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Unformatted text preview: “errata” — 2005/8/23 — 15:21 — page 1 — #1 a105 a105 a105 a105 a105 a105 a105 a105 Errata Goodman, Algebra: Abstract and Concrete, 2nd ed. • Page 6: Comments on the paragraph following figure 1.2.5: The “centroid” of the square is the center of mass; it is the intersection of the two diagonals. Consider an axis joining two opposite vertices of the square, or the centers of two opposite edges. The figure can be rotated by by 180 degrees ( π radians) around such an axis. Such a rotation flips the figure over the axis, exchanging top and bottom. If you flip the figure twice over the same axis, you return it to its original position. It would be convenient to refer to these motions as “flips”. • Page 26: Add two properties to the list of known properties of the integers: 1. The product of two nonzero integers is nonzero. 2. For all integers a , b ,  ab  ≥ max { a  ,  b } . • Page 35, Exercise 1.6.3: Suppose that the natural number p > 1 has the property ... • Page 38, last line: This shows that (d) implies (a). • Page 49, proof of 1.8.12: Write p = a n x n + a n 1 x n 1 + ··· + a . • Page 51, second line of Remark 1.8.17: . . . to produce an algo rithm . . . • Page 54, exercise 1.8.5: Let h be a nonzero element of I ( f , g ) of least degree. • Page 69, Formally, a product or operation on a set G is a function from G × G to G . For example, the operation of addition on Z is the function on Z × Z whose value at ( a , b ) is a + b . • Page 74, Exercise 1.10.9: Show that the set of affine transforma tions of R n . . . • Page 75, Before definition 1.11.1: Again, it is fruitful . . . • The discussion of the RSA algorithm on pages 8081 is incom plete. Replace Lemma 1.12.1 with the following: Lemma 1.12.1. For all integers a and h , if h ≡ 1 ( mod m ) , then a h ≡ a ( mod n ) . Proof. Write h = tm + 1. Then a h = aa tm , so a h a = a ( a tm 1 ) . We have to show that a h a is divisible by n . If q does not divide a , then a is relatively prime to q , so a q 1 ≡ 1 ( mod q ) , by Fermat’s little theorem, Proposition ?? . Since ( q 1 ) divides tm , it follows that a tm ≡ 1 ( mod q ) ; that is q divides a tm 1. 1 “errata” — 2005/8/23 — 15:21 — page 2 — #2 a105 a105 a105 a105 a105 a105 a105 a105 2 Thus, either q divides a , or q divides a tm 1, so q divides a h a = a ( a tm 1 ) in any case. Similarly, p divides a h a . But then a h a is divisible by both p and q , and hence by n = pq = l.c.m. ( p , q ) . a73 Now Lemma 1.12.2 is valid without the hypothesis that a is rel atively prime to n . Consequently, in the paragraphs following this lemma, we can eliminate the remark restricting a to be less than n , and the comments about a being relatively prime to n : the proce dure is valid for an arbitrary integer a ....
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This note was uploaded on 11/14/2011 for the course MATH 367 taught by Professor Sdd during the Spring '11 term at Middle East Technical University.
 Spring '11
 sdd
 Algebra

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