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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)

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.

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