Hints to Selected Exercises Exercise 1.1 Answer: 2n. Exercise 1.2 The answer to the second question is negative. Indeed, let X Df1; 2g, Y Df2g. Using unions and intersections, we can obtain only the sets X \ Y D Y \ Y D Y [ Y D Y ; X [ Y D X [ X D X \ X D X : Thus, each formula built from X, Y, \,and[ produces either X Df1; 2g or Y Df2g. But X X Y Df1g. Exercise 1.3 There are six surjections f0; 1; 2g f0; 1g and no injections. Symmetrically, there are six injections f0; 1g ,!f0; 1; 2g and no surjections. Exercise 1.5 If X is finite, then a map X ! X that is either injective or surjective is automatically bijective. Every infinite set X contains a subset isomorphic to the set of positive integers N, which admits the nonsurjective injection n 7! .n C 1/ and the noninjective surjection 1 7! 1, n 7! .n 1/ for n > 2. Both can be extended to maps X ! X by the identity action on X X N . Exercise 1.6 Use Cantor’s diagonal argument: assume that all bijections N ⥲ N are numbered by positive integers, run through this list and construct a bijection that sends each k D 1; 2; 3; : : : to a number different from the image of k under the k-th bijection in the list. nCm1 nCm1 .nCm1/Š Exercise 1.7 Answer: m1 D n D nŠ.m1/Š . Hint: the summands are . ; ;:::; / in bijectionP with the numbered collections of nonnegative integers k1 k2 km such that ki D n. Such a collection is encoded by the word consisting of .m 1/ 0 1 letters and n letters : write k1 ones, then zero, then k2 ones, then zero, etc. nCk Exercise 1.8 Answer: k . A diagram is a broken line going from the bottom left-hand corner of the rectangle to its upper right-hand corner and consisting of n horizontal and k vertical edges. © Springer International Publishing AG 2016 519 A.L. Gorodentsev, Algebra I, DOI 10.1007/978-3-319-45285-2 520 Hints to Selected Exercises Exercise 1.9 If z is equivalent to both x and y, then each equivalence u x implies by transitivity and symmetry the equivalence u y and conversely. 0 0 0 0 Exercise 1.10 Let Œx n D Œxn and Œ y n D Œyn,thatis,x D x C nk, y D y C n` for some k;`2 Z.Thenx0 C y0 D x C y C n.k C `/ and x0y0 D xy C n.`x C ky C k`n/ 0 0 are congruent modulo n to x C y and xy respectively, i.e., Œx C y n D Œx C yn and 0 0 Œx y n D Œxyn. Exercise 1.11 Say that x y if there exists a chain satisfying the conditions from Exercise 1.11. Verify that this is an equivalence relation and check that it is a subset of every equivalence relation containing R. Exercise 1.12 Check of transitivity: if . p; q/ .r; s/ and .r; s/ .u; w/, i.e., psrq D 0 D usrw,thenpswrqw D 0 D usqrwq, which forces s. pwuq/ D 0 and pw D uq, i.e., . p; q/ .u; w/. Exercise 1.13 Let ˛ be the smaller of the two angles between `1 and `2.Then reflection in `i followed by reflection in `j is a rotation about O D `i \ `j through the angle 2˛ in the direction from `i to `j. Thus, 12 D 21 if and only if the lines are perpendicular. Exercise 1.14 The table of products gf is as follows: gŸf .1; 2; 3/ .1; 3; 2/ .3; 2; 1/ .2; 1; 3/ .2; 3; 1/ .3; 1; 2/ .1;2;3/ .1; 2; 3/ .1; 3; 2/ .3; 2; 1/ .2; 1; 3/ .2; 3; 1/ .3; 1; 2/ .1;3;2/ .1;3;2/.1;2;3/.3;1;2/.2;3;1/.2;1;3/.3;2;1/ .3;2;1/ .3; 2; 1/ .2; 3; 1/ .1; 2; 3/ .3; 1; 2/ .1; 3; 2/ .2; 1; 3/ .2;1;3/ .2; 1; 3/ .3; 1; 2/ .2; 3; 1/ .1; 2; 3/ .3; 2; 1/ .1; 3; 2/ .2;3;1/ .2; 3; 1/ .3; 2; 1/ .2; 1; 3/ .1; 3; 2/ .3; 1; 2/ .1; 2; 3/ .3;1;2/ .3;1;2/.2;1;3/.1;3;2/.3;2;1/.1;2;3/.2;3;1/ Exercise 1.16 Let x 2 W be the minimal element in the set of w 2 W for which ˙.w/ fails. Since ˙.w/ holds for all w < x,then˙.x/ must hold as well. Contradiction. Exercise 1.19 The axiom of choice allows us to choose an upper bound b.W/ 2 P for every W 2 W.P/.Iff .x/>x for all x 2 P, then the map W.P/ ! P; W 7! f b.W/ ; contradicts Lemma 1.2 on p. 15. Exercise 2.2 Answers: 1 C x and xy C x C y . Exercise 2.3 It is enough to verify the invariance of the equivalence classes of the results only under changes of fractions by means of the generating relations (2.12), a ac i.e., by b 7! bc . Exercise 2.5 Use increasing induction on k starting from k D 0 to verify that all Ek belong to .a; b/ (thus, all Ek are divisible by GCD.a; b/). Then use decreasing Hints to Selected Exercises 521 induction on k starting from k D r C 1 to verify that all Ek are divisible by Er (thus, E0 D a, E1 D b,andGCD.a; b/ D ax C by are divisible by Er). Exercise 2.8 The existence of a factorization can be proved by induction on jnj: If n is prime, its prime factorization is n D n; if not, then n D n1 n2,where jn1j; jn2j < n. Thus, n1, n2 are factorizable by induction. The proof of the uniques is based on the following: for all z 2 Z and prime p 2 Z, either GCD.z; p/ Djpj and p j z or GCD.z; p/ D 1 and p, z are coprime. Given two coinciding products p1p2 pk D q1q2 qm, it follows fromQ Lemma 2.3 on p. 26 that p1 cannot be coprime to each qj, because p1 divides qi. Thus, p1 divides some qi,sayq1.Since q1 q1 p1 q1 p1 is prime, D˙ P. Cancel and and repeat the argument. 1 . 1/k k 1 2 3 Exercise 2.9 a D k>0 a D a C a a C (the sum is finite, because a is nilpotent). pnm . / pn Exercise 2.10 The residue class pn mod p is equal to the coefficient of x in pnm the expansion of the binomial .1 C x/ over the finite field Fp D Z=. p/. Since the p pn pn map a 7! a respects sums over Fp, its n-fold iteration yields .1 C x/ D 1 C x . n n m n Hence .1 C x/p m D 1 C xp D 1 C mxp C higher powers. Exercise 2.12 The axioms are be checked componentwise, and the clearly hold because each Kx is a ring. Exercise 2.13 An element .a; b/ 2 Fp Fq is invertible if and only if a ¤ 0 and b ¤ 0. A nonzero element .a; b/ 2 Fp Fq divides zero if and only if a D 0 or b ¤ 0. Exercise 2.17 Both statements follow from Lemma 2.5 on p. 33, which says that every ring homomorphism to an integral domain sends the unit element to the unit element. Exercise 2.18 Since the polynomial xp x is nonzero and of degree p, it has at most p roots1 in a field F. Hence, the fixed points of the Frobenius endomorphism Fp W F ! F are exhausted by p elements of the prime subfield Fp F. n n/=. / n1 n2 n3 2 n2 n1 Exercise 3.3 y x y x D y C y x C y x C! Cyx C x . P X X k k k Exercise 3.5 f .x/ a x f .x t/ a x t t f.x/ Let D k .Then C D k D , k; where ! X 1 k X k k d k f.x/ D a x D a x : k Š dxk k k> k>0 Exercise 3.6 The right-hand side is nothing but the output of long division of f by x ˛. However, now that it has been written down, the equality can easily be checked by straightforward expansion. Exercise 3.7 Completely similar to Exercise 2.8. 1See Sect. 3.2 on p. 50. 522 Hints to Selected Exercises Exercise 3.8 Completely similar to Exercise 2.5 on p. 25. Exercise 3.9 A reducible polynomial of degree 6 3 has a divisor of degree 1. Exercise 3.10 Uniqueness follows from Corollary 3.3 on p. 50. To construct f , note that Y fi.x/ D .x a/ ¤i vanishes at all points a except for ai,atwhichfi.ai/ ¤ 0. Thus gi.x/ D fi.x/=fi.ai/ satisfies ( 1 for D i; gi.a/ D 0 otherwise : P Q n xa Hence f .x/ D b1g1 C b2g2 CCbngn D D0 bi ¤ solves the problem.