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A Book of Abstract Algebra
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Algebra at the more advanced level is often described as modern or abstract algebra. In fact, bothof these descriptions are partly misleading. Some of the great discoveries in the upper reaches of presentday algebra (for example, the so-called Galois theory) were known many years before the American Civil War ; and the broad aims of algebra today were clearly stated by Leibniz in the seventeenth century. Thus, “modern” algebra is not so very modern, after all! To what extent is it abstract? Well, abstraction is allrelative; one person’s abstraction is another person’s bread and butter. The abstract tendency inmathematics is a little like the situation of changing moral codes, or changing tastes in music: What shocksone generation becomes the norm in the next. This has been true throughout the history of mathematics.For example, 1000 years ago negative numbers were considered to be an outrageous idea. After all,it was said, numbers are for counting: we may have one orange, or two oranges, or no oranges at all; buthow can we have minus an orange? The logisticians, or professional calculators, of those days usednegative numbers as an aid in their computations; they considered these numbers to be a useful fiction, forif you believe in them then every linear equation ax + b = 0 has a solution (namely x = −b/a, provided a ≠0). Even the great Diophantus once described the solution of 4x + 6 = 2 as an absurd number. The idea ofa system of numeration which included negative numbers was far too abstract for many of the learnedheads of the tenth century!

F. Elementary Properties of Homomorphisms Let A and B be rings, and f : A B a homomorphism. Prove each of the following: 1 f(A) = {f(x): x A} is a subring of B. 2 The kernel of f is an ideal of A. 3 f(0) = 0, and for every a A, f(a) = f(a). 4 f is injective iff its kernel is equal to {0}. 5 If B is an integral domain, then either f(l) = 1 or f(l) = 0. If f(l) = 0, then f(x) = 0 for every x A. If f(1) = 1, the image of every invertible element of A is an invertible element of B. 6 Any homomorphic image of a commutative ring is a commutative ring. Any homomorphic image of a field is a field. 7 If the domain A of the homomorphism f is a field, and if the range of f has more than one element, then f is injective. (HINT: Use Exercise D6.)
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G. Examples of Isomorphisms 1 Let A be the ring of Exercise A2 in Chapter 17. Show that the function f(x) = x 1 is an isomorphism from 2 Let A. to A hence be the following subset of 2( ): Prove that the function is an isomorphism from . [REMARK: You must begin by checking that f is a well-defined function; that is, if a + bi = c + di, then f(a + bi) = f(c + di). To do this, note that if a + bi = c + di then a c = (d b)i; this last equation is impossible unless both sides are equal to zero, for otherwise it would assert that a given real number is equal to an imaginary number.] 3 Prove that {(x, x) : x } is a subring of , and show {(x, x) : x } . 4 Show that the set of all 2 2 matrices of the form
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5 Prove that 2 then prove that 2 3 . Finally, explain why if k l, then k l . (REMEMBER: How do you show that two rings, or groups, are not isomorphic?) . H. Further Properties of Ideals Let A be a ring, and let J and K be ideals of A. Prove parts 1-4. (In parts 2-4, assume A is a commutative ring.) 1 If J 2 For any a A, Ia = {ax + j + k : x A, j J, k K} is an ideal of A. # 3 The radical of J is the set rad J = {a A : an J for some n }. For any ideal J, rad J is an ideal K = {0}, then jk = 0 for every j J and k K.
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4 For any a A, {x A : ax = 0} is an ideal (called the annihilator of a). Furthermore, {x A : ax = 0 for every a A} is an ideal (called the annihilating ideal of A). If A is a ring with unity, its annihilating ideal is equal to {0}. 5 Show that {0} and A are ideals of A. (They are trivial ideals; every other ideal of A is a proper ideal.) A proper ideal J of A is called maximal if it is not strictly contained in any strictly larger proper ideal: that is, if J K, where K is an ideal containing some element not in J, then necessarily K = A. Show that the following is an example of a maximal ideal: In ( ), the ideal J = {f : f(0) = 0}. [HINT: Use Exercise D5. Note that if g K and g(0) 0 (that is, g J), then the function h(x) = g(x) g(0) is in J hence h(x) g(x) K. Explain why this last function is an invertible element of ( ).]
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