Let Z n be the set of ordered n-tuples with integerentries. Define addition and

Question

Let Zn be the set of ordered n-tuples with integerentries. Define addition and multiplication on Zncoordinate-wise and show Zn is a commutativering.  Then define Rn for any ring R. Show Rn is a ring and is commutative if R is. We haven't discussed coordinate-wise and the text gives littleinformation. How do I do this? Let Zn be the set of ordered n-tuples with integerentries. Define addition and multiplication on Zncoordinate-wise and show Zn is a commutativering.  Then define Rn for any ring R. Show Rn is a ring and is commutative if R is. We haven't discussed coordinate-wise and the text gives littleinformation. How do I do this?

Explanation / Answer

we shall prove for Z2 and generalizefor n tuples. suppose ( a,b) , (c,d) are any elements of Z2 then a , b , c,d are integers. so, a+c , b+d , ac , bd are integers. so, ( a+c,b+d) and ( ac , bd) are in Z2 ==> ( a,b) + (c,d) and ( a,b) * ( c,d) are inZ2. + and * obey closure law. we know that a + (c + e) = ( a+c) + e and b+ (d+f) = ( ( b+d) + f for any integers a , b , c, d, eand f. so, ( a ,b) + { ( c,d) + ( e,f)} = ( a,b) + ( c+e, d+f) = (a+(c+e) , b+(d+f)) = ( (a+c)+e, (b+d)+f)   from theabove                                            = ( a+c,b+d) + ( e,f)                                             = { ( a,b) + (c,d) }+ ( e,f) + obeys associative law. similarly a(ce) = (ac) e and b(df) = ( bd)f lead to (a,b) *{(c,d)*(e,f)} = { (a,b) * (c,d)}* (e,f) * obeys associative law. 0 is an integer. so, (0,0) is in Z2 suchthat (a,b)+ (0,0)= ( a+0,b+0) = ( a,b) = ( 0,0)+(a,b) (0,0) is the zero element in Z2. (a,b) is in Z2 ==> a , b are integers ==>-a, -b are integers ==> ( -a,-b) is in Z2 suchthat ( a,b) +(-a,-b) = (a-a,b-b) = (0,0) = ( -a+a,-b+b) =(-a,-b)+(a,b) (a,b) and ( -a,-b) are additive inverses inZ2. since addition of integers is abelian , using a+c = c+aand b+d = d+b we can write ( a,b)+(c,d) = ( a+c,b+d) = ( c+a,d+b) =( c,d) + ( a,b) addition is abelian in Z2. since a(ce) = (ac) e and b(df) = (bd) f ----(2) integersa , b , c, d, e and f , we get (a,b)* {(c,d)*(e,f)} =      (a,b)*(ce,df) = ( a(ce), b(df)) = ((ac)e,(bd)f) by (2)                         = (ac,bd) * ( e,f)                         = ( (a,b)*(c,d))*(e,f) * is associative in Z2 since integers obey distributivity , using that property herewe can write (a,b) * {(c,d)+(e,f)} = (a,b) *( c+e, d+f) = ( a( c+e) , b(d+f)) = ( ac+ae , bd+bf) = ( ac , bd) +( ae,bf) = ( a,b)*(c,d) + (a,b)* ( e,f) distributivity holds in Z2. thus, Z2 is a ring under the component wiseaddition and component wise multiplication. now, keeping the necessary and sufficient condition tobe asubring in view, suppose (a,b) ,(c,d) are two members ofZ2 then a , b , c and d are integers. so, a-c , b-dare integers. so, ( a-c,b-d) is in Z2. ==> ( a,b) - ( b-d) is in Z2. condition (1) satisfied. a , b , c and d are integers. so, ac , bd areintegers. ==> (a,b)*(c,d) is in  Z2. thus, Z2 is a ring under the component wiseaddition and component wise multiplication. now, keeping the necessary and sufficient condition tobe asubring in view, suppose (a,b) ,(c,d) are two members ofZ2 then a , b , c and d are integers. so, a-c , b-dare integers. so, ( a-c,b-d) is in Z2. ==> ( a,b) - ( b-d) is in Z2. condition (1) satisfied. a , b , c and d are integers. so, ac , bd areintegers. condition (2) satisfied. by necessary and sufficient condition to be a subring,Z2 is a subring of R2.

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