MTH 310-3 Abstract Algebra I and Number Theory S17 Homework 3/ Solutions Exercise 1. Prove the following Theorem: Theorem Let R and S be rings. Define an addition and multiplication on R S by for all r, r R and s, s S. Then a R S is a ring. b 0 R S 0 R, 0 S. c r, s r, s for all r R, s S. r, s + r, s r + r, s + s r, sr, s rr, ss d If R and S are both commutative, then so is R S. e If both R and S have an identity, then R S has an identity and 1 R S 1 R, 1 S. a: We need to verify all the eight axioms of a ring. Let a, b, c R and d, e, f S. Ax1 closure of addition By Ax1 for R and S, a + b R and d + e S and so a + b, d + e R S. By definition of addition, a, d + b, e a + b, d + e and so a, d + b, e R S. Ax2 associative addition a, d + b, e + c, f a, d + b + c, e + f by the definition of addition a + b + c, d + e + f by the definition of addition a + b + c, d + e + f Ax2 for R and S a + b, d + e + c, f by the definition of addition a, d + b, e + c, f by the definition of addition. So addition in R S is associative. Ax3 commutative addition a, d + b, e a + b, d + e by the definition of addition b + a, e + d Ax3 for R and S a, d + b, e by the definition of addition. So the addition in R S is commutative. Ax4 additive identity Define 0 R S : 0 R, 0 S. Then
a, d + 0 R S a, d + 0 R, 0 S by the definition of 0 R S a + 0 R, d + 0 S by the definition of addition a, d Ax4 for R and S Since the addition in R S is commutative, we also get 0 R S + a, d a, d and so 0 R S is an additive identity. Ax5 additive inverse Define a, d : a, d Then a, d + a, d a, d + a, d by the definition of a, d a + a, d + d by the definition of addition 0 R, 0 S Ax5 for R and S So a, d is an additive inverse of a, d. Ax6 closure of multiplication By Ax6 for R and S, ab R and de S and so ab, de R S. By definition of multiplication, a, db, e ab, de and so a, db, e R S. Ax7 associative multiplication a, db, ec, f a, dbc, ef by the definition of multiplication abc, def by the definition of multiplication abc, def Ax7 for R and S ab, dec, f by the definition of multiplication a, db, ec, f by the definition of multiplication. So multiplication in R S is associative. Ax8 distributive laws a, db, e + c, f a, db + c, e + f by the definition of addition ab + c, de + f by the definition of multiplication ab + ac, de + df Ax7 for R and S ab, de + ac, df by the definition of addition a, bb, e + a, bd, f by the definition of multiplication and
a, d + b, ec, f a + b, d + ec, f by the definition of addition a + bc, d + ef by the definition of multiplication ac + bc, df + ef Ax7 for R and S ac, df + bc, ef by the definition of addition a, dc, f + b, ec, f by the definition of multiplication. So the distributive laws hold in R S. We verified all the eight axioms and so R S is a ring. b By definition of 0 R S, see *, we have 0 R S 0 R, 0 S. c By definition of r, s, see **, we have r, s r, s. d Suppose that both R and S are commutative. So Ax9 holds in R and S. a, db, e ab, de by the definition of multiplication ba, ed Ax9 for R and S b, ea, d by the definition of multiplication. So the multiplication in R S is commutative. e Suppose that R has an identity. Put 1 R S : 1 R, 1 S. Then a, d1 R S a, d1 R, 1 S by the definition of 1 R S a1 R, d1 S by the definition of multiplication a, d Ax10 for R and S and 1 R S a, d 1 R, 1 S a, d by the definition of 1 R S 1 R a, 1 S d by the definition of multiplication a, d Ax10 for R and S So 1 R S is an multiplicative identity in R S, and 1 R S 1 R, 1 S. Exercise 2. Let R be a ring and a, b, c, d R. Prove that a bc d ac ad + bd bc In each step of your proof, quote exactly one Axiom, Definition or Theorem. a bc d ac d bc d Theorem 2.2.9j ac ad bc bd Theorem 2.2.9j, twice ac ad + bc bd Definition of, see 2.2.7 ac ad + bd bc Theorem 2.2.9h ac ad + bd + bc Definition of ac ad + bd + bc Ax 2 ac ad + bd bc Definition of
