Solutions to Homework Problems from Chapter 3

1 Solutions to Homework Problems from Chapter 3 The following subsets ofZ(with ordinary addition and multiplication) satisfy all but one of theaxioms for a ring. In each case, which axiom fails.(a) The setSof odd integers. The sum of two odd integers is a even integer. Therefore, the setSis not closed under , Axiom 1 is violated.(b) The set of nonnegative integers. Ifais a positive integer, then there is no solution ofa+x= 0 that is also positive. Hence, Axiom5 is (a) Show that the setRof all multiples of 3 is a subring ofZ.(b) Letkbe a fixed integer. Show that the set of all multiples ofkis a subring ofZ. Clearly, (b) implies (a); so let us just prove (b). LetS={z Z|z=nkfor somen Z}.In general, to show that a subsetSof a ringR, is a subring ofR, it is sufficient to show that(i)Sis closed under addition inR(ii)Sis closed under multiplication inR;(iii) 0R S;(iv) whena S, the equationa+x= 0 Rhas a solution , b, c S Zwitha=rk,b=sk,c=tk.

2 (i)a+b=rk+sk= (r+s)k S(ii)ab= (rk)(sk) = (rsk)k S(iii)0Z= 0 = 0 k S(iv)a=rs S x= rs Sis a solution ofa+x= 0 SThus,Sis a subring LetR={0, e, b, c}with addition and multiplication defined by the tables below:+0e b c 0e b c00e b c0 0 0 0 0ee0c be0e b cbb c0eb0b e ccc b e0c0c c0 Assume distributivity and associativity and show thatRis a ring with identity. IsRcommutative?Axioms (1) and (6) are satisfied by virtue of the tables above. We are also allowed to assume that Axioms (2),(7) and (8) hold. Axiom (3), commutatively of addition, is also evident from the symmetry of the additiontable. Similarly, the symmetry of the multiplication table implies that multiplication is commutative for12this set. From the addition table it is also clear that 0 +a=afor anya R; so Axiom (4) is Itremains to verify Axiom (5).

3 Thus, we need to find a solution inRofa+x=Ofor eacha R. From the addition table we havea= 0, x= 0 a+x= 0a=e , x=e a+x= 0a=b , x=b a+x= 0a=c , x=c a+x= 0,so Axiom (5) is also LetF={0, e, a, b}with addition and multiplication defined by the tables below:+0e a b 0e a b00e a b0 0 0 0 0ee0b ae0e a baa b0ea0a b ebb a e0b0b e aAssume distributivity and associativity and show thatRis a first show thatFis a ring:Axioms (1) and (6) are satisfied by virtue of the tables above. We are also allowed to assume that Axioms (2),(7) and (8) hold. Axiom (3), commutatively of addition, is also evident from the symmetry of the additiontable. Similarly, the symmetry of the multiplication table implies that multiplication is commutative forthis set.

4 From the addition table it is also clear that 0 +s=afor anys R; so Axiom (4) is Itremains to verify Axiom (5). Thus, we need to find a solution inRofs+x=Ofor eacha R. From the addition table we haves= 0, x= 0 s+x= 0s=e , x=e s+x= 0s=a , x=a s+x= 0s=b , x=b s+x= 0,so Axiom (5) is also verified. We also note thats e=sfor anys6= 0 inF. Thus,Fis a commutative ringwith show thatFis a field we need to show further thatFis a division ring; , for eachs6= 0 FinF, theequationssx= 1F ehas a solution inF. From the multiplication table we sees=e , x=e sx=es=a , x=b sx=es=b , x=a sx=e ,SoFis a commutative division ring; , a Which of the following five sets are subrings ofM(R). Which ones have an identity?(a)A={(0r0 0)|r Q}3 This is a subring since(0r0 0)+(0r 0 0)=(0r+r 0 0) A(0r0 0)(0r 0 0)=(0rr 0 0) AIt does not have an identity, however.

5 (b)B={(a b0c)|a, b, c Z}This is a subring since(a b0c)+(a b 0c )=(a+a b+b 0c+c ) B(a b0b)(a b 0c )=(aa ab +bc 0bc ) BIt does have an identity, namely,I=(1 00 1).C={(a ab b)|a, b R}This is a subring since(a ab b)+(a a b b )=(a+a a+a b+b b+b ) C(a ab b)(a a b b )=(aa +ab aa +ab ba +bb ba +bb ) CIt does not have an identity, however.(c)D={(a0a0)|a R}This is a subring since(a0a0)+(a 0a 0)=(a+a a+a ) D(a0a0)(a 0a 0)=(aa 0aa 0) DIt does have an identity however. For(1 01 0) Dand(1 01 0)(a0a0)=(a0a0)=(a0a0)(1 01 0)D={(a00a)|a R}This is a subring since(a00a)+(a 00a )=(a+a 00a+a ) D(a00a)(a 00a )=(aa 00aa ) D4It does have an identity, namely,I=(1 00 1). LetRandSbe rings. Show that the subset R={(r,0S)|r R}is a subring ofR S.

6 Do the samefor the set S={(0R, s)|s S}. , Ris a subset ofR S. By Theorem ,R Sis a ring. To show that Ris a subring ofR Swe must verify(i) Ris closed under addition(ii) Ris closed under multiplication(iii) 0R S R(iv) Whena R, the equationa+x= 0R Shas a solution in RLeta= (r,0S) andb= (t,0S) be arbitrary elements of R.(i)a+b= (r,0S) + (t,0S) = (r+t,0S+ 0S) = (r+t,0S) R(ii)ab= (r,0S)(t,0S) = (rt,0S 0S) = (rt,0S) R(iii)0R S= (0R,0S) R(iv) Leta= (r,0S) be an arbitrary element of Rand letube a solution ofr+x= 0 RinR. If we set u= (u,0S), then u Rand we havea+ u= (r,0S) + (u,0S) = (r+u,0S+ 0S) = (0R,0S) = 0R ifais any element of R, there is a solution ofa+x= 0R Sin proof that Sis a subring ofR Sis IfRis a ring, show thatR ={(r, r)|r R}is a subring ofR verify the four properties of a subring (as listed in the previous problem).

