APP. B] ALGEBRAIC SYSTEMS 465
B.94. Consider the integral domainD={a+b√
13 |a, bintegers}. (See Example B.15(b).) Ifα=a+√
13, we define
N(α)=a^2 − 13 b^2. Prove:
(i) N(αβ)=N(α)N(β). (iii) Among the units ofDare±1, 18± 5√
13; and− 18 ± 5√
13.
(ii) αis a unit if and only ifN(α)=+1. (iv) The numbers 2, 3 −√
13 and− 3 −√
13 are irreducible.POLYNOMIALS OVER A FIELD
B.95. Find the roots off(t)assumingf(t)has an integer root: (a)f(t)=t^3 − 2 t^2 − 6 t−3; (b)f(t)=t^3 −t^2 − 11 t−10;
(c)f(t)=t^3 + 2 t^2 − 13 t−6.
B.96. Find the roots off(t)assumingf(t)has a rational root: (a)f(t)= 2 t^3 − 3 t^2 − 16 t−7; (b)f(t)= 2 t^3 −t^2 − 9 t+9.
B.97. Find the roots off(t)=t^4 − 5 t^3 + 16 t^2 − 9 t−13, given thatt= 2 + 3 iis a root.
B.98. Find the roots off(t)=t^4 −t^3 − 5 t^2 + 12 t−10, given thatt= 1 −iis a root.
B.99. For any scalara∈K, define theevaluation mapψa:K[t]→Kbyψa(f (t))=f(a). Show thatψais a ring
homomorphism.B.100. Prove: (a) Proposition B.14. (b) Theorem B.26.
Answers to Supplementary Problems
B.39. (a) 12, 15, 18, 6; (b) Yes, yes; (c) 1; (d) Only 1 and it
is its own inverse.
B.40. (a) 17,−32, 29/2; (b)Yes, yes; (c) Zero; (d) Ifa= 1 /2,
thenahas an inverse which is−a/( 1 + 2 a).
B.41. (a) Yes; (b) No; (c) No; (d) It is meaningless to talk
about inverses when no identity element exists.
B.42. (a) Sixteen, since there are two choices,aorb, for
each of the four productsaa,ab,ba, andbb. (b) Let
aa=b,ab=a,ba=b,bb=a. Thenab=ba.
Also,(aa)b=bb=a, buta(ab)as=b.
B.43. (a)A,D; (b) none.
B.44. (a) Yes; (b) yes; (c) yes; (d) no.
B.45. (a) {1,−1, 0}; (b) There is no set.
B.46. (a) Yes, yes, no; (b) Yes, no, no.
B.47. (a) (3, 10), (−5, 1); (b) yes, no; (c) (1, 0); (d) The
element (a, b) has an inverse ifa = 0, and its
inverseis(1/a,−b/a).
B.48. (a) (19, 20), (18, 7). (d) (a,b)∼(c,d)ifad=bc.
(e)S/∼is isomorphic to the positive rational numbers
under addition. ThusS/∼has no identity element and
no inverses.
B.49. (a) (4, 9), (6, 8); (d) (a,b)∼(c,d)ifa+d=b+c.
(e)S/∼is isomorphicZsince every integer is the
difference of two positive integers. ThusS/∼has an
identity element, and every element has an inverse.
B.50. (a)H = l{ 0 , 5 , 10 , 15 }and|H| = 4. (b)H,
1 +H ={ 1 , 6 , 11 , 16 },2+H ={ 2 , 7 , 12 , 17 },
3 +H={ 3 , 8 , 13 , 18 },4+H={ 4 , 9 , 14 , 19 }.
B.51. (a)x^2 =1ifx=1. (b) No. (c) {1}, {1, 5}, {1, 7},
{1, 11},G.
B.52. (a) See Fig. B-9(a). (b) 11, 13, 17; (c) (i)|%|=6,
gp(5)= G; (ii)| 13 | =3,gp( 13 ) ={ 1 , 7 , 13 };
(d) Yes, sinceG=gp(5).
B.53. (a) See Fig. B-9(b). (b) 4, 3, 4.Fig. B-9