Homework 1/Solutions. Graded Exercises

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1 MTH Abstract Algebra I and Number Theory S18 Homework 1/Solutions Graded Exercises Exercise 1. Below are parts of the addition table and parts of the multiplication table of a ring. Complete both tables. + w x y z w w x y z y z w x z w y w x y z w x y y z Note that w + w = w and so the Additive Identity Law implies w = 0 R. Thus Ax 4 implies that (I) w + a = a and a + w = a for all a in the ring. By the given part of the addition table: (II) z + x = w and y + z = x Using Ax 3 we compute (III) x + z Ax 3 = z + x (II) = w and z + y Ax 3 = y + z (II) = x By (I),(II) and (III) we conclude that the addition table is ( ) + w x y z w w x y z x x y z w y y z w x z z w x y By the given part of the multiplication table we have We compute xx = y (1) Since w = 0 R, Theorem 1.2.9(c) gives xy (*) =x(x + x) Ax 8 = xx + xx (1) = y + y (*) = w (2) yx (*) =(x + x)x Ax 8 = xx + xx (1) = y + y (*) = w (3) xz (*) =x(x + y) Ax 8 = xx + xy (1),(2) = y + w (*) = y (4) zx (*) =(x + y)x Ax 8 = xx + yx (1),(3) = y + w (*) = y (5) yy (*) =y(x + x) Ax 8 = yx + yx (3) = w + w (*) = w (6) yz (*) =y(x + y) Ax 8 = yx + yy (3),(6) = w + w (*) = w (7) zy (*) =z(x + x) Ax 8 = zx + zx (5) = y + y (*) = w (8) zz (*) =z(x + y) Ax 8 = zx + zy (5),(8) = y + w (*) = y (9) wa = 0 R a = 0 R = w and aw = a0 R = 0 R = w (10)

2 for all a in R. Thus the multiplication table is: w x y z w w w w w x w y w y y w w w w z w y w y Exercise 2. Let R be a ring and a, b, c, d R. Prove that (a b)(c d) = ( (ac ad) + bd ) bc In each step of your proof, quote exactly one Axiom, Definition or Theorem. (a b)(c d) = a(c d) b(c d) Theorem 1.2.9j = (ac ad) (bc bd) Theorem 1.2.9j, twice = (ac ad) + ( (bc bd)) Definition of, see = (ac ad) + (bd bc) Theorem 1.2.9h = (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. The ring R = {0 R } in Example (f) is a counterexample. 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 + b)(a + b) a + b R by Ax 1 and (*) for x = a + b = a(a + b) + b(a + b) Ax 8 = (aa + ab) + (ba + bb)) Ax 8, twice = (0 R + ab) + (ba + 0 R ) (*) for x = a and x = b = ab + ba Ax 4, twice We proved that ab + ba = 0 R. Thus ab = (ba) by the Additive Inverse Law. Not Graded Exercises Exercise 5. Let R = {0, e, b, c} with addition and multiplication defined by the following tables. Assume associativity and distributivity and show that R is a ring with identity. Is R commutative? + 0 e b c 0 0 e b c e e 0 c b b b c 0 e c c b e 0 0 e b c e 0 e b c b 0 b b 0 c 0 c 0 c (Ax1) (closure of additions) All entrees in the addition table are from R. So a + b R for all a, b R.

3 (Ax2) (associative addition) Luckily we are allowed to assume this without proof. (Ax3) (commutative addition) The addition table is symmetric along the diagonal through +. Note that this means that a + b = b + a for all a, b R. (Ax4) (additive identity) The second row of the addition table equals the first. Thus 0 + r = r for all r R. The second column equals the first. So r + 0 = r for all r R. Hence 0 R := 0 is an additive identity. (Ax5) (additive inverse) All entrees in the diagonal through + equal 0. So r + r = 0 = 0 = 0 R for all r R. Thus every element has an additive inverse, namely itself. (Ax6) (closure of multiplication) All entrees in the multiplication table are from R. a, b R. So ab R for all (Ax7) (associative multiplication) Given to us for granted. (Ax8) (distributive laws) Given to us for granted. (Ax9) (commutative multiplication) The multiplication table is symmetric along the diagonal through. Note that this means that ab = ba for all a, b R. (Ax10) (multiplicative identity) The third row of the multiplication table equals the first. Thus e r = r for all r R. The third column equals the first. So r e = r for all r R. Hence 1 R := e is a multiplicative identity. Since Axiom 1-8 hold, R is a ring. Since Axiom 9 holds, R is commutative and since Axiom 10 holds, R has an identity. Exercise 6. 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, s)(r, 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.

4 (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, d)(b, e) = (ab, de) and so (a, d)(b, e) R S. (Ax7) (associative multiplication) (a, d)((b, e)(c, f)) = (a, d)(bc, ef) (by the definition of multiplication) = (a(bc), d(ef) (by the definition of multiplication) = ((ab)c, (de)f) ((Ax7) for R and S) = (ab, de)(c, f) (by the definition of multiplication) = ((a, d)(b, e))(c, f) (by the definition of multiplication). So multiplication in R S is associative.

5 (Ax8) (distributive laws) (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 multiplication) = (ab + ac, de + df) ((Ax7) for R and S) = (ab, de) + (ac, df) (by the definition of addition) = (a, b)(b, e) + (a, b)(d, f) (by the definition of multiplication) and ((a, d) + (b, e))(c, f) = (a + b, d + e)(c, f) (by the definition of addition) = ((a + b)c, (d + e)f) (by the definition of multiplication) = (ac + bc, df + ef) ((Ax7) for R and S) = (ac, df) + (bc, ef) (by the definition of addition) = (a, d)(c, f) + (b, e)(c, 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, d)(b, e) = (ab, de) (by the definition of multiplication) = (ba, ed) ((Ax9) for R and S) = (b, e)(a, 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, d)1 R S = (a, d)(1 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 ).

6 Exercise 7. 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 1.2.9(i) = 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 + b)(a + b) a + b R by Ax 1 and (*) for x = a + b = a(a + b) + b(a + 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 = (aa + bb) + (ab + ba) Ax 2 = (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.

Homework 3/ Solutions

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,