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Remarks and Errata to the Third Edition of
"A Concrete Introduction to Higher Algebra", by Lindsay N. Childs, published by
Springer-Verlag New York, 2008 (first edition, 1979, second edition, 1995).

My thanks to Ted Turner and Marco Varisco of Albany, Herb Enderton of UCLA, Keith Conrad
and John Watterlond
of the Univ.
of Connecticut, John McKay of Concordia, Dmitry Gokhman of the Univ. of Texas at San
Antonio, Dr. Andrew Mauer-Oats of Evanston Township High School, Julian Wilson of the
Hanze Institute of Technology (Groningen), UAlbany
students Heather Cavelius, Anh Ngoc Le, Matt Patrick and Yueyi Zheng, and reviewers Ken
Brown (MR) and Rade Dimitric (ZB),
for providing errata for this edition.
Errors that if unnoticed could lead to significant confusion are starred.
p. x, line -2: "A course in Number Theory could follow chapters..." (Rade Dimitric)
*p. 27, in Theorem 1, replace "q > 0" by "q \ge 0" (that is, "q greater than or equal to 0"). (Ted Turner)
p. 56, Exercise 5. Replace "< sqrt(n)" by "less than or equal to sqrt(n)". (The square of a prime is a counterexample to <.) (Dmitry Gokhman)
*pp. 78-79. In example 6, replace the number 3589 by 3489 everywhere, and on page 79, line 11, adjust the congruence accordingly by
replacing 5 by 4. (Ken Brown ("the Glasgow one"))
*p. 113, Exercise 54, the domain of the function f_b should be Z/mZ. The function may be one-to-one on U_m even if b is not a unit, as Dmitry Gokhman pointed out--for
example, m = 6, b= 2. It is a nice problem to try to find all moduli m for which there exists a non-unit b so that f_b is one-to-one on U_m.
p, 121, line 20: "1194673 is a multiple of 53, 1194689 is a multiple of 23, ..."
p. 121, line -9: In Chapter 10 we will give a method..."
*p. 138, line 4. "Theorem 6" should be "Theorem 12". (Herb Enderton)
*p. 142, line -11. f(0) = 0_R is required by property (iv). (Herb Enderton)
*p. 146, line 2, Z//6Z should be Z/6Z. (Herb Enderton)
*p. 165-6, Example 4 is totally defective, since the matrix A has determinant 0 modulo 26. (John McKay)
p. 172, line -8. Rade Dimitric complained about the statement, "the notion of order is similar to that of least common multiple".
Perhaps I should have explained better what I meant. The order of an element of a finite group and the least common multiple of
two numbers are both least elements of non-empty sets of natural numbers. The existence of each is known from the
well ordering principle. In practice, to find the least common multiple of b and c, it
doesn't suffice to find a common multiple m of b and c -- we must also check that m is the least number that is a common multiple of b
and c. Similarly, to find the order of an element a, it doesn't suffice to find an exponent e > 0 so that a^e = 1 -- we
must also check that e is the least positive integer so that a^e = 1.
Students sometimes forget to check the leastness when looking for the order of an element.
p. 174, Exercise 8, line 2: "and the order of a modulo s...": the "a" should be in italics. (Herb Enderton)
p. 178, Exercise 33: U(1) should be U(p). (Marco Varisco)
*p. 194, lines 2, 12, 13: replace 179 by 177: we expand the exponent 177 = 128 + 32 + 16 + 1, not the modulus 179. (Herb Enderton)
p. 201, line 7: L Adleman (1977) should be L. Adleman (1978) (Marco Varisco)
*p. 202, lines 11, 15: the equalities should be congruences mod m. (Herb Enderton)
p. 207, line -9: either delete "n divides" or make it read "verify that n divides 2^{n-1}
-1, then..." (Marco Varisco)
*p. 235. Exercise 21 should read: "Show that every element [a] in the coset [b]U_m(m-1) satisfies [a]^{m-1} = [c]."
(Anh Ngoc Le)
*p. 238,
Exercise 27, (ii). Since 14 is not coprime to 35, the solutions to x^2 congruent to 14 modulo 35 are not units,
so do not lie in a coset of H = U_{35}(2). But this is an
instructive error. Since 14 is congruent to 49 modulo 35, two solutions to the congruence are x = 7 and -7 modulo 35.
(One can show, using the Chinese Remainder Theorem, that those are the only solutions of the congruence.)
