Homework 2 January 19, 2006 Math 522. Direction: This homework is due on January 26, In order to receive full credit, answer

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1 Homework 2 Jauary 9, 26 Math 522 Directio: This homework is due o Jauary 26, 26. I order to receive full credit, aswer each problem completely ad must show all work.. What is the set of the uits (that is, the multiplicative iverses) of the rig (Z, +, )? Call this set U(). Show that U() is a group uder multiplicatio modulo. Aswer: The elemets of Z that have multiplicative iverse are:, 3, 7, 9. Hece we have U() = {, 3, 7, 9}. This set U() forms a group uder multiplicatio modulo sice (a) idetity i U() is, (b) each elemet of U() has a iverse: =, 3 = 7, 7 = 3, 9 =, (c) the biary operatio multiplicatio mudulo is associative. 2. Let x ad y be two elemets of the commutative rig with uity (R, +, ). By iductio show that the biomial theorem (x + y) = x k y k holds o this commutative rig k (R, +, ). k= Aswer: Let N be a atural umber. The biomial theorem says that (x + y) = k= x k y k = k x + x y + + We wat to prove the above statemet by iductio o. Sice (x + y) + y. = (x + y)(x + y) [ ( ] = (x + y) x + x y + + )y = x + + x y + + xy + x y + x y [] [] = x x y x + y + y = x + + x y + + x + y + y +, + the formula holds for all positive itegers. The last lie of the above formula follows from the facts =, =, ad + = + k k k for all k =, 2, 3,...,. y +

2 Remark: If the rig is ot commutative, the xy yx, ad we ca ot have such a biomial theorem for ocommutative rigs. 3. A elemet x of a rig (R, +, ) is said to be a ilpotet elemet if there exists a positive iteger such that x =. Show that if x is a ilpotet elemet of the rig (R, +, ) with uity, the x ad + x both have multiplicative iverse. [Hit: Use the factorizatio x = ( x) ( + x + x x ) for x ad a similar formula for x +.] Aswer: Sice x is a ilpotet elemet of the rig (R, +, ), we have x = for some positive iteger. We wat to show that x has a multiplicative iverse. From the idetity x = ( x) ( + x + x x ), usig x = we get = ( x) ( + x + x x ). Therefore ( x) has a multiplicative iverse ad it is give by +x+x 2 + +x. Similarly from the idetity + ( ) x = ( + x) ( x + x 2 x ( ) x ), we get = ( + x) ( x + x 2 x ( ) x ), sice x =. Therefore, ( + x) has a iverse. 4. Usig biomial theorem show that if x ad y are two ilpotet elemets i the commutative rig with uity (R, +, ), the x + y is also a ilpotet elemet i the rig (R, +, ). Aswer: Sice x ad y are ilpotet elemets, therefore there exist positive itegers m ad such that x m = ad y =. We wat to show that x + y is ilpotet. That is, we wat to fid a positive iteger k such that (x + y) k =. We guess this k to be m + ad compute (x + y) m+ m + m + m + = x m+ + x m+ y + + y m+ m + m + m + m + = x m x + x m x y + + y m y m + = = Hece x + y is ilpotet.

3 5. Fid two zero-divisors i the rig (M 2 (Z), +, ) of two-by-two matrices with iteger etries. Aswer: Two zero-divisors i the rig (M 2 (Z), +, ) are ad sice or ( ( ) = ) =. Remark: Note that (M 2 (Z), +, ) is a ocommutative rig. If X ad Y are two elemets i M 2 (Z), the XY Y X. I have selected the above matrices i such a way that o matter which way you multiply them you will get a zero matrix. 6. Let Z be the set of itegers. Let P(Z) be the power set of Z, that is the collectio of all subsets of Z. Let ad be two biary operatios o P(Z) defied as X Y = (X Y )\(X Y ) ad X Y = X Y. Show that the algebraic system (P(Z),, ) is a rig. What is the characteristic of this rig? Aswer: Let Z be the set of itegers ad P(Z) be the set of all subsets of Z. We wat to show (P(Z),, ) is a rig. That is (P(Z), ) is a abelia group, the biary operatio is associative, ad distributive over o the set P(Z). (i) We show P(Z) is closed uder. Let X ad Y i P(Z). The X Y = (X Y ) \ (X Y ) P(Z). Hece X Y belogs to P(Z) wheever X ad Y are i P(Z). Thus P(Z) is closed uder. (ii) Next we show is the additive idetity. This is true because X = (X ) \ (X ) = X \ = X. Similarly X = X. Hece serves as the idetity i P(Z). (iii) X is the iverse of itself i (P(Z), ) sice X X = (X X) \ (X X) = X \ X =. Hece each set i (P(Z)) has a iverse with respect to. (iv) Next we show is associative. Let X, Y, W be three sets i P(Z). We have to show that (X Y ) W = X (Y W ). (You have to do a lot of book keepig for this part. I leave that to you.)

4 (iv) Next we show is commutative. Sice for ay three sets X, Y, W i P(Z) X Y = (X Y ) \ (X Y ) = (Y X) \ (Y X) = Y X the biary operatio is commutative. Hece we have show that (P(Z), ) is a abelia group. Next we show that the biary operatio is associative. Let X, Y, W be three sets i the set P(Z). Sice (X Y ) W = (X Y ) W = X Y W = X (Y W ) = X (Y W ) = X (Y W ) the biary operatio is associative. Fially, we show the biary operatio is distributive over. Sice for ay three sets X, Y, W i P(Z), we have X [Y W ] = X [(Y W ) \ (Y W )] = X (Y W ) \ X (Y W ) = (X Y ) (X W ) \ (X Y ) (X W ) = (X Y ) (X W ) = (X Y ) (X W ), the biary operatio is left distributive over o the set P(Z). Similarly, it ca be show that the biary operatio is right distributive over o the set P(Z). Hece (P(Z),, ) is a rig. The characteristic of this rig is 2 sice for ay X P(Z), we see that X X = which is 2X =. 7. Give a example of a fiite field (F, +, ) that have two ozero elemets a ad b such that a 2 + b 2 =. Aswer: Cosider the field (Z 5, +, ). Let a = 3 ad b = 4. The a 2 + b 2 = = 25 mod 5 =. 8. Let x ad y be ay two elemets i the itegral domai (Z 7, +, ) whose characteristic is 7. Prove that (x + y) 7 = x 7 + y 7. Aswer: Usig the biomial theorem, we see that (x + y) 7 = x 7 + x 6 y + x 5 y = x 7 + 7x 6 y + (3)7x 5 y xy 6 + y 7 = x y 7 = x 7 + y 7. 7 xy y 7 7

5 9. The algebraic structure (Z 3 Z 6, +, ), where the biary operatios + ad are defied as (a, b) + (c, d) = (a + c, b + d) ad (a, b) (c, d) = (ac, bd), forms a rig (similar to the exteral direct product of groups you have see i chapter seve). What is the characteristic of this rig (Z 3 Z 6, +, )? Aswer: Char(Z 3 Z 6 ) = lcm(3, 6) = 6.. Prove that the rig of Gaussia itegers (Z[i], +, ), where Z[i] = { a + bi a, b Z }, has o zero-divisors. Aswer: Suppose (Z[i], +, ) has a zero-divisor, say a + bi. Sice a + bi is a zero divisor, there exists a ozero elemet c + di such that (a + bi)(c + di) = which is Therefore ac bd + (bc + ad)i =. ac bd = bc + ad = which is a b c =. b a d a b Sice det, solvig the above equatio, we get c = ad d = ad this is a b a cotradictio to the fact that c + di is a ozero elemet.