclotomic polynomials II
18. Cyclotomic polynomials II 18.1 Cyclotomic polynomials over Z 18.2 Worked examples Now that we have Gauss lemma in hand we can look at cyclotomic polynomials again, not as polynomials with coefficients
More information(Rgs) Rings Math 683L (Summer 2003)
(Rgs) Rings Math 683L (Summer 2003) We will first summarise the general results that we will need from the theory of rings. A unital ring, R, is a set equipped with two binary operations + and such that
More informationIUPUI Qualifying Exam Abstract Algebra
IUPUI Qualifying Exam Abstract Algebra January 2017 Daniel Ramras (1) a) Prove that if G is a group of order 2 2 5 2 11, then G contains either a normal subgroup of order 11, or a normal subgroup of order
More informationSolutions to oddnumbered exercises Peter J. Cameron, Introduction to Algebra, Chapter 3
Solutions to oddnumbered exercises Peter J. Cameron, Introduction to Algebra, Chapter 3 3. (a) Yes; (b) No; (c) No; (d) No; (e) Yes; (f) Yes; (g) Yes; (h) No; (i) Yes. Comments: (a) is the additive group
More informationTheorem 5.3. Let E/F, E = F (u), be a simple field extension. Then u is algebraic if and only if E/F is finite. In this case, [E : F ] = deg f u.
5. Fields 5.1. Field extensions. Let F E be a subfield of the field E. We also describe this situation by saying that E is an extension field of F, and we write E/F to express this fact. If E/F is a field
More informationAlgebra SEP Solutions
Algebra SEP Solutions 17 July 2017 1. (January 2017 problem 1) For example: (a) G = Z/4Z, N = Z/2Z. More generally, G = Z/p n Z, N = Z/pZ, p any prime number, n 2. Also G = Z, N = nz for any n 2, since
More informationHonors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35
Honors Algebra 4, MATH 371 Winter 2010 Assignment 4 Due Wednesday, February 17 at 08:35 1. Let R be a commutative ring with 1 0. (a) Prove that the nilradical of R is equal to the intersection of the prime
More information2 Lecture 2: Logical statements and proof by contradiction Lecture 10: More on Permutations, Group Homomorphisms 31
Contents 1 Lecture 1: Introduction 2 2 Lecture 2: Logical statements and proof by contradiction 7 3 Lecture 3: Induction and WellOrdering Principle 11 4 Lecture 4: Definition of a Group and examples 15
More informationSome practice problems for midterm 2
Some practice problems for midterm 2 Kiumars Kaveh November 14, 2011 Problem: Let Z = {a G ax = xa, x G} be the center of a group G. Prove that Z is a normal subgroup of G. Solution: First we prove Z is
More information(1) A frac = b : a, b A, b 0. We can define addition and multiplication of fractions as we normally would. a b + c d
The Algebraic Method 0.1. Integral Domains. Emmy Noether and others quickly realized that the classical algebraic number theory of Dedekind could be abstracted completely. In particular, rings of integers
More information1. multiplication is commutative and associative;
Chapter 4 The Arithmetic of Z In this chapter, we start by introducing the concept of congruences; these are used in our proof (going back to Gauss 1 ) that every integer has a unique prime factorization.
More informationSUMMARY OF GROUPS AND RINGS GROUPS AND RINGS III Week 1 Lecture 1 Tuesday 3 March.
SUMMARY OF GROUPS AND RINGS GROUPS AND RINGS III 2009 Week 1 Lecture 1 Tuesday 3 March. 1. Introduction (Background from Algebra II) 1.1. Groups and Subgroups. Definition 1.1. A binary operation on a set
More informationCOURSE SUMMARY FOR MATH 504, FALL QUARTER : MODERN ALGEBRA
COURSE SUMMARY FOR MATH 504, FALL QUARTER 20178: MODERN ALGEBRA JAROD ALPER Week 1, Sept 27, 29: Introduction to Groups Lecture 1: Introduction to groups. Defined a group and discussed basic properties
More information* 8 Groups, with Appendix containing Rings and Fields.
