458 ALGEBRAIC SYSTEMS [APP. B
(c) Substitute each of the ten elements ofZ 10 intof(x)to see which elements yield 0. We have:
f( 0 )= 4 ,f( 2 )= 0 ,f( 4 )= 2 ,f( 6 )= 0 ,f( 8 )= 4
f( 1 )= 0 ,f( 3 )= 4 ,f( 5 )= 4 ,f( 7 )= 0 ,f( 9 )= 2
Thus the roots are 1, 2, 6, and 7. (This example shows that a polynomial of degreencan have more thannroots
over an arbitrary ring. This cannot happen if the ring is a field.)B.21.Prove that in a ringR: (i)a· 0 = 0 ·a=0; (ii)a(−b)=(−a)b=−ab; (iii)(− 1 )a=−a(whenRhas
an identity element 1).
(i) Since 0= 0 +0, we have
a· 0 =a( 0 + 0 )=a· 0 +a· 0
Adding−(a· 0 )to both sides yields 0=a·0. Similarly 0·a=0.
(ii) Usingb+(−b)=(−b)+b=0, we have
ab+a(−b)=a(b+(−b))=a· 0 = 0
a(−b)+ab=a((−b)+b)=a· 0 = 0
Hencea(−b)is the negative ofab; that is,a(−b)=−ab. Similarly,(−a)b=−ab.
(iii) We have
a+(− 1 )a= 1 ·a+(− 1 )a=( 1 +(− 1 ))a= 0 ·a= 0
(− 1 )a+a=(− 1 )a+ 1 ·a=((− 1 )+ 1 )a= 0 ·a= 0
Hence(− 1 )ais the negative ofa; that is,(− 1 )a=−a.B.22.LetDbe an integral domain. Show that ifab=acwitha=0 thenb=c.
Sinceab=ac, we have
ab−ac=0 and so a(b−c)= 0
Sincea=0, we must haveb−c=0, sinceDhas no zero divisors. Henceb=c.B.23.SupposeJandKare ideals in a ringR. Prove thatJ∩Kis an ideal inR.
SinceJandKare ideals, 0∈Jand 0∈K. Hence 0∈J∩K. Now leta,b∈J∩Kand letr∈R. Then
a,b∈Janda,b∈K. SinceJandKare ideals,
a−b, ra, ar∈J and a−b, ra, ar∈K
Hencea−b,ra,ar∈J∩K. ThereforeJ∩Kis an ideal.B.24.LetJbe an ideal in a ringRwith an identity element 1. Prove: (a)If1∈JthenJ=R;(b) If any unit
u∈JthenJ=R.
(a)If1∈Jthen for anyr∈Rwe haver· 1 ∈Rorr∈J. HenceJ=R.
(b)Ifu∈Jthenu−^1 ·u∈Jor 1∈J. HenceJ=Rby part (a).B.25.Prove: (a) A finite integral domainDis a field. (b)Zpis a field wherepis a prime number.
(c) (Fermat) Ifpis prime, thenap≡a(modp) for any integera.
(a) SupposeDhasnelements, sayD={a 1 ,a 2 ,...,an}. Letabe any nonzero element ofD. Consider thenelements
aa 1 ,aa 2 ,..., an
Sincea=0, we haveaai=aakimpliesai=ak(Problem B.22). Thus the abovenelements are distinct, and so
they must be a rearrangement of the elements ofD. One of them, sayaak, must equal the identity element 1 of
D; that is,aak=1. Thusakis the inverse ofa. Sinceawas any nonzero element ofD, we have thatDis a field.
(b) RecallZp={ 0 , 1 , 2 ,...,p− 1 }. We show thatZphas no zero divisors. zero divisors. Suppose a∗b=0inZp;
that is, 0 (modp). Thenpdividesab. Sincepis prime,pdividesaorpdividesb. Thusa≡0 (modp)or
b≡0 (modp); that is,a=0orb=0inZp. Accordingly,Zphas no zero divisors and henceZpis an integral
domain. By part (a),Zpis a field.
(c)Ifpdividesa, thena≡0 (modp) and soap≡a≡0 (modp). Supposepdoes not dividea, thenamay be
viewed as a nonzero element ofZpis a field, its nonzero elements form a groupGunder multiplication of order
p−1. By Problem B.45,ap−^1 =1inZp.
In other words,ap−^1 ≡1 (modp). Multiplying by a givesap≡a(modp), and the theorem is proved.