Math 430 { Problem Set 5 Solutions

1 Math 430 Problem Set 5 SolutionsDue April 1, Find all of the abelian groups of order 200 up to abelian group is a direct product of cyclic groups. Using the fact thatZmn =Zm Znif and only if gcd(m,n) = 1, the list of groups of order 200 is determined by the factorization of 200into primes: Z8 Z25 Z4 Z2 Z25 Z2 Z2 Z2 Z25 Z8 Z5 Z5 Z4 Z2 Z5 Z5 Z2 Z2 Z2 Z5 Show that the infinite direct productG=Z2 Z2 is not finitely that every element ofGhas order 2 and thatGis abelian. The group generated byany finite set ofkelements thus has at order at most 2k, whileGhas infinite order. ThusGcannotbe finitely Which of the following sets are rings with respect to the usual operations of addition and multiplication?If the set is a ring , is it also a field?(a) is a subring ofZand thus a ring : (7n) + (7m) = 7(m+n) so it is closed under addition; (7n)(7m) = 7(7mn) so it is closed under multiplication; (7n) = ( 7)(n), so it is closed under is not a field since it does not have an identity.

2 (b) is a ring : the operations of arithmetic modulo 18 are well defined. It is not afield, since 2 9 = 0 gives a pair of zero divisors.(c)Q( 2) is a subfield ofRand thus a field (in addition to being a ring : (a+b 2) + (c+d 2) = (a+c) + (b+d) 2 so it is closed under addition; (a+b 2)(c+d 2) = (ac+ 2bd) + (ad+bc) 2 so it is closed under multiplication; (a+b 2) = ( a) + ( b) 2 so it is closed under negation; (a+b 2) 1=aa2 2b2 ba2 2b2 2 anda2 2b26= 0 fora,b Q(unless both are zero)1(f)R={a+b3 3 :a,b Q} is not a ring since3 3 3 3 is not inR.(h)Q(3 3) is a subfield ofRand thus a field: (a+b3 3 +c3 9) + (d+e3 3 +f3 9) = (a+d) + (b+e)3 3 + (c+f)3 9 so it is closed underaddition; (a+b3 3 +c3 9)(d+e3 3 +f3 9) = (ad+ 3bf+ 3ce) + (ae+bd+ 3cf)3 3 + (af+be+cd)3 9so it is closed under multiplication; (a+b3 3 +c3 9) = ( a) + ( b)3 3 + ( c)3 9 so it is closed under negation.)

3 Closure under inverses takes a bit more work, since there is no single conjugate. We give In our first approach, we show directly that each element has an inverse. Given (a+b3 3+c3 9) we prove that there is some (d+e3 3+f3 9) with (a+b3 3+c3 9)(d+e3 3+f3 9) = 1using the formula for multiplication above. This requires solvingad+ 3ce+ 3bf= 1bd+ae+ 3cf= 0cd+be+af= system will have a solution as long as the determinantdet a3c3bb a3cc b a =a3+ 3b3+ 9c3 9abcis nonzero. Multiplying all ofa,bandcby a common rational value we may assume thatthey are all integers and share no common factor. Ifa3+ 3b3+ 9c3 9abc= 0 thena3isa multiple of 3, and thusais by unique factorization (saya= 3a ). Then9a 3+b3+ 3c3 9a bc= this forbandc, we see that all are divisible by 3, contradicting the fact thatwe chose them to have no common The other approach uses the extended Euclidean algorithm for polynomials.

4 Since3 3 isirrational,x3 3 is irreducible and thus gcd(x3 3,a+bx+cx2) = 1. Therefore, thereexistf(x),g(x) Q[x] withf(x)(x3 3) +g(x)(a+bx+cx2) = this equation atx=3 3 shows thatg(3 3) is the inverse ofa+b3 3 +c3 LetRbe the ring of 2 2 matrices of the form(a b0 0),wherea,b R. Show that althoughRis a ring that has no identity, we can find a subringSofRwithan that (a b0 0) is an identity forR. Then(1 10 0)(a b0 0)=(1 10 0)(a b0 0)=(1 10 0),2soa= 1 andb= 1. But (1 10 0) is not an identity, since(1 00 0)(1 10 0)=(1 10 0).ThusRhas no the subring of matrices of the form (a00 0). Then (1 00 0) is an identity forS, since(1 00 0)(a00 0)=(a00 0),(a00 0)(1 00 0)=(a00 0). Find all homomorphisms :Z/6Z is a ring homomorphism, it must also be a group homomorphism (of additivegroups).