7 (i)(r, r) + (r , r ) = (r+r , r+r ) R (ii)(r, r)(r , r ) = (rr , rr ) R (iii)0R R= (0R,0R) R (iv)Ifuis a solution ofr+x= 0 RinR, thenu = (u, u) is a solution of (r, r) +u = 0R Rlying inR ; for(r, r) +u = (r, r) + (u, u) = (r+u, r+u) = (0R,0R) = 0R Is{1, 1, i, i}a subring ofC?No, sincei+i= 2i / {1, 1, i, i}, this subset is not closed under addition; hence it is not a subring. Also,0 = 0C/ {1, 1, i, i}. Letpbe a positive prime and letRbe the set of all rational numbers that can be written in the formrpiwithr, i Z. Show thatRis a subring again verify properties (i) - (iv) of a subring.(i)rpi+r pj=rpj+pir pipj=rpj+r pipi+j R(ii)rpi r pj=rr pipj=rr pi+j R(iii)0Q= 0 =0p R(iv) There is always a solution ofrpi+x= 0Q= 0 inR; namelyx= LetTbe the ring of continuous functions fromRtoRand letf, gbe given byf(x) ={0 ifx 2x 2 if 2< x,g(x) ={2 xifx 20 if 2< thatf, g Tand thatf g= 0T, and therefore thatTis not an integral ,fandgare certainly continuous on the intervals ( ,2) and (2,+ ) since they are prescribed bypolynomial functions also continuous atx= 2 sincelimx 2 f(x) = limx 2+f(x) = 0 = limx 2 g(x) = limx 2+g(x).}}

8 Therefore,f, g T. The product offandgis then the function defined by(f g)(x) =f(x)g(x) ={(0)(2 x) = 0 ifx 2(x 2)(0) = 0 if 2< g(x) = 0 for allx. But this function is just the additive identity ofT. Thus,f g= 0T. Sincef, g6= 0T, butf g= 0T,Tcan not be an integral LetQ( 2) ={r+s 2|r, s Q}.Show thatQ( 2) is a subfield prove that a subsetSof a fieldFis a subfield, one must show thatSis a subring ofFand that(v) 1F S(vi) Ifs S, then the equationsx= 1 Falways has a solution ( 2) is certainly a subset ofR. Consider two arbitrary elementsr+s 2,r +s 2 Q( 2). We have(i)(r+s 2)+(r +s 2)= (r+r ) + (s+s ) 2 Q( 2)(ii)(r+s 2)(r +s 2)= (rr + 2ss ) + (rs +sr ) 2 Q( 2)(iii)0R= 0 = 0 + 0 2 QAlso,(iv)r+s 2 Q( 2) r+ ( s) 2 Q( 2)so every equation of the forma+x= 0 Rwitha Q( 2) has a solution inQ( 2).}

9 We also have(v)1R= 1 = 1 + 0 2 Q( 2)S .Finally, we have (recalling that ifr, s Qthenr2 2s26= 0 unlessr=s= 0)06=r+s 2 Q( 2) rr2 2s2 sr2 2s2 2 Q( 2)6and(vi)(r+s 2)(rr2 2s2 sr2 2s2 2)=(r+s 2)(r s 2)r2 2s2=r2 2s2r2 2s2= 1,so every non-zero element ofQ( 2) is a LetHbe the set of real quaterions and 1,i,j, andkthe matrices1 =(1 00 1),i=(i00 i),j=(0 1 1 0),k=(0ii0).(a) Prove thati2=j2=k2= 1jk= kj=iij= ji=kki= ik=j These identities are all proved by direct calculation; ,i2=(i00 i)(i00 i)=( 1 00 1)= 1(b) Show thatHis a noncommutative ring with identity. Let us parmeterizeHas {(a+ibc+id c+id a ib)|a, b, c, d R}={(zw w z)|z, w C}.The latter parameterization displaysHas a subset of the setM2(C) of 2 2 complex matrices. SinceM2(C)is known to be a ring (under the usual operations of matrix addition and matrix multiplication) to showthatHis a ring, it suffices to verify that the subsetHofM2(C) i is closed under addition; ii is closed undermultiplication iii contains the element 0 =(0 00 0).

10 Iv contains the solution ofa+x= 0 ifa H. Theseproperties are easily confirmed by direct calculation.(c) Show thatHis a division ring. The multiplicative inverse of 2 2 complex matrixMexists whenever the determinant ofMdoes notvanish. IfM H, thendet(M) =det(a+ibc+id c+id a ib)= (a+ib) (a ib) (c+id) ( c+id)=a2+b2+c2+d2= 0 iffa=b=c=d= every nonzero element ofHhas a multiplicative inverse; soHis a division ring.(d) Show that the equationx2= 1 has infinitely many Solutions Prove Theorem IfRandSare rings, then we can give the Cartesian productR Sthe structureof a ring by setting(r, s) + (r , s ) = (r+r , s+s )(r, s)(r , s ) = (rr , ss )0R S= (0R,0S).IfRandSare both commutative, then so isR S. IfRandSeach have an identity, then so doesR must confirm that the 8 axioms of a ring are satisfied.