Exercise 3. Prove or give a counterexample: If R is a ring with identity, then 1 R 0 R. We will give a counterexample. Let a be any object. Put R {a} and define an addition and multiplication on R by a + a a and a a a. Then R is closed under addition and multiplication. Observe that the addition and multiplication is associative and commutative, and also the distributive law holds, indeed in each case both sides of the relevant equations evaluate to a. Since a + a a and a is the only element in R, we conclude that a is an additive identity and so Ax 4 holds with 0 R a. As a + a a 0 R, Ax 5 holds with a a. Thus R is a ring. Since aa a and a is the only element in R, we conclude that a is multiplicative identity. Thus R is a ring with identity 1 R a. From 1 R a and 0 R a we get 1 R 0 R. This shows that there does exist a ring with an identity such that 1 R 0 R. Remark: The counterexample given above is the only counterexample. Indeed suppose that R is a ring with identity and that 1 R 0 R. Let r R. Then r 1 R r Ax 10 0 R r Since 1 R 0 R 0 R Theorem 2.2.9c We showed that r 0 R for all r R. Thus R {0 R }, and R is a ring with just one element. Proof. For the next two exercise we will prove an easy theorem. Theorem A: Let R be a ring and a, b R. Then a + ba + b aa + bb + ab + ba. a + ba + b aa + b + ba + b Ax 8 aa + ab + ba + bb Ax 8, twice aa + ab + bb + ba Ax 3 aa + ab + bb + ba Ax 2 aa + ab + bb + ba Ax 2 aa + bb + ab + ba Ax 3 aa + bb + ab + ba Ax 2 Exercise 4. Let R be a ring such that a a 0 R for all a R. Show that ab ba for all a, b R. By hypothesis we have x x 0 R for all x R Let a, b R. We compute 0 R a + ba + b * for x a + b aa + bb + ab + ba Theorem A 0 R + 0 R + ab + ba * for x a and x b 0 R + ab + ba Ax 4 ab + ba Ax 4 We proved that ab + ba 0 R. Thus ab ba by the Additive Inverse Law. Exercise 5. Let R be a ring such that a a a for all a R. Show that
a a + a 0 R for all a R b R is commutative. By hypothesis we have xx x for all x R a Let a R. Then a aa * for x a a a Theorem 2.2.9i a * for x a We proved that a a for all a R Thus a + a a + a 0 R, by Axiom 5. b Let a, b R. Then a + b a + ba + b * for x a + b aa + bb + ab + ba Theorem A a + b + ab + ba * for x a and x b We proved that a + b a + b + ab + ba and so the Additive Identity Law shows that ab + ba 0 R. Hence the Additive Inverse Law implies that ab ba. By ** applied with ba in place of a, we have ba ba. As ab ba, this gives ab ba. Hence Ax 9 holds and so R is commutative. Exercise 6. Prove or give a counterexample: Let R be a ring and a, b R. Then a + b 2 a 2 + 2ab + b 2. Note here that according to Definition 2.3.5b 2d d + d for any d in R. Solution 1: a + b 2 a 2 + 2ab + b 2 a + ba + b aa + ab + ab + bb definition of d 2 and 2d aa + ab + ba + bb aa + ab + ab + bb GAL, GDL ba + bb ab + bb Additive Cancellation Law ba ab Additive Cancellation Law So a + b 2 a 2 + 2ab + b 2 for all a, b R if and only if ba ab for all a, b R and if and only of R is commutative. Since there do exist non-commutative rings for example M 2 R, the statement is false. Solution 2: Let R M 2 R, a : [ ] 0 1 and b : [ ] 1 0 0.
and Then a + b 2 [ ] 0 1 + [ ] 2 1 0 [ ] 2 1 1 [ ] 1 1 a 2 + 2ab + b 2 [ ] 2 [ ] [ ] 0 1 0 1 1 0 + 2 + [ ] 2 1 0 [ ] + 2 Thus a + b 2 a 2 + 2ab + b 2. Hence the statement is false in general. [ ] + [ ] 1 0 Exercise 7. Let R be a commutative ring with identity. Suppose that 1 R + 1 R 0 R. Prove that [ ] 1 0. for all a, b R. a + b 2 a 2 + b 2. We commpute a + b 2 a + ba + b definition of x 2 aa + ab + ba + bb GDL aa + ab + ba + bb GAL aa + ab + ab + bb Since R is commutative aa + ab1 R + ab1 R + bb Ax 10 aa + ab1 R + 1 R + bb Ax 8 aa + ab0 R + bb 1 R + 1 R 0 R by hypothesis aa + 0 R + bb Theorem 2.2.9c aa + bb Ax 4 a 2 + b 2 definition of x 2 Exercise 8. Let S : {a, b, c, d} and let + be the addition on S defined by + a b c d a b a d a b c d b a c d a c c d b d b c Compute all possible sums of a, b, c, d, where sum is defined as in 2.3.1. As in Example 2.3.2 the sums of a, b, c, d are a + b + c + d, a + b + c + d, a + b + c + d, a + b + c + d, a + b + c + d. We will now compute each of these sums using the addition table: a + b + c + d a + b + c a + b a, a + b + c + d a + b + d a + a b, a + b + c + d a + c + d a + c d, a + b + c + d a + b + d a + d a, a + b + c + d a + c + d d + d c.