5 Thuso 6 (1) = (0) = 0, and therefore (1) = 0,5 or 10 (and is determined by (1)). If (1) = 5, then (1) = (1 1)= (1) (1)= 5 5= 10,which is a contradiction. So the only two possibilities are 1(n) = 0 for alln Z6. 2(n) = 10nfor alln show that these are both well defined ring both cases, adding a multiple of 6 tonchanges the result by a multiple of 15 (0 in the first case and60 in the second), so they are well defined. They are additive group homomorphisms by the distributivelaw inZ15. They are multiplicative since0 0 = 0(10n) (10m) = 100nm= Define a map :C M2(R) by (a+bi) =(a b b a).Show that is an isomorphism ofCwith its image inM2(R). first show that is a homomorphism. ((a+bi) + (c+di)) = ((a+c) + (b+d)i)=(a+c b+d b d a+c) (a+bi) + (c+di) =(a b b a)+(c d d c)=(a+c b+d b d a+c) ((a+bi)(c+di)) = ((ac bd) + (ad+bc)i)=(ac bd ad+bc ad bc ac bd) (a+bi) (c+di) =(a b b a)(c d d c)=(ac bd ad+bc ad bc ac bd)Finally, we show that is injective and thus an isomorphism onto its image.

6 If (a+bi) = (0 00 0) thena= 0 andb= Letabe any element in a ringRwith identity. Show that ( 1)a= have( 1)a+a= ( 1)a+ (1)a= ( 1 + 1)a= (0)a= result now follows from the uniqueness of additive Prove the Correspondence Theorem: LetIbe an ideal of a ringR. ThenS S/Iis a one-to-onecorrespondence between the set of subringsScontainingIand the set of subrings ofR/I. Furthermore,the ideals ofRcorrespond to the ideals We first show that the functionS7 S/Isends subrings ofRto subrings ofR/I. Ifs,t Sthen(s+I) + (t+I) = (s+t) +I S/IsinceSis closed under addition, (s+I)(t+I) = (st) +I S/IsinceSis closed under multiplication and (s+I) = ( s) +I S/IsinceSis closed undernegation. ThusS/Iis a subring. We now show that this function is surjective. LetT R/Ibe a subring and setS={x R:x+I T}.

7 ThenSis a subring ofR: ifs,t Sthens+t Ssince (s+t) +I= (s+I) + (t+I)andTis closed under addition,st Ssince (st) +I= (s+I)(t+I) andTis closed undermultiplication, and s Ssince ( s) +I= (s+I) andTis closed under negation. Moreover,ScontainsIsincei+I= 0 +I Tfor everyi I. Finally,S/I=Tby construction. Next, we show that this function is injective. SupposeS1andS2are two subrings ofRthatcontainIand thatS1/I=S2/IinsideR/I. We show thatS1 S2(the opposite inclusion isanalogous). Supposex S1. SinceS1/I=S2/I, there is somey S2withx+I=y+I. Thusthere is ani Iwithx=y+i. SinceI S2, bothyandiare inS2and thusxis as Next, suppose thatSis actually an ideal ofR. To show thatS/Iis an ideal ofR/I, we take anarbitrarys+I S/Iandx+I R/I. Thenxs+I S/Iandsx+I S/IsinceSis an idealofR.

8 Finally, suppose thatT R/Iis an ideal. LetS={x R:x+I T}as before. We showthatSis an ideal ofR, proving that the correspondence restricts to a correspondence on Sandx Rthenxs+I= (x+I)(s+I) S/IsinceS/Iis an ideal; similarly an LetRbe an integral domain. Show that if the only ideals inRare{0}andRitself,Rmust be a order to show thatRis a field it suffices to prove that every nonzeroa Rhas aninverse. Leta Rbe nonzero, and consider the ideal a generated bya. Sincea6= 0, this ideal isnonzero and thus is all ofRby assumption. Therefore it contains 1; by the definition of a principalideal there is someb Rwithab= 1, providing the inverse LetRbe a ring such that 1 = 0. Prove thatR={0}. Rthena= (1)a= (0)a= 0, soR={0}. Letpbe a prime. Prove thatZ(p)={a/b:a,b Zand gcd(b,p) = 1}is a show that this is a subring ofQand thus a ring .

9 Ifa/b,c/d Z(p)to show thata/b+c/d= (ad+bc)/(bd) and (a/b)(c/d) =ac)/(bd) Z(p)it suffices to show that gcd(bd,p) = follows from the fact thatpis prime: if it does not dividebordthen it cannot dividebd. Negationis even easier: (a/b) = ( a)/b Z(p)since it has the same An elementxin a ring is called an idempotent ifx2=x. Prove that the only idempotents in anintegral domain are 0 and 1. Find a ring with an idempotentxnot equal to 0 or (x 1)x=x2 x= 0. An integral domain has no zero divisors, so the onlypossibilities forxare 0 and Z6is an example of an idempotent that is neither 0 nor