Now if one looks at the "coset" [7]H, one finds that [7]H = {[7], [7], [-7], [-7]},
hence contains only two elements of Z/35Z,
not four elements. Critical to the proof of Lagrange's Theorem is the result that every coset bH of a subgroup H of G
contains the same number of elements as H. This
example shows that if one is too casual about what is meant by a "coset", then that result is false.
*p. 247, line 5. "the left coset a*H and the right coset H*a are equal." (Ken Brown)
*p. 254. In the EEA matrix for solving 74r - 63s = 2, the row containing 66 should contain 6 and -6. (Yueyi Zheng)
p. 273, line -11: "as noted in the following exercises." (Marco Varisco)
p. 277, line 13. "The proof of Theorem 7 relates to the Chinese Remainder Theorem" (Herb Enderton)
p. 290, line -4. "...in formula (13.2) above. To illustrate..." (Rade Dimitric)
p. 292, first line of Example 1. "Let Func(R, R) be the ring of functions..." (Herb Enderton)
*p. 296, line -11. (x^2 -1)(x^2+1) = x^4-1. (Heather Cavelius)
p. 298, last line of Exercise 3, add a period before the ). (Herb Enderton has sharp eyes for proofreading!)
*p. 300, second line of Theorem 8: "... written as d = rf + sg for..." (Herb Enderton)
*p. 305, Exercise 45. "then f(x) has an irreducible factor of degree \le n/2" (\le means "less than or equal to"). (Herb Enderton)
p. 309, Theorem 1. The letter r is used for the remainder on line 7, so should preferably not be used as the number of prime factors of
g in the theorem. So replace r by k in the statement and proof of Theorem 1 (except in Lemma 2). (Ted Turner)
*p. 329, line -11: the displayed polynomial should be p(x) = x^2 -2ax + (a^2 + b^2).
(Marco Varisco)
p. 336, line 6. Section 12E should be section 9E.
p. 347, line 2 of Example 10: Phi(x) = x^{p-1} + x^{p-2} + ... . (Marco Varisco)
p. 347, line 4 of Example 10: the phi should be Phi (Ted Turner)
p. 355, line 9 should read "if m divides f - g". (John Watterlond)
*p. 357, line -10 (Example 1). f(x) = m(x)(x^3 + 2x^2 + 2x) + (x+1) (an example where detaching the coefficients
when dividing polynomials led to error!). (Matt Patrick)
p. 416, line 14. "An idea like this could only..." (Keith Conrad)
*p. 450, Exercise 21: show that (((p-1)/2)!)^2 is congruent to (-1)^{(p-1)/2} (mod p).
(students of Andrew Mauer-Oats)
p. 451, line 4: three dots ... are missing in the displayed formula. (Marco Varisco)
p. 463, line 6. Keith Conrad's web page address should start www.math.uconn.edu/~conrad/ ... .
p. 473, line l6. "Suppose we wish to factor a number N of 100 digits": the "N" should be in italics. (Keith Conrad)
*p. 489, Example 13 is defective: the first remainder in Euclid's Algorithm is x^3 + x^2
+ x.
*p. 536, line 14 (last line of the proof): replace r by c: "is < 0 for 0 < x < c, and is
> 0 for x > c".
p. 569ff. Often the hints refer to parts a), b), ... of an exercise when the reference
shold be to parts i), ii), ..., for example, for exercise 9 of Chapter 4 (p. 572). (Marco
Varisco)
p. 569, hint for Chapter 1, Exercise 2 d): reflexivity also fails (Julian Wilson)
*p. 570. The last three hints for Chapter 2 should be for Exercises 36, 37 and 38. (Herb Enderton)
p. 577, Chapter 9, Exercise 14: the answers to part (ii) is 1 and to part (iii) is 5. (Herb Enderton)
*p. 578. Chapter 9. The hints on page 578 should be to Exercises 70 (which is correct in the text), 75 (not 71), 76 (not 73), 79 (not
76), 82 (not 80), 85 (not 82), 88 (not 85) and 89 (not 86). (Herb Enderton)
p. 582, Chapter 14, #4 (iii): Let y = x^4. (Marco Varisco)
p. 584, Chapter 15, #42: the proof of Theorem 10 ... applies. (Marco Varisco)
p. 597, "Lenstra, A. K. and Lenstra, H. W. Jr., The development of the number field sieve ..." (Keith Conrad)
p. 601, the term "Logarithm table" is on p. 107 and p. 403. (Ted Turner)
p. 603, the term "Roots of unity" is on page 318 and on p. 347. (Herb Enderton)
Last update, April 13, 2012