* 8 Groups, with Appendix containing Rings and Fields Binary Operations Definition We say that is a binary operation on a set S if, and only if, a, b, a b S Implicit in this definition is the idea that
More informationEighth Homework Solutions
Math 4124 Wednesday, April 20 Eighth Homework Solutions 1. Exercise 5.2.1(e). Determine the number of nonisomorphic abelian groups of order 2704. First we write 2704 as a product of prime powers, namely
More informationGroups, Rings, and Finite Fields. Andreas Klappenecker. September 12, 2002
Background on Groups, Rings, and Finite Fields Andreas Klappenecker September 12, 2002 A thorough understanding of the Agrawal, Kayal, and Saxena primality test requires some tools from algebra and elementary
More informationFermat s Last Theorem for Regular Primes
Fermat s Last Theorem for Regular Primes S. M.C. 22 September 2015 Abstract Fermat famously claimed in the margin of a book that a certain family of Diophantine equations have no solutions in integers.
More informationRings If R is a commutative ring, a zero divisor is a nonzero element x such that xy = 0 for some nonzero element y R.
Rings 10262008 A ring is an abelian group R with binary operation + ( addition ), together with a second binary operation ( multiplication ). Multiplication must be associative, and must distribute over
More informationAlgebraic Structures Exam File Fall 2013 Exam #1
Algebraic Structures Exam File Fall 2013 Exam #1 1.) Find all four solutions to the equation x 4 + 16 = 0. Give your answers as complex numbers in standard form, a + bi. 2.) Do the following. a.) Write
More informationMT5836 Galois Theory MRQ
MT5836 Galois Theory MRQ May 3, 2017 Contents Introduction 3 Structure of the lecture course............................... 4 Recommended texts..................................... 4 1 Rings, Fields and
More informationHomework problems from Chapters IVVI: answers and solutions
Homework problems from Chapters IVVI: answers and solutions IV.21.1. In this problem we have to describe the field F of quotients of the domain D. Note that by definition, F is the set of equivalence
More informationWHY WORD PROBLEMS ARE HARD
WHY WORD PROBLEMS ARE HARD KEITH CONRAD 1. Introduction The title above is a joke. Many students in school hate word problems. We will discuss here a specific math question that happens to be named the
More informationAlgebra Homework, Edition 2 9 September 2010
Algebra Homework, Edition 2 9 September 2010 Problem 6. (1) Let I and J be ideals of a commutative ring R with I + J = R. Prove that IJ = I J. (2) Let I, J, and K be ideals of a principal ideal domain.
More informationQuasireducible Polynomials
Quasireducible Polynomials Jacques Willekens 06Dec2008 Abstract In this article, we investigate polynomials that are irreducible over Q, but are reducible modulo any prime number. 1 Introduction Let
More informationCSIR  Algebra Problems
CSIR  Algebra Problems N. Annamalai DST  INSPIRE Fellow (SRF) Department of Mathematics Bharathidasan University Tiruchirappalli 620024 Email: algebra.annamalai@gmail.com Website: https://annamalaimaths.wordpress.com
More informationEXERCISES. a b = a + b l aq b = ab  (a + b) + 2. a b = a + b + 1 n0i) = oii + ii + fi. A. Examples of Rings. C. Ring of 2 x 2 Matrices
/ rings definitions and elementary properties 171 EXERCISES A. Examples of Rings In each of the following, a set A with operations of addition and multiplication is given. Prove that A satisfies all the
More informationSolutions to oddnumbered exercises Peter J. Cameron, Introduction to Algebra, Chapter 2
Solutions to oddnumbered exercises Peter J Cameron, Introduction to Algebra, Chapter 1 The answers are a No; b No; c Yes; d Yes; e No; f Yes; g Yes; h No; i Yes; j No a No: The inverse law for addition
More informationDefinitions. Notations. Injective, Surjective and Bijective. Divides. Cartesian Product. Relations. Equivalence Relations
Page 1 Definitions Tuesday, May 8, 2018 12:23 AM Notations " " means "equals, by definition" the set of all real numbers the set of integers Denote a function from a set to a set by Denote the image of
More informationTROPICAL SCHEME THEORY
TROPICAL SCHEME THEORY 5. Commutative algebra over idempotent semirings II Quotients of semirings When we work with rings, a quotient object is specified by an ideal. When dealing with semirings (and lattices),
More information8. Prime Factorization and Primary Decompositions
70 Andreas Gathmann 8. Prime Factorization and Primary Decompositions 13 When it comes to actual computations, Euclidean domains (or more generally principal ideal domains) are probably the nicest rings
More informationTHE GROUP OF UNITS OF SOME FINITE LOCAL RINGS I
J Korean Math Soc 46 (009), No, pp 95 311 THE GROUP OF UNITS OF SOME FINITE LOCAL RINGS I Sung Sik Woo Abstract The purpose of this paper is to identify the group of units of finite local rings of the
More informationModern Algebra Prof. Manindra Agrawal Department of Computer Science and Engineering Indian Institute of Technology, Kanpur
Modern Algebra Prof. Manindra Agrawal Department of Computer Science and Engineering Indian Institute of Technology, Kanpur Lecture  05 Groups: Structure Theorem So, today we continue our discussion forward.
More informationMath 451, 01, Exam #2 Answer Key
Math 451, 01, Exam #2 Answer Key 1. (25 points): If the statement is always true, circle True and prove it. If the statement is never true, circle False and prove that it can never be true. If the statement
More informationMath 429/581 (Advanced) Group Theory. Summary of Definitions, Examples, and Theorems by Stefan Gille
Math 429/581 (Advanced) Group Theory Summary of Definitions, Examples, and Theorems by Stefan Gille 1 2 0. Group Operations 0.1. Definition. Let G be a group and X a set. A (left) operation of G on X is
More informationMath Introduction to Modern Algebra
Math 343  Introduction to Modern Algebra Notes Rings and Special Kinds of Rings Let R be a (nonempty) set. R is a ring if there are two binary operations + and such that (A) (R, +) is an abelian group.
More information1. Factorization Divisibility in Z.
8 J. E. CREMONA 1.1. Divisibility in Z. 1. Factorization Definition 1.1.1. Let a, b Z. Then we say that a divides b and write a b if b = ac for some c Z: a b c Z : b = ac. Alternatively, we may say that
More informationSUMMARY ALGEBRA I LOUISPHILIPPE THIBAULT
SUMMARY ALGEBRA I LOUISPHILIPPE THIBAULT Contents 1. Group Theory 1 1.1. Basic Notions 1 1.2. Isomorphism Theorems 2 1.3. Jordan Holder Theorem 2 1.4. Symmetric Group 3 1.5. Group action on Sets 3 1.6.
More informationECEN 5022 Cryptography
Elementary Algebra and Number Theory University of Colorado Spring 2008 Divisibility, Primes Definition. N denotes the set {1, 2, 3,...} of natural numbers and Z denotes the set of integers {..., 2, 1,
More informationYale University Department of Mathematics Math 350 Introduction to Abstract Algebra Fall Midterm Exam Review Solutions
Yale University Department of Mathematics Math 350 Introduction to Abstract Algebra Fall 2015 Midterm Exam Review Solutions Practice exam questions: 1. Let V 1 R 2 be the subset of all vectors whose slope
More information5 Group theory. 5.1 Binary operations
5 Group theory This section is an introduction to abstract algebra. This is a very useful and important subject for those of you who will continue to study pure mathematics. 5.1 Binary operations 5.1.1
More informationA connection between number theory and linear algebra
A connection between number theory and linear algebra Mark Steinberger Contents 1. Some basics 1 2. Rational canonical form 2 3. Prime factorization in F[x] 4 4. Units and order 5 5. Finite fields 7 6.
More information1 2 3 style total. Circle the correct answer; no explanation is required. Each problem in this section counts 5 points.
1 2 3 style total Math 415 Examination 3 Please print your name: Answer Key 1 True/false Circle the correct answer; no explanation is required. Each problem in this section counts 5 points. 1. The rings
More informationMath 121 Homework 5: Notes on Selected Problems
Math 121 Homework 5: Notes on Selected Problems 12.1.2. Let M be a module over the integral domain R. (a) Assume that M has rank n and that x 1,..., x n is any maximal set of linearly independent elements
More informationIntroduction to Groups
Introduction to Groups HongJian Lai August 2000 1. Basic Concepts and Facts (1.1) A semigroup is an ordered pair (G, ) where G is a nonempty set and is a binary operation on G satisfying: (G1) a (b c)
More information0 Sets and Induction. Sets
0 Sets and Induction Sets A set is an unordered collection of objects, called elements or members of the set. A set is said to contain its elements. We write a A to denote that a is an element of the set
More informationFactorization in Polynomial Rings
Factorization in Polynomial Rings Throughout these notes, F denotes a field. 1 Long division with remainder We begin with some basic definitions. Definition 1.1. Let f, g F [x]. We say that f divides g,
More informationMODEL ANSWERS TO HWK #10
MODEL ANSWERS TO HWK #10 1. (i) As x + 4 has degree one, either it divides x 3 6x + 7 or these two polynomials are coprime. But if x + 4 divides x 3 6x + 7 then x = 4 is a root of x 3 6x + 7, which it
More informationCDM. Finite Fields. Klaus Sutner Carnegie Mellon University. Fall 2018
CDM Finite Fields Klaus Sutner Carnegie Mellon University Fall 2018 1 Ideals The Structure theorem Where Are We? 3 We know that every finite field carries two apparently separate structures: additive and
More informationCourse 2316 Sample Paper 1
Course 2316 Sample Paper 1 Timothy Murphy April 19, 2015 Attempt 5 questions. All carry the same mark. 1. State and prove the Fundamental Theorem of Arithmetic (for N). Prove that there are an infinity
More informationRing Theory Problem Set 2 Solutions
Ring Theory Problem Set 2 Solutions 16.24. SOLUTION: We already proved in class that Z[i] is a commutative ring with unity. It is the smallest subring of C containing Z and i. If r = a + bi is in Z[i],
More informationFROM GROUPS TO GALOIS Amin Witno
WON Series in Discrete Mathematics and Modern Algebra Volume 6 FROM GROUPS TO GALOIS Amin Witno These notes 1 have been prepared for the students at Philadelphia University (Jordan) who are taking the
More informationMath Introduction to Modern Algebra
Math 343  Introduction to Modern Algebra Notes Field Theory Basics Let R be a ring. M is called a maximal ideal of R if M is a proper ideal of R and there is no proper ideal of R that properly contains
More informationMATH 3030, Abstract Algebra FALL 2012 Toby Kenney Midyear Examination Friday 7th December: 7:0010:00 PM
MATH 3030, Abstract Algebra FALL 2012 Toby Kenney Midyear Examination Friday 7th December: 7:0010:00 PM Basic Questions 1. Compute the factor group Z 3 Z 9 / (1, 6). The subgroup generated by (1, 6) is
More informationGroups and Symmetries
Groups and Symmetries Definition: Symmetry A symmetry of a shape is a rigid motion that takes vertices to vertices, edges to edges. Note: A rigid motion preserves angles and distances. Definition: Group
More information32 Divisibility Theory in Integral Domains
3 Divisibility Theory in Integral Domains As we have already mentioned, the ring of integers is the prototype of integral domains. There is a divisibility relation on * : an integer b is said to be divisible
More information9. Finite fields. 1. Uniqueness
9. Finite fields 9.1 Uniqueness 9.2 Frobenius automorphisms 9.3 Counting irreducibles 1. Uniqueness Among other things, the following result justifies speaking of the field with p n elements (for prime
More informationRings. EE 387, Notes 7, Handout #10
Rings EE 387, Notes 7, Handout #10 Definition: A ring is a set R with binary operations, + and, that satisfy the following axioms: 1. (R, +) is a commutative group (five axioms) 2. Associative law for
More informationModern Algebra I. Circle the correct answer; no explanation is required. Each problem in this section counts 5 points.
1 2 3 style total Math 415 Please print your name: Answer Key 1 True/false Circle the correct answer; no explanation is required. Each problem in this section counts 5 points. 1. Every group of order 6
More informationFinite Fields. Sophie Huczynska. Semester 2, Academic Year
Finite Fields Sophie Huczynska Semester 2, Academic Year 200506 2 Chapter 1. Introduction Finite fields is a branch of mathematics which has come to the fore in the last 50 years due to its numerous applications,
More information5.1 Commutative rings; Integral Domains
5.1 J.A.Beachy 1 5.1 Commutative rings; Integral Domains from A Study Guide for Beginner s by J.A.Beachy, a supplement to Abstract Algebra by Beachy / Blair 23. Let R be a commutative ring. Prove the following
More informationAbstract Algebra, Second Edition, by John A. Beachy and William D. Blair. Corrections and clarifications
1 Abstract Algebra, Second Edition, by John A. Beachy and William D. Blair Corrections and clarifications Note: Some corrections were made after the first printing of the text. page 9, line 8 For of the
More informationRecall, R is an integral domain provided: R is a commutative ring If ab = 0 in R, then either a = 0 or b = 0.
Recall, R is an integral domain provided: R is a commutative ring If ab = 0 in R, then either a = 0 or b = 0. Examples: Z Q, R Polynomials over Z, Q, R, C The Gaussian Integers: Z[i] := {a + bi : a, b
More informationRUDIMENTARY GALOIS THEORY
RUDIMENTARY GALOIS THEORY JACK LIANG Abstract. This paper introduces basic Galois Theory, primarily over fields with characteristic 0, beginning with polynomials and fields and ultimately relating the
More information